text:
User:Egm6321.f10.team2.oztekin/HW3

= Problem 1: Derive EOM for the particle =

From the lecture slide Mtg 13-1.

Given
The picture was given in the MT13-1 for the moving particle :

For the dynamic objects Newton's Second law cites that: The total force applied on a body is equal to the time derivative of linear momentum of the body.See Newton's Law of Motion According to this law :
 * {| style="width:100%" border="0"

$$  \displaystyle \vec{F}=\frac{\mathrm{d} (m\vec{v})}{\mathrm{d} t} $$ (Eq 1.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

If we manipulate this equation for the x and y axis;
 * {| style="width:100%" border="0"

$$  \displaystyle \sum F_{x}=ma_{x} $$     (Eq 1.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \sum F_{y}=ma_{y} $$     (Eq 1.3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

m: mass of the particle

k:air resistence constant

g:acceleration of gravity

Find
Case 1:Derive the EOM's for the moving particle with air resistance.

Case 2:k=0 no air resistance.

Case 3:Exist air resistance.

Case 1
For the x axis:
 * {| style="width:100%" border="0"

$$  \displaystyle a_{x}=\frac{\mathrm{d} v_{x}}{\mathrm{d} t} $$ (Eq 1.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

 * {| style="width:100%" border="0"

$$  \displaystyle F_{air_x}=kv^{n}cos(\alpha ) $$     (Eq 1.5) Then now we can say that according to Newton's Law: Total force acting on the particle=Particle momentum change. Since there is only force that acting on the particle on x axis is just the x component of air resistance.By tis way we can concluded that:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle m\frac{\mathrm{d} v_{x}}{\mathrm{d} t}=-kv^{n}cos(\alpha ) $$     (Eq 1.6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Similarly if we write EOM for the y axis
 * {| style="width:100%" border="0"

$$  \displaystyle F_{air_y}=kv^{n}sin(\alpha ) $$     (Eq 1.7)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

 * {| style="width:100%" border="0"

$$  \displaystyle a_{y}=\frac{\mathrm{d} v_{y}}{\mathrm{d} t} $$ (Eq 1.8)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

And total forces acting on the particle are y companent of air resistance and gravity force.

 * {| style="width:100%" border="0"

$$  \displaystyle m\frac{\mathrm{d} v_{y}}{\mathrm{d} t}=-kv^{n}sin(\alpha )-mg $$     (Eq 1.9)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Case 2
For k=0;
 * {| style="width:100%" border="0"

$$  \displaystyle m\frac{\mathrm{d} v_{x}}{\mathrm{d} t}=0 $$     (Eq 1.10)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

 * {| style="width:100%" border="0"

$$  \displaystyle m\frac{\mathrm{d} v_{y}}{\mathrm{d} t}=-mg $$     (Eq 1.11)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \int dv_{x}=\int dt $$ (Eq 1.12)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

 * {| style="width:100%" border="0"

$$  \displaystyle v=V_{xo} $$     (Eq 1.13)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

 * {| style="width:100%" border="0"

$$  \displaystyle x(t)=\int V_{x0}dt+x(0) $$     (Eq 1.14)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle x(t)=V_{x0}t+x(0) $$     (Eq 1.15)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \int dv_{y}=-\int gdt $$     (Eq 1.16)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

 * {| style="width:100%" border="0"

$$  \displaystyle v_{y}=-gt+V_{y0} $$     (Eq 1.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle x(t)=\int (-gt+V_{y0})+x(0) $$     (Eq 1.18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="0"

$$  \displaystyle x(t)=-\frac{gt^{2}}{2}+V_{y0}t+y(0) $$     (Eq 1.19)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Case 3.1: constant mass
To alleviate notion, we define,

 * $$\displaystyle z\equiv {{v}_{y}}$$

Then the original equation can be written as,

 * {| style="width:100%" border="0"

$$ \displaystyle m{z}'=-k{{z}^{n}}-mg$$ (Eq 1.21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

This nonlinear first order ODE can be solved by the separation of variables. By some manipulation, we can write Eq 1.21 as,

 * {| style="width:100%" border="0"

$$ \displaystyle -\frac{m}{mg+k{{z}^{n}}}dz=dt$$ (Eq 1.22)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Then we integrate Eq 1.22, we have,

 * {| style="width:100%" border="0"

$$ \displaystyle \int{\frac{m}{mg+k{{z}^{n}}}dz=C-t}$$ (Eq 1.23)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

where $$C$$ is some constant.

If you want to use Euler interation factor method, we can still get the same answer as Eq 1.23.

The original equation can be written as,

 * {| style="width:100%" border="0"

$$ \displaystyle k{{z}^{n}}+mg+m{z}'=0$$ (Eq 1.24)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

It's easy to see that the Eq 1.24 is not exact. So we expect we can multiple an integrating factor $$\displaystyle h$$ to make it exact.

Define:

 * $$\displaystyle \left\{ \begin{align}

& M\equiv k{{z}^{n}}+mg \\ & N\equiv m \\ \end{align} \right.$$

The prerequisite that Eq 1.24 being exact is,

 * {| style="width:100%" border="0"

$$ \displaystyle \frac{\partial (hM)}{\partial z}=\frac{\partial (hN)}{\partial t}$$ (Eq 1.25)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Thus,

 * {| style="width:100%" border="0"

$$ \displaystyle {{h}_{t}}N-{{h}_{z}}M+h({{N}_{t}}-{{M}_{z}})=0$$ (Eq 1.26)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

If we assume $$\displaystyle h$$ is a function only of $$\displaystyle z$$, meaning $$\displaystyle {{h}_{t}}=0$$, then substituting into Eq 1.26 $$\displaystyle M$$ and $$\displaystyle N$$, we have

 * {| style="width:100%" border="0"

$$ \displaystyle \frac{h}=\frac{1}{M}({{N}_{t}}-{{M}_{z}})=\frac{-kn{{z}^{n-1}}}{k{{z}^{n}}+mg}$$ (Eq 1.27)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

By integrating Eq 1.27, we have,

 * {| style="width:100%" border="0"

$$ \displaystyle h=\frac{1}{k{{z}^{n}}+mg}$$ (Eq 1.28)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We can see the assumption "$$\displaystyle h$$ is a function only of $$\displaystyle z$$" is satisfied.

Then we mutiple $$\displaystyle h$$ to both sides of Eq 1.24, we have,

 * {| style="width:100%" border="0"

$$ \displaystyle 1+\frac{m}{k{{z}^{n}}+mg}{z}'=0$$ (Eq 1.29)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We can easily see that Eq 1.29 is indeed exact. Because it's exact we can rewrite Eq 1.29 as,

 * {| style="width:100%" border="0"

$$ \displaystyle \frac{d}{dt}(t+\int{\frac{m}{mg+k{{z}^{n}}}dz})=0$$ (Eq 1.30)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

That means,

 * {| style="width:100%" border="0"

$$ \displaystyle t+\int{\frac{m}{mg+k{{z}^{n}}}dz}=C$$ (Eq 1.31)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

or

 * {| style="width:100%" border="0"

$$ \displaystyle \int{\frac{m}{mg+k{{z}^{n}}}dz=C-t}$$ (Eq 1.32)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

where $$C$$ is some constant.

Case 3.2: varying mass
If mass of the rocket becomes variable, then the original equation can be written as:
 * {| style="width:100%" border="0"

$$ \displaystyle k{{z}^{n}}+m(t)g+m(t){z}'=0$$ (Eq 1.33)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

I believe this equation is unlikely to be made exact because the assumption "$$\displaystyle h$$ is a function only of $$\displaystyle z$$" won't be satisfied neither will "$$\displaystyle h$$ is a function only of $$\displaystyle t$$". Without any of these two crucial assumptions, it is very difficult to find an appropriate Euler integrating factor. And also the variable mass makes the ODE's variables unseparable. I think we should turn to other method(numerical method?) for the solution.

= Problem 2 =

From the lecture slide Mtg 13-3

Given
The pendulum problem was shown in MT13-3 as

Angular momentum of a particle is gives as :
 * {| style="width:100%" border="0"

$$  \displaystyle \vec{L}=\vec{r}\times \vec{p} $$     (Eq 2.1) Where r=the position of the particle from the origin
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

p=The linear momentum of a particle.

Find
Derive the EOM.

Solution
For deriving EOM's we think in angular basis. Since we are interested in angular momentum we can enlarge Newton's Law for linear momentum into angular momentum.According to Newton's Law Total moment acting on the particle=Total angular momentum chance of particle.So

 * {| style="width:100%" border="0"

$$  \displaystyle \vec{r}\times \frac{\mathrm{d}(m\vec{v)}}{\mathrm{d} t}=\sum \vec{r}\times \vec{F} $$     (Eq 2.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \sum \vec{M}=\sum \vec{r}\times \vec{F} $$     (Eq 2.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Total moment acting on the particle mass of m1=Spring moment+gravity force moment+control force moment.

Spring moment

 * {| style="width:100%" border="0"

$$  \displaystyle \vec{M}_{spring}=\vec{a}\times k\vec{a}(sin\theta _{1}-sin\theta _{2}) $$     (Eq 2.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Gravity force moment

 * {| style="width:100%" border="0"

$$  \displaystyle \vec{M}_{gravity}=\vec{l}sin\theta _{1}\times m\vec{g} $$     (Eq 2.5) For small angular displacements
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle sin\theta _{1}\approx \theta _{1} $$     (Eq 2.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Control Force moment
 * {| style="width:100%" border="0"

$$  \displaystyle \vec{M}_{control}=\vec{l}\times \vec{u_{1}} $$     (Eq 2.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

If we subtitute all term in angular momentum equation we concluded

 * {| style="width:100%" border="0"

$$  \displaystyle m_{1}l^{2}\ddot{\theta _{1}}=-ka^{2}(\theta _{1}-\theta _{2})-m_{1}gl\theta_{1}+u_{1}l $$     (Eq 2.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Similarly for the second pendulum

 * {| style="width:100%" border="0"

$$  \displaystyle m_{2}l^{2}\ddot{\theta _{2}}=-ka^{2}(\theta _{2}-\theta _{1})-m_{2}gl\theta_{2}+u_{2}l $$     (Eq 2.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We can write this equations in matrix form for the SC_L1_ODEs.The form can be shown as basically:

 * {| style="width:100%" border="0"

$$  \displaystyle \dot{X}_{4\times 1}=A(t)_{4\times 4}.X_{4\times 1}+B(t)_{4\times 2}.U_{2\times 1} $$     (Eq 2.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle X=\begin{bmatrix} \theta _{1}\\ \dot{\theta _{1}}\\ \theta _{2}\\ \dot{\theta _{2}}\\
 * style="width:95%" |
 * style="width:95%" |

\end{bmatrix} $$     (Eq 2.11)
 *

 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \left[ \begin{matrix} \overset{.}{\mathop}\, \\ \overset{..}{\mathop}\, \\ \overset{.}{\mathop}\, \\ \overset{..}{\mathop}\, \\ \end{matrix} \right]=\left[ \begin{matrix} 0 \\   -\frac{k{{a}^{2}}}-\frac{g}{l}  \\ 0 \\   \frac{k{{a}^{2}}}  \\ \end{matrix}\begin{matrix} 1 \\   0  \\   0  \\   0  \\ \end{matrix}\begin{matrix} 0 \\   \frac{k{{a}^{2}}}  \\ 0 \\   -\frac{k{{a}^{2}}}-\frac{g}{l}  \\ \end{matrix}\begin{matrix} 0 \\   0  \\   1  \\   0  \\ \end{matrix} \right]\left[ \begin{matrix} {{\theta }_{1}} \\ \overset{.}{\mathop}\, \\ {{\theta }_{2}} \\ \overset{.}{\mathop}\, \\ \end{matrix} \right]+\left[ \begin{matrix} 0 \\   \frac{1}{{{m}_{1}}l}  \\ 0 \\   0  \\ \end{matrix}\begin{matrix} 0 \\   0  \\   0  \\   \frac{1}{{{m}_{2}}l}  \\ \end{matrix} \right]\left[ \begin{matrix} {{u}_{1}} \\ {{u}_{2}} \\ \end{matrix} \right] $$     (Eq 2.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

= Problem 3 =

From the lecture slide Mtg 14-1

Given
Linear first order ODE:

 * {| style="width:100%" border="0"

$$ \displaystyle \dot{x}(t)=a(t)x(t)+b(t)u(t)$$ (Eq 3.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Initial Condition:

 * $$\displaystyle {{t}_{0}}$$ and $$\displaystyle x({{t}_{0}})$$.

Find
$$\displaystyle x(t)$$ when $$\displaystyle a(t)$$ and $$\displaystyle b(t)$$ are constants and varying coefficients respetively.

Case of Varying Coefficients
First assume $$\displaystyle a(t)$$ and $$\displaystyle b(t)$$ are varying coefficients since constants are only a particular case.

The Eq 3.1 can be written as:

 * {| style="width:100%" border="0"

$$ \displaystyle \dot{x}(t)-a(t)x(t)=b(t)u(t)$$ (Eq 3.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Thus we have the Euler integrating factor:

 * {| style="width:100%" border="0"

$$ \displaystyle h(t)=\exp \int_ – ^{t}{-a(s)ds}$$ (Eq 3.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Or:

 * {| style="width:100%" border="0"

$$ \displaystyle h(t)=\exp \int_^{t}{-a(s)ds}+{{k}_{1}}$$ (Eq 3.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where $$\displaystyle {{t}_{M}}$$ is the lower limit of integral with respect to integration constant $$\displaystyle {{k}_{1}}$$.

Then we have $$\displaystyle x(t)$$:

 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} x(t) & =\exp \int_ – ^{t}{a(s)ds*\int_ – ^{t}{\left[ \exp \int_ – ^{\tau }{-a(s)ds} \right]}}*bu(\tau )d\tau \\ & =\exp \int_^{t}{a(s)ds}*\left\{ \int_^{t}{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau +{{k}_{2}} \right\} \end{align}$$ (Eq 3.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where $$\displaystyle {{t}_{N}}$$ is the lower limit of integral with respect to integration constant $$\displaystyle {{k}_{2}}$$. Note that $$\displaystyle {{k}_{1}}$$ disappeared because Patterson and Rasul have proved that it is unnecessary. In fact $$\displaystyle {{t}_{M}}$$ will also be canceled in the following procedure.

To derive $$\displaystyle {{t}_{N}}$$ and $$\displaystyle {{k}_{2}}$$, we should substituting the initial condition into Eq 3.5:

 * {| style="width:100%" border="0"

$$ \displaystyle x({{t}_{o}})=\exp \int_^{a(s)ds}*\left\{ \int_^{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau +{{k}_{2}} \right\}$$ (Eq 3.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

then we get $$\displaystyle {{k}_{2}}$$:

 * {| style="width:100%" border="0"

$$ \displaystyle {{k}_{2}}=\frac{x({{t}_{0}})}{\exp \int_^{a(s)ds}}-\int_^{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau $$ (Eq 3.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substituting $$\displaystyle {{k}_{2}}$$ back into Eq 3.5, we derive:

 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} x(t) & =\exp \int_^{t}{a(s)ds}*\left\{ \int_^{t}{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau +\frac{x({{t}_{0}})}{\exp \int_^{a(s)ds}}-\int_^{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau \right\} \\ & =x({{t}_{0}})\exp \left[ \int_^{t}{a(s)ds}-\exp \int_^{a(s)ds} \right]+\exp \int_^{t}{a(s)ds}*\left\{ \int_^{t}{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau -\int_^{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau \right\} \\ & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\exp \int_^{t}{a(s)ds}*\int_^{t}{\left[ \exp \int_{\tau }^{a(s)ds} \right]}*b(\tau )u(\tau )d\tau \\ & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\exp \int_^{t}{a(s)ds}*\int_^{t}{\left[ \exp (\int_{\tau }^{t}{a(s)ds+\int_{t}^{a(s)ds)}} \right]}*b(\tau )u(\tau )d\tau \\ & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\exp \int_^{t}{a(s)ds}*\exp \int_{t}^{a(s)ds*}\int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*b(\tau )u(\tau )d\tau \\ & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*b(\tau )u(\tau )d\tau \end{align}$$ (Eq 3.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

So we have got the $$\displaystyle x(t)$$ when $$\displaystyle a(t)$$ and $$\displaystyle b(t)$$ are varying coefficients:

 * {| style="width:100%" border="0"

$$ \displaystyle x(t)=x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*b(\tau )u(\tau )d\tau $$ (Eq 3.9)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Verifying of result
Note that the derivation rule of $$\displaystyle \frac{d}{dt}\int_^{t}{f(\tau )d\tau }$$ between that of $$\displaystyle \frac{d}{dt}\int_^{t}{f(\tau ,t)d\tau }$$ are totaly different:

 * {| style="width:100%" border="0"

$$ \displaystyle \frac{d}{dt}\int_^{t}{f(\tau )d\tau =}f(t)$$, (Eq 3.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

However,

 * {| style="width:100%" border="0"

$$ \displaystyle \frac{d}{dt}\int_^{t}{f(\tau ,t)d\tau =}f(t,t)+\int_^{t}{\frac{\delta f(\tau ,t)}{\delta t}d\tau }$$ (Eq 3.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

More generally,
 * {| style="width:100%" border="0"

$$ \displaystyle \frac{d}{dt}\int_{b(t)}^{a(t)}{f(\tau ,t)d\tau =}\frac{da(t)}{dt}f\left( a(t),t \right)-\frac{db(t)}{dt}f\left( b(t),t \right)+\int_{b(t)}^{a(t)}{\frac{\delta f(\tau ,t)}{\delta t}d\tau }$$ (Eq 3.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Thus by Eq 3.11,

 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\begin{align} \dot{x}(t) & ={{\left[ x({{t}_{0}})\exp \int_^{t}{a(s)ds} \right]}^{\prime }}+{{\left\{ \int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*b(\tau )u(\tau )d\tau \right\}}^{\prime }} \\ & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}*a(t)+\exp \int_{t}^{t}{a(s)ds}*b(t)u(t)+\int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*a(t)b(\tau )u(\tau )d\tau \\ & =a(t)\left\{ x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*b(\tau )u(\tau )d\tau \right\}+b(t)u(t) \\ & =a(t)x(t)+b(t)u(t) \end{align}$$ (Eq 3.13)
 *

 * }

Thus the result is veryfied. See also Team2 discussion on Mtg 14.

Case of Constant Coefficients
As constant coefficients is only a particular case of varying coefficients, we can just replace $$\displaystyle a(s)$$ by $$\displaystyle a$$ in the Eq 3.9,

 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} x(t) & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*b(\tau )u(\tau )d\tau \\ & =x({{t}_{0}})[\exp a(t-{{t}_{0}})]+\int_^{t}{\left[ \exp a(t-\tau ) \right]}*b(\tau )u(\tau )d\tau \end{align}$$ (Eq 3.14)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Then this is the answer to the case of constant coefficent.

Another way to solve the Case of Constant Coefficients(by Team2. Yaakov)
The problem is time independent; therefore, we can rewrite the initial equation as:

 * {| style="width:100%" border="0"

$$ \displaystyle \dot{x}(t)-ax(t)=bu(t)$$ (Eq 3.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

To calculate the integrating factor we find

 * {| style="width:100%" border="0"

$$ \displaystyle \exp (\int{-adt)=\exp (-at})$$ (Eq 3.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now we can multiply the entire equation 3.16 with the integrating factor

 * {| style="width:100%" border="0"

$$ \displaystyle \exp (-at)\dot{x}(t)-\exp (-at)ax(t)=\exp (-at)bu(t)$$ (Eq 3.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

It is noteworthy to realize that the LHS of the equation is equal to $$\displaystyle [\exp (-at)x(t)]'$$

Integrating both sides of the new equation from $$\displaystyle {{t}_{0}}$$ to $$\displaystyle t$$ and changing the variables from $$\displaystyle t$$ to $$\displaystyle \tau $$ for integration will yield us the following equation:

 * {| style="width:100%" border="0"

$$ \displaystyle \int\limits_^{t}{[\exp (-a\tau )x(\tau )]'=}\int\limits_^{t}{\exp (-a\tau )bu(\tau )dt}$$ (Eq 3.18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$ \displaystyle \exp (-at)x(t)-\exp (-a{{t}_{0}})x({{t}_{0}})=\int\limits_^{t}{\exp (-a\tau )bu(\tau )dt}$$ (Eq 3.19)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solving the equation for $$x(t)$$ will yield us to the proof:

 * {| style="width:100%" border="0"

$$ \displaystyle x(t)=[\exp a(t-{{t}_{0}})]x({{t}_{0}})+\int\limits_^{t}{[\exp a(t-\tau )]bu(\tau )d\tau }$$ (Eq 3.20)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Exponantial Of Matrix
Taylor series are consist of infinite components of function and its derivatives.The series are centered on any number that is interested in.

 * {| style="width:100%" border="0"

$$  \displaystyle f(a)+\frac{{f}'(a)}{1!}\left ( x-a \right )+\frac{{f}''(a)}{2!}\left ( x-a \right )^{2}+\frac{f^{3}(a)}{3!}\left ( x-a \right )^{3}........... $$     (Eq 3.21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For f(x)=expx can be characterized as the same way.

 * {| style="width:100%" border="0"

$$  \displaystyle e^{x}=\sum_{n=0}^{\infty }\frac{x^{n}}{n!}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+........... $$     (Eq 3.22)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

And if we substitute matrix A into x we concluded that:

 * {| style="width:100%" border="0"

$$  \displaystyle e^{A}_{n\times n}=\sum_{n=0}^{\infty }\frac{x^{n}}{n!}=I_{n\times n}+A_{n\times n}+\frac{A_{n\times n}^{2}}{2!}+\frac{A_{n\times n}^{3}}{3!}+........... $$     (Eq 3.23)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Same observation for Bryson, A.E., and Ho, Y.C Book's Eq.
The solution of SC_N1_ODE is given in the book as follows ;

 * {| style="width:100%" border="0"

$$  \displaystyle X(t)=\phi (t,t_{0})X(t_{0})+\int_{t_{0}}^{t}\phi (t,t_{0})B(\tau )U(\tau )d\tau $$     (Eq 3.24)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

So we can see that

 * {| style="width:100%" border="0"

$$  \displaystyle \phi (t,t_{0})_{n\times n}=expA(t-t_{0})_{n\times n} $$ (Eq 3.25)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \dot{X(t)}=A\phi (t,t_{0})x(t_{0})+\frac{\mathrm{d} t}{\mathrm{d} t}\phi (t,t)B(t)U(t)+\int_{t_{0}}^{t}\frac{\mathrm{d} }{\mathrm{d} t}\phi (t,\tau )B(\tau )U(\tau )d\tau $$     (Eq. 3.26)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \dot{X(t)}=A\phi (t,t_{0})x(t_{0})+B(t)U(t)+A\int_{t_{0}}^{t}\varphi (t,\tau )B(\tau )U(\tau )d\tau $$     (Eq 3.27)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We concluded that
 * {| style="width:100%" border="0"

$$  \displaystyle \dot{X}(t)-AX(t)=B(t )U(t) $$     (Eq. 3.28)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Since zou has drived comprehensively in his discussion page we concluded briefly just to show that book's equation yields exactly the same equtiation we drived in our lecture.

= Problem 4 =

From the lecture slide Mtg 14-2

Given
Solution of system of coupled linear first order ODE of constant coefficient:

 * {| style="width:100%" border="0"

$$ \displaystyle \underline{x}(t)=[\exp (t-{{t}_{0}})\underline{A}]\underline{x}({{t}_{0}})+\int_^{t}{\left[ \exp (t-\tau )\underline{A} \right]\underline{B}\underline{u}(\tau )}d\tau $$ (Eq 4.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Find
Solution of System of coupled linear first order ODE of varying coefficient

Solution
Just turn constant cofficient matrices $$\underline{A}$$ and $$\underline{B}$$ in Eq 4.1 into varying coefficient matrices case, and via similarity to Eq 3.9, we have,

 * {| style="width:100%" border="0"

$$ \displaystyle \underline{x}(t)=[\exp \int_^{t}{\underline{A}(s)ds}]\underline{x}({{t}_{0}})+\int_^{t}{\left[ \exp \int_{\tau }^{t}{\underline{A}(s)ds} \right]\underline{B}(\tau )\underline{u}(\tau )}d\tau $$ (Eq 4.2)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

= Problem 5 =

From the lecture slide Mtg 15-1

Given
System of Linear first order ODE:

 * {| style="width:100%" border="0"

$$ \displaystyle \underline(t)=\underline{A}\underline{x}(t)+\underline{B}\underline{u}(t)$$ (Eq 5.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where $$\displaystyle \underline{A}$$ and $$\displaystyle \underline{B}$$ are constant matrices.

Initial condition:

 * $$\displaystyle {{t}_{0}}$$ and $$\displaystyle \underline{x}({{t}_{0}})$$.

Find
$$\displaystyle \underline{x}(t)$$

Solution
From Eq 5.1 we derive Euler integrating factor,

 * {| style="width:100%" border="0"

$$ \displaystyle \underline{h}(t)=\exp \int_ – ^{t}{-\underline{A}ds}=\exp (\underline{M}-t\underline{A})$$ (Eq 5.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where $$\displaystyle \underline{M}$$ is a matrix consists of constants.

Thus we get $$\displaystyle \underline{x}(t)$$,

 * {| style="width:100%" border="0"

$$ \displaystyle \underline{x}(t)=\exp (t\underline{A}-\underline{M})*\left\{ \int_^{t}{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )d\tau +\underline{N}} \right\}$$ (Eq 5.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where $$\displaystyle {{t}_{N}}$$ is lower limit of integral with respect to $$\displaystyle \underline{N}$$ which is also a matrix consists of constants.

To derive $$\displaystyle \underline{N}$$ we should use the initial condition,

 * {| style="width:100%" border="0"

$$ \displaystyle \underline{x}({{t}_{0}})=\exp (t\underline{A}-\underline{M})*\left\{ \int_^{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )d\tau +\underline{N}} \right\}$$ (Eq 5.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Since,

 * {| style="width:100%" border="0"

$$ \displaystyle \exp (t\underline{A}-\underline{M})*\exp (M-t\underline{A})=1$$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Then $$\displaystyle \exp (M-t\underline{A})$$ are the inverse matrix of $$\displaystyle \exp (t\underline{A}-\underline{M})$$.

Then, by manipulation of Eq 5.4, we have,

 * {| style="width:100%" border="0"

$$ \displaystyle \underline{N}=\exp (M-{{t}_{0}}\underline{A})\underline{x}({{t}_{0}})-\int_^{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )}d\tau $$ (Eq 5.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substituting back to Eq 5.4, we have

 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} \underline{x}(t) & =\exp (t\underline{A}-\underline{M})*\left\{ \int_^{t}{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )d\tau +\exp (M-{{t}_{0}}\underline{A})\underline{x}({{t}_{0}})-\int_^{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )}d\tau } \right\} \\ & =\exp (t\underline{A}-\underline{M}+M-{{t}_{0}}\underline{A})\underline{x}({{t}_{0}})+\exp (t\underline{A}-\underline{M})*\int_^{t}{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )}d\tau \\ & =[\exp (t-{{t}_{0}})\underline{A}]\underline{x}({{t}_{0}})+\int_^{t}{\left[ \exp (t-\tau )\underline{A} \right]\underline{B}\underline{u}(\tau )}d\tau \end{align}$$ (Eq 5.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Thus we have derive $$\displaystyle \underline{x}(t)$$:

 * {| style="width:100%" border="0"

$$ \displaystyle \underline{x}(t)=[\exp (t-{{t}_{0}})\underline{A}]\underline{x}({{t}_{0}})+\int_^{t}{\left[ \exp (t-\tau )\underline{A} \right]\underline{B}\underline{u}(\tau )}d\tau $$ (Eq 5.7)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

=Problem 6:Discuss h=

Given
Example given in MT15-3 as

 * {| style="width:100%" border="0"

$$  \displaystyle x({y}')^{2}+y{y}'+xy{y}''=0 $$     (Eq 6.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle g(x,y,p)=x({y}')^{2}+y{y}' $$     (Eq 6.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle f(x,y,p)=xy $$     (Eq 6.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle h_{x}+h_{y}p=0 $$     (Eq 6.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Find
Without assuming h=const. discuss other solutions.

Solution
For the condition given Eq.6.4 both terms must be zero. So

 * {| style="width:100%" border="0"

$$  \displaystyle h_{x}=0 $$     (Eq 6.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle h_{y}p=0 $$     (Eq 6.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

And we know that h=h(x,y).Possible outcomes for the derivations;

 * {| style="width:100%" border="0"

$$  \displaystyle h_{x}=h(x,y) $$     (Eq 6.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle h_{y}=h(x,y) $$     (Eq 6.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

By this way we can not find common h for stisfy Eq.6.4

=Problem 7:SC-L1-ODE-CC=

Given

 * {| style="width:100%" border="0"

$$ \displaystyle \dot{\phi}=\omega $$ (Eq 7.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \dot{\omega}=-\frac{1}{\tau }\omega+\frac{Q}{\tau }\delta $$ (Eq 7.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \dot{\delta}=u $$ (Eq 7.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \dot{x}(t)=\underline{A} \underline{x}(t)+\underline{B} \underline{u}(t) $$ (Eq 7.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Find
Make 7.1-7.3 in the form of 7.4

Solve

 * {| style="width:100%" border="0"

$$ \displaystyle \begin{bmatrix} \dot{\phi }\\ \dot{\omega }\\ \dot{\delta } \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0\\ 0 & -\frac{1}{\tau } & \frac{Q}{\tau }\\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} \phi \\ \omega \\ \delta \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

+ \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} u(t) $$ (Eq 7.5)
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \underline{A}= \begin{bmatrix} 0 & 1 & 0\\ 0 & -\frac{1}{\tau } & \frac{Q}{\tau }\\ 0 & 0 & 0 \end{bmatrix}$$ (Eq 7.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \underline{B}= \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$$ (Eq 7.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

=Problem 8:Exactness Condition for Second Order=

Given
Second order non-linear function can be shown as follows which was given in MT15-2 ;

 * {| style="width:100%" border="0"

$$  \displaystyle F\left ( x,y,{y}',{y}'' \right )=0 $$     (Eq 8.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

F is a function that derivative of function that is non-linear first order equation equals constant.

 * {| style="width:100%" border="0"

$$  \displaystyle F\left ( x,y,{y}',{y}'' \right )=\frac{\mathrm{d} \phi \left ( x,y,{y}' \right )}{\mathrm{d} x} $$ (Eq 8.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

F is exact if there is exist a fuunction $$ \phi (x,y,{y}')$$

If we want to show that an arbitrary second order function to be exact, this arbitrary function has to provide two exactness conditions.

First condition:An arbitrary second order function has to provide this form:

 * {| style="width:100%" border="0"

$$  \displaystyle F\left ( x,y,{y}',{y} \right )=g\left ( x,y,p \right )+f\left ( x,y,p \right ){y} $$     (Eq 8.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where $$p={y}' $$

Second Condition:has two parts

 * {| style="width:100%" border="0"

$$  \displaystyle f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y} $$     (Eq 8.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle f_{xp}+pf_{yp}+2f_{y}=g_{pp} $$     (Eq 8.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where
 * {| style="width:100%" border="0"

$$  \displaystyle g\left ( x,y,p \right )=\phi _{x}+\phi _{y}{y}' $$     (Eq 8.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle f\left ( x,y,p \right )=\phi _{p}^{} $$     (Eq 8.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Find
Given function in lecture mtg 17-2 as follows ;

 * {| style="width:100%" border="0"

$$  \displaystyle \left ( 15p^{4}cosx^{2} \right ){y}''+6xy^{2}{y}'+\left [ -6xp^{5}sinx^{2}+2y^{3} \right ]=0 $$     (Eq 8.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For this equation check exactness.

Solution

 * {| style="width:100%" border="0"

$$  \displaystyle f= 15p^{4}cosx^{2} $$     (Eq 8.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle g=6xy^{2}{y}'+\left [ -6xp^{5}sinx^{2}+2y^{3} \right ] $$     (Eq 8.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle f_{xx}=-30p^{4}sinx^{2}-60p^{4}x^{2}cosx^{2} $$     (Eq 8.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle f_{xy}=0 $$     (Eq 8.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle f_{yy}=0 $$     (Eq 8.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle g_{xp}=6y^{2}-30p^{4}sinx^{2}-60p^{4}x^{2}cosx^{2} $$     (Eq 8.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle g_{yp}=12xy $$     (Eq 8.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle g_{y}=12xyp+6y^{2} $$     (Eq 8.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle f_{xp}=-120p^{3}xsinx^{2} $$     (Eq 8.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle f_{yp}=0 $$     (Eq 8.18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle f_{y}=0 $$     (Eq 8.19)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle g_{pp}=-120p^{3}xsinx^{2} $$     (Eq 8.20)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

By substituting this tools into two equations (8.4 and 8.5) we observe that ;

 * {| style="width:100%" border="0"

$$  \displaystyle -30p^{4}sinx^{2}-60p^{4}x^{2}cosx^{2}+2p.0+p^{2}.0=6y^{2}-30p^{4}sinx^{2}-60p^{4}x^{2}cosx^{2}+p12xy-12xyp-6y^{2} $$     (Eq 8.21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle -30p^{4}sinx^{2}-60p^{4}x^{2}cosx^{2}=-30p^{4}sinx^{2}-60p^{4}x^{2}cosx^{2} $$     (Eq 8.22)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle -120p^{3}xsinx^{2}+p.0+0=-120p^{3}xsinx^{2} $$     (Eq 8.23)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle -120p^{3}xsinx^{2}=-120p^{3}xsinx^{2} $$     (Eq 8.24)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

=Problem 9=

Given
For this problem, the equation is the same as which was given in problem 8. We can find first integral $$ \phi \left ( x,y,{y}' \right ) $$ by taking first derivative of f.

 * {| style="width:100%" border="0"

$$  \displaystyle \phi \left ( x,y,{y}' \right )=h\left ( x,y \right )+\int f\left ( x,y,{y}' \right )dp $$     (Eq 9.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \phi \left ( x,y,{y}' \right )=h\left ( x,y \right )+3p^{5}cosx^{2} $$     (Eq 9.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle g=\phi _{y}{y}'+\phi _{x} $$     (Eq 9.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

By substituting $$\phi _{x}$$ and$$ \phi _{y}$$ in the definition of g we concluded ;

 * {| style="width:100%" border="0"

$$ \displaystyle (6xy^2)y^{'}+2y^3=h_yy^{'}+h_x $$ (Eq 9.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The following was also given:
 * {| style="width:100%" border="0"

$$ \displaystyle h_yy{'}=2y^3 $$ (Eq 9.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Find
If  $$ \displaystyle h_{x}=6xy^{2}{y}'$$ then what is $$ \displaystyle \phi (x,y,p)$$ (Meeting 18-3)

Solution
For the second order ODEs we assume that :

 * {| style="width:100%" border="0"

$$  \displaystyle \phi =h(x,y)+\int f(x,y,p)dp $$     (Eq 9.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

So h(x,y) is only function of x and y. For the second option

 * {| style="width:100%" border="0"

$$  \displaystyle h_{x}=6xy^{2}{y}' $$     (Eq 9.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

tells us that after takind partial derivative of h over x we would find that another function depends on y prime.

 * {| style="width:100%" border="0"

$$  \displaystyle z(x,y,{y}')=6xy^{2}{y}' $$     (Eq 9.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

So the choice of derivation of h must be depend on x,y,and $${y}'$$ if we assume the choice is true.(?) However we can not expect that. Because h only depends on x and y.The derivation of h over x must yield a function depends on x and y.

=Contributions Of Team Members=

 * Problem 1:    Solved by:Oztekin,Sahin,Zou

Author:Oztekin,Zou Author:Oztekin
 * Problem 2:    Solved by:Oztekin,Sahin

Author:Zou
 * Problem 3-4-5: Solved by:Zou,Yaakow,Oztekin
 * Problem 6-7 : Solved by:Rasul,Patterson,Dube

Author:Dube,Oztekin
 * Problem 8-9 : Solved by:Dube,Oztekin

Author:Dube,Oztekin

= References =

title: User:Egm6321.f10.team2.oztekin/HW3


text:
User:Jtneill/Twitter

http://twitter.com/jtneill

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Jtneill/Twitter


text: File:2010 10 08 17 07 22.djvu

title: File:2010 10 08 17 07 22.djvu


text:
User:Leutha/Social Enterprise glossary

This is a glossary of concepts which may be useful for those dealing with Social Enterprises.

 * Hyperbolic Discounting
 * Wicked Problem

title: User:Leutha/Social Enterprise glossary


text: Template:Colbegin

title: Template:Colbegin


text: Template:Div col end

title: Template:Div col end


text:
User:Egm6321.f10.team2.vrasul/HW3

Given
Variable Definition:

$$\delta$$ = Aileron Angle (Deflection)

$$\phi$$ = Roll Angle

$$\omega$$ = Velocity of Roll Angle

$$Q$$ = Aileron Effectiveness

$$\tau$$ = Roll-Time Constant

 * {| style="width:100%" border="0"

$$\dot{\phi}=\omega$$ (Eq 6.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |

$$\dot{\omega}=-\tfrac{1}{\tau} \omega + \tfrac{Q}{\tau} \delta$$ (Eq 6.2)
 *

 * }

 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |

$$\dot{\delta}=u$$

(Eq 6.3)
 *

 * }

Find
Put above equations (Eq. 6.1 to 6.3) in the matrix form of:

 * {| style="width:100%" border="0"

$$\dot{x}\left ( t \right)=\underline{A} \underline{x}\left(t \right) + \underline{B} \underline{u}\left(t \right) $$     (Eq 6.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solution
The above expressions (Eq. 6.1 to 6.3) must resemble Eq. 6.4; therefore, it is imperative that they are dependent on time ‘t.’

To bring Eq. 6.2 to this form $$\delta$$ will be expressed in terms of u(t).

Integrate Eq. 6.3 with respect to (wrt) time t yields:

 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |

$$\delta=ut$$

(Eq 6.5)
 *

 * }

Eq. 6.2 can now be written in its intended form.

 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |

$$\dot{\omega}=-\tfrac{1}{\tau} \omega + \tfrac{Q}{\tau} ut$$ (Eq 6.6)
 *

If Eq. 6.1 is true then the integral of $$ \dot{\omega}$$ will yield the solution to $$ \dot{\phi}$$
 * }

Take the integral of Eq. 6.6:

 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |

$${\omega}=-\tfrac{t}{\tau} \omega + \tfrac{Q}{2\tau} ut^2$$ (Eq 6.7)
 *

 * }

Therefore,

 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |

$$\dot{\phi}=-\tfrac{t}{\tau} \omega + \tfrac{Q}{2\tau} ut^2$$ (Eq 6.8)
 *

 * }

Eq. 6.3 is simply a control signal and has no dependence on the roll time constant $$\tau$$. With this assumption it can be said that Eq 6.3 needs no further manipulation.

Eq. 6.1, 6.6 and 6.8 can now be written in matrix form of Eq. 6.4.

 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |

$$

\begin{bmatrix} \dot{\phi}\\ \dot{\omega}\\ \dot{\delta} \end{bmatrix} = \underbrace{ \begin{bmatrix} \tfrac{1}{\tau} 0 0\\

0 \tfrac{1}{\tau} 0\\ 0 0 0

\end{bmatrix} }_{\underline{A}}

\underbrace{

\begin{bmatrix} \phi\\ \omega\\ \delta \end{bmatrix} }_{\underline{x}}

+ \underbrace{ \begin{bmatrix} \tfrac{Q t^2}{2\tau}\\ \tfrac{Q}{\tau}\\ 1 \end{bmatrix} }_{\underline{B}}

\underbrace{ \begin{bmatrix} u \end{bmatrix} }_{\underline{u}}

$$     (Eq 6.9)
 *

 * }

In conclusion, the form of Eq 6.1, 6.2 and 6.3 has been written in the form of 6.4. As stated previously the expression $$ \dot{\delta} $$ has no components associated to the roll time constant.

Given

 * {| style="width:100%" border="0"

$$h_x + h_y y^\prime = 0 $$ (Eq 7.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Find
Find the solution of Eq. 7.1.

Solution
For there to be an other solution to Eq. 7.1 the following must hold true:

 * {| style="width:100%" border="0"

$$h_{xy} = h_{yx} $$ (Eq 7.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

That is, the derivative of,
 * {| style="width:100%" border="0"

$$h_x = -h_y y^\prime$$ (Eq 7.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

with respect to y must yield $$h_{yx}$$.

Perform the above operation to Eq 7.3.

 * {| style="width:100%" border="0"

$$h_{xy} = -h_y y^{\prime\prime}+ h_{yy} y^\prime$$ (Eq 7.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Clearly Eq 7.4 does not equal Eq. 7.2. As state previously, for Eq 7.1 to hold true Eq 7.2 must be met. This can only be attained if h was indeed a constant thereby giving $$h_x$$ and $$h_y$$ a value of 0.

title: User:Egm6321.f10.team2.vrasul/HW3


text:
User:Egm6321.f2010.team5.riveros/hw3p6

= Problem 6 - Obtaining SC-L1-ODE-CC with int. factor method = From Meeting 15, p. 15-1

Given
We are given a system of coupled linear first order ODEs with constant coefficients,
 * {| style="width:100%" border="0"

A part of the solution to this system is the state transition matrix, $$\underline{\Phi}$$, which has the properties
 * style="width:95%" | $$\displaystyle \underline{\dot{x}}(t)=\underline{A}\underline{x}(t)+\underline{B}\underline{u}(t)$$
 * (6.1)
 * }
 * }
 * {| style="width:100%" border="0"

and
 * style="width:95%" | $$\displaystyle \frac{d}{dt}\underline{\phi}(t,t_0)=\underline{A}\underline{\phi}(t,t_0)$$
 * (6.2)
 * }
 * }
 * {| style="width:100%" border="0"

where $$I$$ is the identity matrix.
 * style="width:95%" | $$\displaystyle \underline{\phi}(t_0,t_0)=I,$$
 * (6.3)
 * }
 * }

Find
Use equations (6.1), (6.2), and (6.3), to obtain the following equation:
 * {| style="width:100%" border="0"

$$\displaystyle \underline{x}(t)=\underbrace{\left[{\rm exp}\underline{A}(t-t_0)\right]}_{\underline{\phi}(t,t_0)}\underline{x}(t_0)+\int\limits_{t_0}^{t}{\underbrace{\left[{\rm exp}\underline{A}(t-\tau)\right]}_{\underline{\phi}(t,\tau)}\underline{B}\underline{u}(\tau)\,d\tau} $$
 * style="width:95%" |
 * style="width:95%" |
 * (6.4)
 * }

Solution
We can rearrange equation (6.1),
 * {| style="width:100%" border="0"

$$\displaystyle N\dot{x}(t)-M=\dot{x}(t)-Ax(t)=Bu(t)$$ (6.5) Now we substitute $$\displaystyle x(t)$$ for $$\underline{\phi}(t,t_0)$$,
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle \dot{\underline{\phi}}(t,t_0)-A\underline{\phi}(t,t_0)=Bu(t)$$ (6.6) to show that it meets the first condition of exactness. The equation, however, does not satisfy the second condition of exactness. We must apply the Euler integrating factor method to make the equation exact by multiplying both sides of equation (6.5) by an integrating factor $$\displaystyle h(t)$$. The integrating factor $$\displaystyle h(t)$$ is found to be,
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle h(t)=e^{-\underline{A}(t-t_0)}$$ (6.7) We multiply $$h(t)$$ to both sides of equation (6.6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle e^{-\underline{A}(t-t_0)}\dot{\underline{\phi}}(t,t_0)-e^{-\underline{A}(t-t_0)}A\underline{\phi}(t,t_0)=e^{-\underline{A}(t-t_0)}Bu(t)$$ (6.8) Integrating both sides of equation (6.8) from $$\displaystyle t_0$$ to $$\displaystyle t$$ and solving for $$\underline{\phi}(t,t_0)$$ yeilds,
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle \underline{\phi} (t,t_0)=e^{\underline{A}(t-t_0)}\underline{\phi}(t_0,t_0)+\int\limits_{t_0}^{t}e^{\underline{A}(t-\tau)}\underline{B}\underline{u}(t)d\tau $$ (6.9) Substituting $$\displaystyle x(t)$$ into $$\underline{\phi}(t,t_0)$$ yeilds equation (6.4).
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

title: User:Egm6321.f2010.team5.riveros/hw3p6


text: File:Egm6321 f10 team6 HW3 problem1 N2.pdf


text:
User talk:Gtimbe

 Hello Gtimbe, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 16:23, 6 October 2010 (UTC)

title: User talk:Gtimbe


text:
User talk:ARONE

 Hello ARONE, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 16:24, 6 October 2010 (UTC)

title: User talk:ARONE


text:
User talk:Abdurramanemane

 Hello Abdurramanemane, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 16:24, 6 October 2010 (UTC)

title: User talk:Abdurramanemane


text:
User talk:Kate312

 Hello Kate312, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 16:24, 6 October 2010 (UTC)

title: User talk:Kate312


text:
User talk:Suleiz

 Hello Suleiz, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 16:24, 6 October 2010 (UTC)

title: User talk:Suleiz


text:
User talk:MJDabrowska

 Hello MJDabrowska, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 16:24, 6 October 2010 (UTC)

title: User talk:MJDabrowska


text:
User talk:Ash Menon

 Hello Ash Menon, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 16:24, 6 October 2010 (UTC)

title: User talk:Ash Menon


text:
User talk:Pejunior

 Hello Pejunior, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 16:28, 6 October 2010 (UTC)

title: User talk:Pejunior


text:
User:Egm6321.f10.team2.oztekin/sandbox

Same observation for Bryson, A.E., and Ho, Y.C Book's Eq.
The solution of SC_N1_ODE is given in the book as follows ;

 * {| style="width:100%" border="0"

$$  \displaystyle X(t)=\phi (t,t_{0})X(t_{0})+\int_{t_{0}}^{t}\phi (t,t_{0})B(\tau )U(\tau )d\tau $$     (N)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

So we can see that

 * {| style="width:100%" border="0"

$$  \displaystyle \phi (t,t_{0})_{n\times n}=expA(t-t_{0})_{n\times n} $$ (N)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \dot{X(t)}=A\phi (t,t_{0})x(t_{0})+\frac{\mathrm{d} t}{\mathrm{d} t}\phi (t,t)B(t)U(t)+\int_{t_{0}}^{t}\frac{\mathrm{d} }{\mathrm{d} t}\phi (t,\tau )B(\tau )U(\tau )d\tau $$     (N)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \dot{X(t)}=A\phi (t,t_{0})x(t_{0})+B(t)U(t)+A\int_{t_{0}}^{t}\varphi (t,\tau )B(\tau )U(\tau )d\tau $$     (N)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

We concluded that
 * {| style="width:100%" border="0"

$$  \displaystyle \dot{X}(t)-AX(t)=B(t )U(t) $$     (N)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Since zou has drived comprehensively in his discussion page we concluded briefly just to show that book's equation yields exactly the same equtiation we drived in our lecture.

title: User:Egm6321.f10.team2.oztekin/sandbox


text:
Talk:Ideas in Geometry/Instructive examples/Proving the Area of a Trapezoid

I really liked this proof. I though it was clever to make a rectangle out of the two remaining triangles and it works out great. My only suggestion would be to further prove that the two triangles are in fact the same triangle with equal bases. I didn't see anywhere that it says that they are the same but they certainly look the same in the book.

But I think that it would still work out even if they weren't the same length, the algebra would just be much more complicated. ________________________________________________________________________________________________________________________________

I think you all did a good job with this proof by picture. To think about it differently, maybe one consideration can be the way you labeled your trapezoid. I think what is meant by b1 and b2 might be the 2 different bases (the top and bottom parallel lines with different lengths). By starting with this label, you might be able to prove the formula differently.

Overall good job. But, in you're first line of algebra, shouldn't the Area=(b1*h)+(1/2xh)+(1/2xh) rather than multiplying these terms together? However it didn't seem to matter because this was irrelevant to the rest of your proof. Also, I think you could have spent more time explaining the picture and the manipulations in the pictures rather than focusing on and reiterating the algebra that was already shown. Hynes3 21:29, 7 October 2010 (UTC)

_________________________________________________________________________________________________________

I kind of struggled to understand the part where you figure out the equations. I understand the part where you plug in x, but wasn't the equation you needed to plug in h(b1+x)? how did you get b1-x? when plugging it in. Shouldnt it be h(b1+((b2-b1)/2))im just also confused on how you got to h(2b1+b2-b1). Smoo1244

___________________________________________________________________________________________________________________________

I think that you guys did a great job with this proof. At first I was not sure where your variables were coming from; however, once I looked at your picture they made sense. I thought your picture really helped clarify the equations that you were using and it was drawn out very well. Maybe downsize it a little bit so we can see it all on one page? I also liked that you used the two triangles to create a another rectangle. The only real suggestion that I have is that you relate your written description more to your drawing, instead of leaving it up to the reader to do that. I found that I had to read your description a few times before the calculations made sense. Maybe just reference your picture more when you are describing the process. Overall, you guys did a great job of simplifying the problem and making it easier to understand. Good Job! Katelin_Barker90


text: File:Mass vs time.JPG

title: File:Mass vs time.JPG


text:
User talk:Jgonsalves428

 Hello Jgonsalves428, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 20:38, 6 October 2010 (UTC)

title: User talk:Jgonsalves428


text:
User talk:Bforbes~enwikiversity

 Hello Bforbes, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 20:38, 6 October 2010 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Bforbes. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Bforbes~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:28, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:04, 19 April 2015 (UTC)

title: User talk:Bforbes~enwikiversity


text:
User talk:Eck1

 Hello Eck1, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 20:38, 6 October 2010 (UTC)

title: User talk:Eck1


text:
Scholarship

Scholarship is the process of discovering, creating, developing, debating and sharing scholarly knowledge. It is closely related to academia.

Scholarship in general assume a financial aid for students and researchers so that they go for further higher education.

title: Scholarship


text:
User talk:Shiyuanzhuang

 Hello Shiyuanzhuang, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 01:02, 7 October 2010 (UTC)

title: User talk:Shiyuanzhuang


text:
User talk:Orange crow

 Hello Orange crow, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 01:02, 7 October 2010 (UTC)

title: User talk:Orange crow


text:
User:Egm6321.f10.team5.abc/f09.h1

= Problem 1 - Maglev First and Second Derivative Derivations =

Given
The equation of motion for a Maglev train can be modeled by the function :

 Nice contribution Maglev train. Egm6321.f09 12:47, 22 September 2009 (UTC)

 * {| style="width:100%" border="0" align="left"

f(Y^1 (t),t) $$
 * $$\displaystyle
 * $$\displaystyle

 * }
 * }

Find
Derive the first and second time derivatives for the given equation shown by Equation 1 and 2 respectively.

 * {| style="width:100%" border="0" align="left"

\frac{d}{dt} f \left(Y^1(t),t\right) = \frac{\partial f}{\partial s} \left(Y^1(t),t\right) \dot Y^1(t) + \frac{\partial f}{\partial t} \left(Y^1(t),t\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)

 * }
 * }

 * {| style="width:100%" border="0" align="left"

\frac{d^2}{dt^2} f\left(Y^1(t),t\right) = f_{,s} \left(Y^1(t),t\right) \ddot Y^1 + f_{,ss} \left(Y^1(t),t\right) (\dot Y^1)^2 + 2 f_{,st} \left(Y^1(t),t\right) \dot Y^1 + f_{,tt} \left(Y^1(t),t\right)
 * $$\displaystyle
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 2)

 * }
 * }

Solution
 Solving for Equation 1

Taking the time derivative:

\frac{d}{dt} f \left(Y^1(t),t\right) = \frac{d}{dt} f \left(s,t\right) $$

For ease, a dummy variable
 * {| style="width:100%" border="0" align="left"

s = Y^1(t) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

is used and substuted into the previous equation. The chain rule is now applied:
 * {| style="width:100%" border="0" align="left"

\frac{d}{dt} f \left(s,t\right) = \frac{\partial f}{\partial s} \frac{\partial s}{\partial t}+ \frac{\partial f}{\partial t} \underbrace{\frac{\partial t}{\partial t}}_{=1} $$ $$
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Note that
 * $$\frac{\partial s}{\partial t}=\dot Y^1(t)$$

Equation 3 can be rewriten to the form:

 * {| style="width:60%" border="0"

$$ \frac{d}{dt} f \left(Y^1(t),t\right) = \frac{\partial f}{\partial s} \left(Y^1(t),t\right) \dot Y^1(t) + \frac{\partial f}{\partial t} \left(Y^1(t),t\right) $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }

Solving for Equation 2

Note that

\frac{d^2 }{dt^2} f= \frac{d}{dt}\left(\frac{d}{dt}f\right) $$ Substituting in Eq. 1 yields

\frac{d^2}{dt^2} f\left(Y^1(t),t\right) = \frac{d}{dt}\left(\frac{\partial f}{\partial s} \frac{\partial s}{\partial t}+ \frac{\partial f}{\partial t} \right) $$ Now, we can apply the chain rule so that

\frac{d^2}{dt^2} f\left(Y^1(t),t\right) =\frac{\partial^2 f}{\partial s^2} \left(\frac{\partial s}{\partial t}\right)^2 + \frac{\partial^2 f}{\partial s \partial t} \frac{\partial s}{\partial t} \frac{\partial t}{\partial t} + \frac{\partial f}{\partial s} \underbrace{\frac{\partial^2 s}{\partial s \partial t}}_{=0}\frac{\partial s}{\partial t} + \frac{\partial f}{\partial s} \frac{\partial^2 s}{\partial t^2}\frac{\partial t}{\partial t} + \frac{\partial^2 f}{\partial s\partial t}\frac{\partial s}{\partial t} + \frac{\partial^2 f}{\partial t^2} \frac{\partial t}{\partial t} $$ Collecting terms in the above equation

\frac{d^2}{dt^2} f\left(Y^1(t),t\right) =\frac{\partial^2 f}{\partial s^2} \left(\frac{\partial s}{\partial t}\right)^2 + 2\frac{\partial^2 f}{\partial s \partial t} \frac{\partial s}{\partial t} + \frac{\partial f}{\partial s} \frac{\partial^2 s}{\partial t^2} +  \frac{\partial^2 f}{\partial t^2} $$ To compare with the given notation, the derivates in the equations above are replaced with subscripts. For example, dX/dy would be simply X,y. The following equation shows this transformation.

\frac{d^2}{dt^2} f\left(Y^1(t),t\right) =f_{,ss} \left(\frac{\partial s}{\partial t}\right)^2 + 2f_{,st} \frac{\partial s}{\partial t} + f_{,s}\frac{\partial^2 s}{\partial t^2} + f_{,tt} $$ Finally, the dummy variable is replaced and the desired solution remains.
 * {| style="width:60%" border="0"

$$ \frac{d^2}{dt^2} f\left(Y^1(t),t\right) =f_{,ss} \left(\dot Y^1\right)^2 + 2f_{,st} \dot Y^1 + f_{,s}\ddot Y^1 + f_{,tt} $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }

= Problem 2 - Method of Integrating Factor =

Given
A Linear 1st order ODE is given by

 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle y' + y = x$$
 * $$\displaystyle (Eq. 4)
 * $$\displaystyle (Eq. 4)

 * }
 * }

Find
Solve Eq. 4 for y(x) and show
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle y(x) = A e^{-x} + x - x$$
 * $$\displaystyle y(x) = A e^{-x} + x - x$$

 * }
 * }

Solution
To find the solution, we need to perform the exactness test which requires two steps:

1. Writing the ODE in the form of
 * {| style="width:100%" border="0"

 * style="width:95%" |
 * $$\displaystyle M(x,y)+N(x,y)y'=0$$
 * $$\displaystyle M(x,y)+N(x,y)y'=0$$

$$
 * $$\displaystyle (Eq. 5)
 * style= |
 * }

2. Checking the condition given by
 * {| style="width:100%" border="0"

 * style="width:95%" |
 * $$\displaystyle \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$
 * $$\displaystyle \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$

$$
 * $$\displaystyle (Eq. 6)
 * style= |
 * }.

Checking Condition 1

Re-writing Eq. 4 in the form of Eq. 5 gives

 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle (y-x)+ y' = 0$$
 * $$\displaystyle (y-x)+ y' = 0$$

 * style= |
 * }

where $$\displaystyle M=(y-x)$$ and $$\displaystyle N=1$$. It is shown that condition 1 is met.

Checking Condition 2

Now, we have to test if $$\displaystyle M_y=N_x$$

 * $$\displaystyle M_y = 1$$ and $$\displaystyle N_x = 0$$

Therefore, Eq. 4 is not an exact differential equation and an integrating factor is needed to force the ODE to be exact. The definition of the integrating factor is

 * {| style="width:100%" border="0"

 * style="width:95%" |
 * $$\displaystyle h(x)=\exp \int^x f(x)dx$$
 * $$\displaystyle h(x)=\exp \int^x f(x)dx$$

 * style= |
 * }

where

 * $$\displaystyle f(x)=\frac{1}{N}\left[ M_p - N_x \right ]$$

Substiuting the values of M and N yield

 * $$\displaystyle f(x)== \frac{1}{1}\left[ 1 - 0 \right ] = 1 $$

Therefore the integrating factor is
 * {| style="width:100%" border="0"

 * style="width:95%" |
 * $$\displaystyle h(x)= e^x$$
 * $$\displaystyle h(x)= e^x$$

 * style= |
 * }

Multiplying Eq. 4 by the integrating factor gives

 * {| style="width:100%" border="0"

 * style="width:95%" |
 * $$\displaystyle e^x y + e^x y' = xe^x$$
 * $$\displaystyle e^x y + e^x y' = xe^x$$

 * style= |
 * }

Now it can be seen that the left hand side is in the form of a derivative product. Recognizing this yields:

 * {| style="width:100%" border="0"

 * style="width:95%" |
 * $$\displaystyle (e^xy)' = xe^x$$
 * $$\displaystyle (e^xy)' = xe^x$$

 * style= |
 * }

Integrating both sides results

 * {| style="width:100%" border="0"

 * style="width:95%" |
 * $$\displaystyle (e^xy) = e^x(x-1)+A$$
 * $$\displaystyle (e^xy) = e^x(x-1)+A$$

Finally, dividing by e^x results in
 * style= |
 * }
 * {| style="width:60%" border="0"

$$\displaystyle y = Ae^{-x}+(x-1)$$
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }

Problem #3: Show that the first order ODE $$(2x^2+\sqrt{y})+x^5y^3y^{\prime}=0$$ is nonlinear

Solution

Define a linear operator $$D\left ( \cdot \right )$$ such that $$D(\cdot )=2x^2+\sqrt{\left (\cdot  \right ) }+x^5\left (\cdot  \right ) ^3\frac{d}{dx}\left ( \cdot  \right )$$

The linearity condition then becomes$$\forall\alpha,\beta \in\mathbb{R}$$, $$D(\alpha u+\beta v)=\alpha D(u)+\beta D(v)$$ where u and v are any functions of x

Substituting in for the left side of the linearity condition gives:

$$D(\alpha u + \beta v)=2x^2+\sqrt{\alpha u + \beta v}+x^5\left ( \alpha u + \beta v \right )^3\frac{d}{dx}\left ( \alpha u+\beta v \right )$$

Substituting for the right side of the linearity condition:

$$\alpha D(u)+\beta D(v)=\alpha\left (2x^2+\sqrt{u}+ x^5u^3\frac{du}{dx} \right )+\beta\left ( 2x^2+\sqrt{v}+x^5v^3\frac{dv}{dx} \right )$$

Simplifying and comparing both sides of the linearity condition:

$$2x^2+\sqrt{\alpha u+\beta v}+\alpha x^5(\alpha u+ \beta v)^3\frac{du}{dx}+\beta x^5(\alpha u + \beta v)^3\frac{dv}{dx}\neq2x^2\alpha+\alpha\sqrt{u}+\alpha x^5u^3\frac{du}{dx}+2x^2\beta+\beta\sqrt{v}+\beta x^5v^3\frac{dv}{dx}$$

The condition is therefore not met, and $$(2x^2+\sqrt{y})+x^5y^3y^{\prime}=0$$ is a nonlinear 1st Order ODE.

Good solution.--Egm6321.f09.TA 03:14, 24 September 2009 (UTC)

Problem 4:
Show that $$F(x,y,y') = 0$$ is a nonlinear, 1st-order ordinary differential equation, where $$F(x,y,y') = x^2 y^5 + 6(y')^2=0$$.

Order of the ODE:

Since, $$F(x,y,y') = x^2 y^5 + 6(y')^2=0$$, the highest order is $$y'$$, the function is a 1st-order ordinary differential equation.

Proof of Nonlinearity:

First, we define the differential operator, $$D(\cdot)$$ as:

$$D(\cdot)=x^2 (\cdot)^5 + 6\left(\frac{d(\cdot)}{dx}\right)^2 = 0$$

Where $$u$$ and $$v$$ are functions of $$x$$, and $$\alpha$$ and $$\beta$$ are scalars. Next, we define the equation $$D(\alpha u + \beta v)$$ as:

$$D(\alpha u + \beta v) = x^2(\alpha u + \beta v)^5 + 6\left[\alpha\frac{du}{dx} + \beta\frac{dv}{dx}\right]^2$$

We then solve for $$\alpha D(u) + \beta D(u)$$, which is given as

$$\alpha D(u) + \beta D(v) = \alpha\left[x^2 u^5 +6\left(\frac{du}{dx}\right)^2 \right] + \beta \left[x^2 v^5 + 6 \left(\frac{dv}{dx}\right)^2\right]$$

Since, $$x^2(\alpha u + \beta v)^5 + 6\left[\alpha\frac{du}{dx} + \beta\frac{dv}{dx}\right]^2 \neq \alpha\left[x^2 u^5 +6\left(\frac{du}{dx}\right)^2 \right] + \beta \left[x^2 v^5 + 6 \left(\frac{dv}{dx}\right)^2\right]$$

It follows that

$$D(\alpha u + \beta v) \neq \alpha D(u) + \beta D(v)$$

In order for $$F(x,y,y') = 0$$ to be linear, the equation $$D(\alpha u + \beta v) = \alpha D(u) + \beta D(u)$$  $$ \forall u, v,$$   and    $$\forall \alpha, \beta \in \Re$$. This is known as the superposition principle and it must be satisfied in order for a differential equation to be linear.

Since the superposition principle is not satisfied, the function is nonlinear.

 Good solution.--Egm6321.f09.TA 03:41, 24 September 2009 (UTC)

Problem #5:

Part 1. Create an exact nonlinear 1st Order ODE of the form $$\Phi_x\left(x,y\right)+\Phi_y\left(x,y\right)y^{\prime}=0$$ using the equation $$\Phi\left(x,y\right)=6x^4+2y^{3/2}$$

Solution:

If,

$$\Phi\left(x,y\right)=6x^4+2y^{3/2}$$

Then

$$\Phi_{x}\left(x,y\right)=24x^{3}=M$$

and

$$\Phi_{y}\left(x,y\right)=3y^{1/2}=N$$

For a nonlinear, 1st order, ODE:

$$M + Ny^{\prime}=0$$

Therefore,

$$24x^3 + 3y^{1/2}y^{\prime}=0$$

We have forced exactness

Part 2. Create three more exact nonlinear 1st Order ODEs by inventing new $$\Phi\left(x,y\right)$$ functions.

Ex. 1

Choose $$\Phi\left(x,y\right)=x^{1/2}+y^{3}\quad\text{:}$$

$$ \begin{align} \Phi_{x}\left(x,y\right)&=0.5x^{-1/2}=M\left(x,y\right) \\ \Phi_{y}\left(x,y\right)&=3y^{2}=N\left(x,y\right) \end{align} $$

This leads to the following exact, first-order ODE:

$$M\left(x,y\right) + N\left(x,y\right)y' = \frac{1}{2}x^{-1/2}+3y^{2}y' = 0$$

Ex. 2

Choose $$\Phi\left(x,y\right)=3y^2+2x^5\quad\text{:}$$

$$\Phi_x\left(x,y\right)=10x^4=M\left(x,y\right)$$

$$\Phi_y\left(x,y\right)=6y=N\left(x,y\right)$$

This leads to the following exact, first-order ODE:

$$M\left(x,y\right) + N\left(x,y\right)y' = 10x^4+(6y)y' = 0$$

Ex. 3

Choose $$\Phi\left(x,y\right)=3x^2y^3\quad\text{:}$$

$$\Phi_x\left(x,y\right)=6xy^3=M\left(x,y\right)$$

$$\Phi_y\left(x,y\right)=9x^2y^2=N\left(x,y\right)$$

This leads to the following exact, first-order ODE:

$$M\left(x,y\right) + N\left(x,y\right)y' = 6xy^3+\left(9x^{2}y^{2}\right)y' = 0$$

Nice work.--Egm6321.f09.TA 03:14, 24 September 2009 (UTC)

title: User:Egm6321.f10.team5.abc/f09.h1


text:
User:Egm6321.f10.team5.abc/f09.h2

= Problem 1 - Integrating Factor Method =

Given
General form of a First Order Nonlinear Ordinary Differential Equations(N1-ODE) is given by,
 * {| style="width:100%" border="0" align="left"

F(x, y(x),\dot y(x)) = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

By using $$\displaystyle F=\frac{d}{dx} \phi(x,y(x))$$ we get

 * {| style="width:100%" border="0" align="left"

F = \underbrace{\frac{\partial \phi}{\partial x}}_{M(x,y)} + \underbrace{\frac{\partial \phi}{\partial y}}_{N(x,y)} \frac{dy}{dx} = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

We can multiply this by an Integrating Factor, $$\displaystyle h(x,y)$$, to get

 * {| style="width:100%" border="0" align="left"

h F = h M + h N \frac{dy}{dx} = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Then it follows that an exactness condition for an N1ODE's is, with equation 2 as the special case where $$h=1$$

 * {| style="width:100%" border="0" align="left"

(hM)_y = (hN)_x
 * $$\displaystyle
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 4)
 * }
 * }

Given that $$\displaystyle h_x N = 0$$, it is found that
 * {| style="width:100%" border="0" align="left"

\Rightarrow -h_y M + h(N_x - M_y) = 0 $$ $$ or
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\Rightarrow \frac{h_y}{h} = \frac{1}{M}(N_x - M_y) = -g(y) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }

Find
Complete the details to obtain Euler's Integrating factor $$\displaystyle h(y)$$.

Solution
Integrate equation 6
 * {| style="width:100%" border="0" align="left"

\int \frac{h_y}{h} dy = - \int g(y) dy $$ $$ This yields
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\ln h = - \int^y g(y) dy $$ $$ The final equation is
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

h = e^{- \int^y g(y) dy} $$ $$  It is a good idea to use a dummy variable, e.g., $$ y'$$ or $$s$$ so that your expression would be $$h(y)=e^{\int^{y}-g(y')dy'}$$. Otherwise your limits of integration are the same as your integrand variable. --Egm6321.f09.TA 01:37, 28 September 2009 (UTC)
 * $$\displaystyle (Eq. 9)
 * }
 * }

Actually, it is best to avoid using $$ y'$$ for a dummy integration variable, since it may be confused with $$\displaystyle \frac{dy}{dx}$$; another symbol such as $$s$$ would be preferred. Egm6321.f09 15:04, 30 September 2009 (UTC)

=Problem 2=

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
A non-homogeous L1-ODE-VC

 * $$ y'+\frac{1}{x}y=x^{2} $$

Find
Show that

 * $$y(x)=\frac{x^{3}}{4}+\frac{C}{x}$$

Solution
The general form for a non-homogenous L1-ODE-VC is:

 * $$y' +a_0(x)y = b(x) \!$$

The one given in the problem can be re-written as:

 * $$\underbrace{1}_{N(x,y)}\dfrac{dy}{dx} + \underbrace{\frac{1}{x}y}_{M(x,y)}=b(x)$$

We know that:

\begin{align} \frac{1}{N}\left(N_x - M_y\right) &= -f(x) \qquad \text{replacing values found above}\\ 1(0 - a_0(x)) &= -f(x)\qquad \text{which yields}\\ f(x) &= a_0(x)\\ \end{align} $$

Then making use of integrating factor knowledge:

\begin{align} a_0(x) &= \frac{1}{x}\\ h(x)&=\exp\int^{x}a_0(s)ds\qquad \text{replacing values for}\quad a_0(x) \quad \text{and evaluating the integral}\\ h(x)&=\exp(\ln x)\\ h(x) &= x \end{align} $$

Now we can solve for $$y(x)$$

\begin{align} y(x) &= \frac{1}{h(x)}\int^{x} h(s)b(s)ds \qquad \text{plugging known values in this expression yields}\\ y(x)&=\frac{1}{x}\int^{x}s s^2 ds \qquad \text{analyzing this integral and expanding terms out we end up with our desire result}\\ y(x)&=\frac{x^3}{4} + \frac{C}{x} \end{align} $$  Nice work.--Egm6321.f09.TA 03:13, 28 September 2009 (UTC)
 * }

=Problem 3=

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
A non-homogenous L1_ODE_VC

 * $$ \frac{1}{2}x^{2}y'+[x^{4}y+10]= 0 $$

Find
Show that the above non-homogenous L1_ODE_VC is exact

Solution
For the non-homogenous L1_ODE_VC to be exact it must staisfy satisfy the 2 conditions of exactness

\begin{align} N(x,y)y' + M(x,y) &= 0 \\ M_y(x,y) &= N_x(x,y) \end{align} $$

So the first condition is met by:

 * $$ \underbrace{\frac{1}{2}x^{2}}_{N(x,y)}y'+\underbrace{[x^{4}y+10]}_{M(x,y)}= 0 $$

Let's take a look at the 2nd condition:

\begin{align} M_y(x,y) & = x^4 \\ N_x(x,y) & = x \end{align} $$

Clearly $$M_y(x,y) \neq N_x(x,y)$$, so we make use of the integrating factor to see if we can make it exactly integrable.

So we re-write the ODE

 * $$ \underbrace{\frac{1}{2}x^{2}}_{N(x,y)}y'+\underbrace{[x^{4}y]}_{M(x,y)}= -10 $$

Looking at case #1 for the integrating factor we obtain:

\begin{align} \frac{1}{N}\left(N_x - M_y\right) &= -f(x) \qquad \text{replacing values found above}\\ \frac{2}{x^2}\left(x - x^4\right) &= -f(x) \qquad \text{applying knowledge from integrating factor we know}\\ h(x)&=\exp \int^{x}f(s)ds \qquad \text{replacing f(x) into the integral and evaluating it, obtain}\\ h(x) &= \frac{1}{x^2}\exp \left(\frac{2x^3}{3}\right) \end{align} $$

Now we can try the 2nd condition of exactness to see if by using integrating factor we can make this ODE exact

\begin{align} \bar{M}_y &= \bar{N}_x\\ \underbrace{(hM)}_{\bar{M}}dx&=\underbrace{(hN)}_{\bar{N}}dy\qquad \text{calculating the respective values we obtain}\\ \bar{N}_x &=x^2 \exp\left(\frac{2x^3}{3}\right)\\ \bar{M}_y &=x^2 \exp\left(\frac{2x^3}{3}\right) \end{align} $$

Hence the non-homogenous L1_ODE_VC is exact.

 Nice work. --Egm6321.f09.TA 03:14, 28 September 2009 (UTC)
 * }

=Problem 4=

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
An aribitrary 1st order ODE:

 * $$ \left(\frac{1}{3}x^{3}\right)(y^{4})y'+(5x^{3}+2)\left(\frac{1}{5}y^{5}\right)=0$$

Find
Show that the above is non-linear and exact

Solution
So let's determine that the above ODE is non-linear by making use of the $$D(.)\!$$ operator

To be a linear ODE the following must be true:

 * $$D\left(\alpha u + \beta v \right) = D\left(\alpha u \right) + D\left(\beta v \right)$$

Now we will define the $$D(.)\!$$ operator for the above ODE:

 * $$D(.)= \left(\frac{1}{3}x^{3}\right)(\cdot)^{4}\frac{d(\cdot)}{dx}+(5x^{3}+2)\left(\frac{1}{5}(\cdot)^{5}\right)$$

Then let's evaluate the linearity condition to see if it's true:

\begin{align} D\left(\alpha u + \beta v \right) &= \left(\frac{1}{3}x^{3}\right)(\alpha u + \beta v)^{4}\frac{d}{dx}(\alpha u + \beta v)+(5x^{3}+2)\left(\frac{1}{5}(\alpha u + \beta v)^{5}\right)\\ D\left(\alpha u \right) + D\left(\beta v \right) &= \left(\frac{1}{3}x^{3}\right)(\alpha u)^{4}\frac{d(\alpha u)}{dx}+(5x^{3}+2)\left(\frac{1}{5}(\alpha u)^{5}\right) +  \left(\frac{1}{3}x^{3}\right)(\beta v)^{4}\frac{d(\beta v)}{dx}+(5x^{3}+2)\left(\frac{1}{5}(\beta v)^{5}\right) \end{align} $$

Clearly $$D\left(\alpha u + \beta v \right) \neq D\left(\alpha u \right) + D\left(\beta v \right)$$ so the ODE is non-linear

 Nonlinearity can be determined by inspection. This would have sufficed for this problem, although your explanation is very complete. --Egm6321.f09.TA 03:18, 28 September 2009 (UTC) To prove that the ODE is exact, let's first look at the 1st condition of exactness:

 * $$ \underbrace{\left(\frac{1}{3}x^{3}\right)(y^{4})}_{N(x,y)}y'+\underbrace{(5x^{3}+2)\left(\frac{1}{5}y^{5}\right)}_{M(x,y)}=0$$

hence $$N(x,y)y' + M(x,y) = 0\!$$ so 1st condition is met.

Looking at the 2nd condition where $$M_y = N_x\!$$. However, from above we can clearly see that:

 * $$M_y \neq N_x$$

So we will try to find an integrating factor to find if our ODE can be exactly integrable

\begin{align} \left(\frac{1}{3}x^{3}\right)(y^{4})y'+(5x^{3}+2)\left(\frac{1}{5}y^{5}\right) &=0 \qquad \text{let us multiply througout  by} \qquad \frac{3}{x^3y^4}\\ y' +\frac{3}{x^3y^4}(5x^{3}+2)\left(\frac{1}{5}y^{5}\right) &=0\qquad \text{re-writing the equation obtain}\\ \underbrace{1}_{N}\cdot y' + \underbrace{\left(3 + \frac{6}{5x^3}\right)}_{a_0}y &=0 \end{align} $$ throughout

Now we will make use of integrating factor knowledge:

\begin{align} h(x,y) &= \exp\int^{x}a_0(s)ds\qquad \text{replacing the value of} \quad a_0\\ h(x,y) &= \exp\int^{x}\left(3 + \frac{6}{5s^3}\right)ds\qquad \text{evaluating the integral obtain}\\ h(x,y) &= \exp\left(3x - \frac{3}{5x^2}\right) \end{align} $$

Now we will try to evaluate the 2nd condition of exactness which is:

\begin{align} (hM)_{y}&=(hN)_x \qquad \text{the condition says:} \\ \bar{M}_y &= \bar{N}_x \qquad \text{evaluating the derivatives we obtain}\\ \dfrac{6\exp(3x)}{\left(\exp^{\frac{1}{x^2}}\right)^{\frac{3}{5}} x^3} + \dfrac{3\exp(3x)}{\left(\exp^{\frac{1}{x^2}}\right)^{\frac{3}{5}} } & = \dfrac{6\exp(3x)}{\left(\exp^{\frac{1}{x^2}}\right)^{\frac{3}{5}} x^3} + \dfrac{3\exp(3x)}{\left(\exp^{\frac{1}{x^2}}\right)^{\frac{3}{5}}} \end{align} $$

Hence the ODE given is then, exact N1-ODE-VC
 * }

= Problem 5 - N2-ODE-VC Exactness =

Given
Consider the following equation
 * {| style="width:100%" border="0" align="left"

xy y'' + x(y')^2 + y y' = 0 $$ $$ Using
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

f(x,y,p) = xy $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

g(x,y,p) = x(y')^2 + y y' = 0 $$ $$ Equation 1 can be written
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

f(x,y,p) y'' + g(x,y,p) = 0 $$ $$ Hence, the first exactness condition is fulfilled.
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * }

Find
Show that the second exactness condition is satisfied for equation 1.

Solution
Note: Equation 1 can be rewritten
 * {| style="width:100%" border="0" align="left"

(xyy')' = 0 $$ $$ First, we write $$f$$ and $$g$$ as
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

f(x,y,p) = xy $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

g(x,y,p) = x p^2 + y p = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }

The second condition of exactness is given by
 * {| style="width:100%" border="0" align="left"

f_{xx} + 2p f_{xy} + p^2 f_{yy} = g_{xp} + p g_{yp} - g_y $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

f_{xp} + p f_{yp} + 2 f_y = g_pp $$ $$ Substituting $$f$$ and $$g$$ into equation 6 and 7 yields
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

0 + 2p \cdot 1 + p^2 \cdot 0 = 2p + p \cdot 1 - p $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

0 + p \cdot 0 + 2 x = 2x $$ $$ or
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

2p = 2p $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

2 x = 2x $$ $$ Hence, the second exactness condition is fulfilled.
 Very good.--Egm6321.f09.TA 02:09, 28 September 2009 (UTC)
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9)
 * }
 * }

= Problem 6: Derive the second exactness condition for N2-ODE=

Given
A general N2-ODE is given in the form:

 * {| style="width:100%" border="0" align="left"

F(x,y,y',y) = 0 = f(x,y,p)y + g(x,y,p) $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

f(x,y,p):=\phi_p(x,y,p) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

g(x,y,p):=\phi_x + \phi_y(x,y,p)p $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }

and
 * {| style="width:100%" border="0" align="left"

p:=y'. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * }

Find
Derive the second condition of exactness (given in Eq. 5)

 * {| style="width:100%" border="0" align="left"

f_{xp}+pf_{yp} + 2f_y = g_{pp} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }

Solution
Differentiate Eq. 3 with respect to p
 * {| style="width:100%" border="0" align="left"

g_p = \phi_{xp} + \phi_y + p\phi_{yp} $$ $$ Note that $$\displaystyle f = \phi_p$$, so
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

g_p = f_{x} + \phi_y + p f_{y} $$ $$ Differentiate a second time
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

g_{pp} = f_{xp} + \phi_{py} + f_{y} + p f_{yp} $$ $$ Substitute $$\displaystyle f = \phi_p$$ again
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

g_{pp} = f_{xp} + p f_{yp} + 2f_{y} $$ $$
 Very nice.--Egm6321.f09.TA 02:20, 28 September 2009 (UTC)
 *
$$\displaystyle (Eq. 9)
 * }
 * }

=Problem 7=

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
The following condition is given

 * $$f_{xx} +2pf_{xy} + p^{2}f_{yy} = g_{xp} + pg_{yp} - g_{y}\!$$

Find
Use $$\phi_{xy}=\phi_{yx}\!$$ to obtain the equation given above

Solution
We know that

\begin{align} \phi_x &= g - p(g_p - f_x) + p^2f_y \qquad \text{and that}\\ \phi_y &= g_p -pf_y - f_x \end{align} $$

So what we will need to do is find $$\phi_{xy}=\phi_{yx} \!$$ respectively and equate them to find the given condition

\begin{align} &\phi_{xy} = g_y - pg_{py} + pf_{xy} + p^2f_{yy} \\ &\phi_{yx} = g_{px} - pf_{yx} - f_{xx} \qquad \text{after re-arranging we obtain the desire given condition}\\ &f_{xx} +2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} \end{align} $$

 Very good. --Egm6321.f09.TA 03:24, 28 September 2009 (UTC)
 * }

Problem 8
$$\ F = 8x^{5}p \qquad \qquad G = 2x^{2}p + 20x^{4}(y^{'})^{2} + 4xy$$

Show that equations (4) & (5) on 10-2 for exactness are satisfied for the above equations?

Solution:

The 2nd exactness condition for a second order ODE is as follows:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. (4) $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. (5) $$

The partial derivatives are identified as,

$$ F_{x}=40x^{4}p  \qquad $$

$$ F_{xx}=160x^{3}p \qquad $$

$$ F_{xy}=0 \qquad $$

$$ F_{xp}=40x^{4}  \qquad $$

$$ F_{y}=0 \qquad $$

$$ F_{yy}=0 \qquad $$

$$ F_{yp}=0 \qquad $$

$$ G_{x}= 4xp + 80x^{3}p^{2} + 4y \qquad $$

$$ G_{xp}= 4x + 160x^{3}p \qquad $$

$$ G_{y}= 4x \qquad $$

$$ G_{yp}= 0 \qquad $$

$$ G_{p}= 2x^{2} + 40x^{4}p \qquad $$

$$ G_{pp}= 40x^{4} \qquad $$

Substituting in Equ.(4), we get

$$ \ 160x^{3}p + 2p (0) + p^{2} (0) = 4x + 160x^{3}p + p (0) - 4x $$

$$ \ 160x^{3}p = 160x^{3}p $$

Eq. (4) is satisfied.

Applying the results in Eq. (5), we get

$$ \ 40x^{4} + p(0) + 2(0) = 40x^{4} $$

$$ \ 40x^{4} = 40x^{4} $$

Eq. (5) is also satisfied.

Since Equation (4) and Equation (5) are both satisfied, then the 2nd exactness condition is satisfied. It is then concluded that the NL.ODE is exact.

 good work. --Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

= Problem 9: Exactness of N2-ODE=

Given
The following N2-ODE,

 * {| style="width:100%" border="0" align="left""

\sqrt{x}y''+2xy'+3y=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1) )
 * }
 * }

Find
Is this ODE exact?

Solution
The 2 exactness conditions are given by,

 * {| style="width:100%" border="0" align="left"

1: f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} $$ $$
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)

 * }
 * }

and

 * {| style="width:100%" border="0" align="left"

2: f_{xp} + pf_{yp} + 2f_{y} = g_{pp} $$ $$
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)

 * }
 * }

By substituting $$\displaystyle p=y'$$, into Eq. 1, the result is:

 * {| style="width:100%" border="0" align="left"

(\sqrt{x})y''+(2xp+3y)=0 $$ $$
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)

 * }
 * }

From Eq. 4 and knowing the form of a N2-ODE (See Problem 7 Eq. 1),

 * {| style="width:100%" border="0" align="left"

f(x,y,p)=\sqrt{x} $$ $$
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

g(x,y,p)=2xp+3y $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }

Taking the partial derivatives of (5) and (6) respectively yields

 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} {f_x = \frac{x^{-0.5}}{2}} & {f_{xx}  = -\frac{x^{-1.5}}{4}} & {g_x  = 2p} & {g_{xp}  = 2}  \\ {} & {f_{xy} = 0} & {g_y  = 3} & {g_{yp}  = 0}  \\ {} & {f_{xp} = 0 } & {g_p  = 2x} & {g_{pp}  = 0 }  \\ {f_y = 0} & {f_{yy}  = 0} & {} & {}  \\ {} & {f_{yp} = 0} & {} & {}  \\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle

 * }
 * }

Substituting the elements of Eq. 2 results in

 * {| style="width:100%" border="0" align="left"

(-\frac{x^{-1.5}}{4}) + 2 p (0) + p^2 (0)  = (2) + p (0) - (3) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * {| style="width:100%" border="0" align="left"

 * style="width:30%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

(-\frac{x^{-1.5}}{4}) = -1 $$ $$
 *
$$\displaystyle (Eq. 7)
 * }
 * }

Obviously, these equations are not equal, therefore given N2-ODE is NOT EXACT.
 Nice work.--Egm6321.f09.TA 02:36, 28 September 2009 (UTC)

title: User:Egm6321.f10.team5.abc/f09.h2