text:
User:Egm4313.s12.team13.r2

=Report 2=

Problem Statement
Find the solution for the general excitation for each of the two systems listed below:

No excitation when $$ r(x)=0\!$$. Solution will then be plotted.

Since the roots of the differential equations have already been given to us, it is simply a matter of transforming it into a general solution from here.

$$ \mathbf{Part\ a}\!$$

$$ \lambda_1=3,\!\qquad\lambda_2= -6\!$$

$$ y(0)=-5,\qquad\! y'(0)=2$$

The roots of the differential equation can be written as

(2.0) $$\color{Red} y(x)=c_1 e^{3x} + c_2 e^{-6x}\!$$

To find the constants, we shall now apply the initial conditions. Since the there are two unknowns in this expression, it will be necessary to take the derivative of equation 2.0 to be able to apply the second initial condition.

(2.1 $$ \frac{d }{dx}\,$$ $$y(x)= \color{Red}y'= 3c_1 e^{3x} - 6c_2 e^{-6x}\!$$

When applying the first initial condition, it results in

(2.2) $$ -5=c_1 e^{0} + c_2 e^{0}\longrightarrow\!\color{Red}-5=c_1+c_2\!$$

Applying the second initial conditions yields

(2.3) $$ 2=3c_1 e^{0} + -6c_2 e^{0}\longrightarrow\!\color{Red}2=3c_1-6c_2\!$$

By solving the system of equations, the constants come out to be

$$\color{Blue}c_1= \frac{-28}{9}\,\qquad c_2=\frac{-17}{9}\,$$

Solution Part a
$$\color{Red}\mathbf{\therefore\! \qquad y(x)=\frac{-28}{9} e^{3x} + \frac{-17}{9}e^{-6x} }$$

$$ \mathbf{Part\ b}\!$$

$$ \lambda_1=-2,\!\qquad\!\lambda_2=+5\!$$

$$ y(0)=1,\!\qquad y'(0)=0$$

Following the exact same procedure in part b as in part a, the general solution is shown below:

(2.4) $$ \color{Red} y(x)=c_1 e^{-2x} + c_2 e^{5x}\!$$

The derivative of equation 2.4 is

(2.3) $$ \frac{d }{dx}\,$$ $$y(x)= \color{Red}y'= -2c_1 e^{-2x} +5c_2 e^{5x}\!$$

Applying the first initial condition yields

(2.5) $$ 1=c_1 e^{0} + c_2 e^{0}\longrightarrow\!\color{Red}1=c_1+c_2\!$$

Applying the second initial condition yields

(2.6) $$ 0=-2c_1 e^{0} + 5c_2 e^{0}\longrightarrow\!\color{Red}0=-2c_1+5c_2\!$$

Solving the system of equations, the constants are figured out to be

$$\color{Blue}c_1= \frac{-5}{2}\,\qquad c_2=\frac{2}{7}\,$$

Solution Part b
$$\color{Red}\mathbf{\therefore\! \qquad y(x)=\frac{-5}{2} e^{-2x} + \frac{2}{7}e^{5x} }$$

Section 3 Notes

Author: Frank Portillo Reviewed : Frank Portillo

Problem Statement
Find and plot the solution for $$ y''-10y'+25=0 \!$$.

Initial Conditions: $$ y(0)=1, y'(0)=0 \!$$

No excitation: $$ r(x)=0 \!$$

Background Information
Finding Homogeneous Solutions in cases where $$ \Delta=a^2-4b=0 \!$$

The two roots are a double root:

$$ \lambda_1=\lambda_2=\lambda=-a/2\!$$ (R2.1)

The first homogeneous solution is:

$$ y_{h,1}(x)= e^{yx} \!$$ (R2.2)

To find the 2nd homogeneous solution we use a trial solution:

$$ y_{h,2}=u(x)y_{h,1}(x) \!$$where u(x) is an unknown function of x (R2.3)

$$ y'_{h,2}=uy'_{h,1}+u'y_{h,1} \!$$ (R2.4)

$$ y_{h,2}=uy_{h,1}+2u'y'_{h,1}+u''y_{h,1}\!$$ (R2.5)

Adding up equations (R2.3), (R2.4), and (R2.5) we get:

$$ 0=u[y_{h,1}+ay'_{h,1}+by_{h,1}]+u'[2y'_{h,1}+ay_{h,1}]+uy_{h,1} \!$$ (R2.6)

We can simplify equation (R2.6) by applying known equations:

$$ 0=u''y_{h,1} \!$$ (R2.7)

Hence:

$$ u''=0 \!$$ (R2.8)

Integrating with respect to x twice yields:

$$ u(x)=xc_3 \!$$
 (R2.9)

So $$ y_{h,2}=(x+c_3)y_{h,1}\!$$
 (R2.10)

The final homogeneous solution is:

$$ y_h=c_1y_{h,1}+c_2xy_{h,1} \!$$
 (R2.11)

Solution
$$ y''-10y'+25y=r(x) \!$$
 (R2.12)

The characteristic equation is:

$$ \lambda^2-10\lambda+25=0\!$$
 (R2.13)

The first homogeneous root is:

$$ \lambda=5 \!$$
 (R2.14)

Hence the homogeneous solution is:

$$ y_h(x)=c_1e^{5x}+c_2xe^{5x} \!$$
 (R2.15)

Apply the initial conditions y(0)=1 and y'(0)=0:

$$ y_h(0)= c_1 =1 \!$$
 (R2.16)

$$ y_h'(x)= 5c_1e^{5x}+5c_2xe^{5x}+c_2 \!$$
 (R2.17)

$$ y_h'(0)=5+c_2 \!$$
 (R2.18)

Hence:

$$c_1=1 \!$$
 (R2.19)

$$ c_2=-5 \!$$
 (R2.20)

The homogeneous equation is:

$$ y_h(x)=e^{5x}-5xe^{5x} \!$$
 (R2.21)

Since r(x)=0 the particular equation is zero. Hence the solution to the problem is:

$$ y(x)=e^{5x}-5xe^{5x} \!$$
 (R2.22)

The plot of the function is:

References:

Section 5 Notes

Author: Kendall Ryser Reviewed : Frank Portillo

Problem Statement
3.) Find the general solution to the following ODE and check the result by substitution.

$$ y'' + 6y' + 8.96y = 0\,\!$$

Solution
The homogeneous characteristic equation for linear 2nd order ODE's with constant coefficients is

$$ \lambda^2 + a\lambda + b = 0\,\!$$
(3.1a)

For the ODE in this problem, the characteristic equation becomes

$$ \lambda^2 + 6\lambda + 8.96 = 0\,\!$$

To solve for the solutions, the discriminant to the quadratic equation must first be calculated.

$$ \Delta = a^2 - 4b\,\!$$
(3.2a)

$$ \Delta = 36 - 4(8.96) = 0.16\,\!$$

Since the discriminant is positive, there will be two distinct real solutions to the ODE. The solutions can be obtained by solving the remainder of the quadratic formula.

$$ \lambda_{1,2} = \frac{-a \pm \sqrt{a^2 - 4b}}{2}\,\!$$
(3.3a)

$$ \lambda_{1,2} = \frac{-6 \pm \sqrt{0.16}}{2}\,\!$$ $$ = \frac{-6 \pm 0.4}{2}\,\!$$ $$ = -3 \pm 0.2\,\!$$

Therefore,

$$ \lambda_{1} = -2.8 \,\!$$
(3.4a)

$$ \lambda_{2} = -3.2 \,\!$$
(3.5a)

The two distinct, linearly independent, homogeneous solutions for the case with two real roots are

$$ y_{h,1}(x) = e^{\lambda_1 x}\,\!$$
(3.6a)

$$ y_{h,2}(x) = e^{\lambda_2 x}\,\!$$
(3.7a)

The homogeneous solution is

$$ y_{h} = c_1 y_{h,1} + c_2 y_{h,2}\,\!$$
(3.8a)

$$ y_{h}(x) = c_1 e^{\lambda_1 x} + c_2 e^{\lambda_2 x}\,\!$$

The final general solution is therefore

$$ y_{h}(x) = c_1 e^{-2.8 x} + c_2 e^{-3.2 x}\,\!$$
(3.9a)

To check this result using substitution, we must take the first and second derivatives of the general solution to substitute back into the original ODE.

$$ y'(x) = -2.8 c_1 e^{-2.8x} - 3.2 c_2 e^{-3.2x}\,\!$$
(3.10a)

$$ y''(x) = 7.84 c_1 e^{-2.8x} + 10.24 c_2 e^{-3.2x}\,\!$$
(3.11a)

Substituting back into the original ODE gives

$$ y'' + 6y' + 8.96y = 0\,\!$$
(3.12a)

$$ 7.84 c_1 e^{-2.8x} + 10.24 c_2 e^{-3.2x} + 6(-2.8 c_1 e^{-2.8x} - 3.2 c_2 e^{-3.2x}) + 8.96(c_1 e^{-2.8 x} + c_2 e^{-3.2 x}) = 0\,\!$$

$$ \cancelto{0}{(7.84 - 16.8 + 8.96)c_1 e^{-2.8x}} + \cancelto{0}{(10.24 - 19.2 + 8.96)c_2 e^{-3.2x}} = 0\,\!$$

Since all terms cancel to 0, $$ y(x) = c_1 e^{-2.8 x} + c_2 e^{-3.2 x}\,\!$$ is a general solution to the original ODE.

Section 3 Lecture Notes

Section 5 Lecture Notes

Author: Gabrielle Steinberg Reviewed : Frank Portillo

Problem Statement
4.) Find the general solution to the following ODE and check the result by substitution.

$$ y'' + 4y' + (\pi^2 + 4)y = 0\,\!$$

Solution
The homogeneous characteristic equation for linear 2nd order ODE's with constant coefficients is

$$ \lambda^2 + a\lambda + b = 0\,\!$$
(3.1b)

For the ODE in this problem, the characteristic equation becomes

$$ \lambda^2 + 4\lambda + (\pi^2 + 4) = 0\,\!$$

To solve for the solutions, the discriminant to the quadratic equation must first be calculated.

$$ \Delta = a^2 - 4b\,\!$$
(3.2b)

$$ \Delta = 16 - 4(\pi^2 + 4)\,\!$$ $$ = 16 - 4\pi^2 - 16\,\!$$ $$ = -4\pi^2\,\!$$

Since the discriminant is negative, there will be two imaginary solutions to the ODE. The solutions can be obtained by solving the remainder of the quadratic formula.

$$ \lambda_{1,2} = \frac{-a \pm \sqrt{a^2 - 4b}}{2}\,\!$$
(3.3b)

$$ \lambda_{1,2} = \frac{-4 \pm \sqrt{-4\pi^2}}{2}\,\!$$ $$ = \frac{-4 \pm 2\pi i}{2}\,\!$$ $$ = -2 \pm \pi i\,\!$$

Therefore,

$$ \lambda_{1} = -2 + \pi i \,\!$$
(3.4b)

$$ \lambda_{2} = -2 - \pi i \,\!$$
(3.5b)

The two distinct, linearly independent, homogeneous solutions for the case with two imaginary roots are

$$ y_{h,1}(x) = e^{a_1 x} \cos b_1 x\,\!$$
(3.6b)

$$ y_{h,2}(x) = e^{a_2 x} \sin b_2 x\,\!$$
(3.7b)

The homogeneous solution is

$$ y_{h} = c_1 y_{h,1} + c_2 y_{h,2}\,\!$$
(3.8b)

$$ y_{h}(x) = c_1 e^{a_1 x} \cos b_1 x + c_2 e^{a_2 x} \sin b_2 x\,\!$$

$$ y_{h}(x) = c_1 e^{-2 x} \cos \pi x + c_2 e^{-2 x} \sin \pi x\,\!$$

The final general solution is therefore

$$ y_{h}(x) = e^{-2 x}(c_1 \cos \pi x + c_2 \sin \pi x)\,\!$$
(3.9b)

To check this result using substitution, we must take the first and second derivatives of the general solution to substitute back into the original ODE.

$$ y'(x) = -2e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) + e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)\,\!$$
(3.10b)

$$ y''(x) = 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - 2e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) - 2e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)\,\!$$
(3.11b)

$$ = 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - 4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)\,\!$$

Substituting back into the original ODE gives

$$ y'' + 4y' + (\pi^2 + 4)y = 0\,\!$$
(3.12b)

$$ 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - \cancelto{0}{4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)} + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x) - 8e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) + \cancelto{0}{4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)} + (\pi^{2} + 4)(e^{-2x})(c_1 \cos \pi x + c_2 \sin \pi x) = 0\,\!$$

$$ \cancelto{0}{-4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x)} + \cancelto{0}{e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)} + \cancelto{0}{(\pi^{2} + 4)(e^{-2x})(c_1 \cos \pi x + c_2 \sin \pi x)} = 0\,\!$$

Since all terms cancel to 0, $$ y(x) = e^{-2 x}(c_1 \cos \pi x + c_2 \sin \pi x)\,\!$$ is a general solution to the original ODE.

Section 5 Lecture Notes

Section 6 Lecture Notes

Section 5 Notes

Author: Gabrielle Steinberg Reviewed : Frank Portillo

Problem Statement
5. Find a general solution. Check by substitution.

$$  \; {y}''+2\pi {y}'+\pi ^{2}y=0 $$

Solution
$$  \; {y}''+2\pi {y}'+\pi ^{2}y= \frac{d^2y}{dx^2}\ + 2 \pi \frac{dy}{dx}\ + \pi ^{2}y=0   $$
(4.0a)

If $$ \; \frac{d}{dx}\ = \lambda $$,

The homogeneous linear ODE with constant coefficients becomes:

$$ \; \lambda ^{2}+2\pi \lambda +\pi ^{2}=0 $$
(4.1a)

By using the quadratic formula, or in this case factoring, we find the solutions of the equation:

Solving for $$ \; \lambda $$,

$$ \; (\lambda +\pi) (\lambda +\pi)=0 $$
(4.2a)

$$ \; \lambda =-\pi, -\pi $$

General Solution:

General solution in the case of distinct real roots:

$$  \; y=c_1e ^ { \lambda _1 x} + c_2xe ^ { \lambda _2 x}  $$
(4.3a)

So the general solution becomes:

$$  \; y=c_1e ^ {-\pi x} + c_2xe ^ {-\pi x}= e ^ {-\pi x} (c_1+c_2x) $$
(4.4a)

Check by Substitution:

To check by substitution, you need to find $$  \; y, y', y'' $$ and plug them into the original ODE.

$$  \; y= e ^ {-\pi x} (c_1+c_2x) $$

$$  \; y'= -\pi c_1e ^ {-\pi x} + c_2 (e ^ {-\pi x} -\pi xe ^ {-\pi x})= - \pi e ^{- \pi x} (c_1 + c_2x) + c_2 e^ {- \pi x} $$
(4.5a)

$$  \; y''= \pi ^{2} c_1e ^ {- \pi x} - \pi c_2e ^ {-\pi x} -\pi c_2(e ^ {- \pi x} - \pi xe^ {- \pi x} )= \pi ^ {2} e^ {- \pi x} (c_1+ c_2x) -2 \pi c_2 e^ {- \pi x} $$
(4.6a)

Plugging in  $$   \; y, y', y $$ into $$   \; {y}+2\pi {y}'+\pi ^{2}y=0 $$ :

$$  \; [ \pi ^ {2} e^ {- \pi x} (c_1 + c_2x) - 2 \pi c_2e^ {- \pi x}] + 2 \pi [ - \pi e^ {- \pi x} (c_1 + c_2x) + c_2e^ {- \pi x} ] + \pi ^ {2} [ e^ {- \pi x} (c_1+c_2x)]=0 $$

When plugging in, the general solution is confirmed because it equals zero.

Section 5 Lecture Notes

Section 5 Notes

Author: Ryan Hagberg Reviewed : Frank Portillo

Problem Statement
6. Find the general solution. Check by substitution.

$$ 10y''-32y'+25.6y=0 $$

Solution
The original problem can be re-written as:
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$$  \displaystyle 10\frac{d^2y}{dx^2}-32\frac{dy}{dx}+25.6y=0 $$     (4.0b)
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then let
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$$  \displaystyle \frac{d}{dx}=\lambda $$     (4.1b) By substituting equation 4.1b into 4.0b we get:
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$$  \displaystyle 10\lambda^2-32\lambda+25.6=0 $$     (4.2b) Solve for $$ \; \lambda $$ using the quadratic equation.
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Quadratic equation:
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$$  \displaystyle \lambda=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$     (4.3b) By using equation 4.3b it is shown that $$ \lambda=\frac{8}{5}$$
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Since $$ \; \lambda$$ is a solution to the ODE then
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$$  \displaystyle y=c_1e ^ { \lambda _1 x} + c_2e ^ { \lambda _2 x} $$ (4.4b) is the solution of the ODE with real and distinct roots.
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By plugging the value of $$ \; \lambda $$ into equation 4.4b we come to the general solution.

General Solution $$ y=e^{\frac{8}{5}x}({c_1+c_2 x}) $$

Checking by Substitution To check by substitution, you need to find $$ \; y, y', y''$$ and plug them into the original ODE.

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$$  \displaystyle y'=\frac{8}{5}e^{\frac{8}{5}x}(c_1+c_2 x)+c_2 e^{\frac{8}{5}x} $$     (4.5b)
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$$  \displaystyle y''=\frac{64}{25}e^{\frac{8}{5}x}(c_1+c_2 x)+\frac{16}{5}c_2 e^{\frac{8}{5}x} $$     (4.6b) Since we already know what $$ \; y $$ is equal to from our general solution we are ready to substitute the general solution, equation 4.5b, and 4.6b back into the original equation.
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After substitution we get:
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$$  \displaystyle \frac{64}{25}e^{\frac{8}{5}x}(c_1+c_2 x)-\frac{64}{25}e^{\frac{8}{5}x}(c_1+c_2 x)+32c_2e^{\frac{8}{5}x}-32c_2e^{\frac{8}{5}x}=?? $$ This equation clearly equals  zero  which proves our general solution is  correct
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Section 5 Lecture Notes

Section 5 Notes

Author: Ryan Hagberg Reviewed : Frank Portillo

Problem R2.5a
Problem 16 on pg 59-Find an ODE $$ \displaystyle y'' + ay' +by=0$$ for the given basis of  $$ \displaystyle e^{2.6x}, e^{-4.3x}$$ This is a case with two distinct Real-Roots. Therefore the general equation for the ODE will be

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$$  \displaystyle y=c_1e^{\lambda_1x}+c_2e^{\lambda_2x} $$     (5.0a)
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Where $$\displaystyle {\lambda_1x}=2.6x $$  and   $$ \displaystyle {\lambda_2x} = -4.3x $$ so that

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$$  \displaystyle y=c_1e^{2.6}+c_2e^{-4.3} $$     (5.1a)
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The auxiliary equation is therefore

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$$  \displaystyle ({\lambda}-2.6)({\lambda}+4.6)=0 $$     (5.2a)
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$$  \displaystyle {\lambda^2} + 1.7{\lambda}-11.18=0 $$     (5.3a)
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which results in

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$$  \displaystyle y'' + 1.7y'-11.18y=0 $$     (5.4a)
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Solution

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$$\displaystyle y'' + 1.7y'-11.18y=0 $$
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Section 5 Notes

Author: Amy Mclarty Reviewed : Frank Portillo

Problem Statement
17. Find an ODE for the given solutions.

$$ e ^{- \sqrt{5} x} $$, $$ e ^{- \sqrt{5} x} $$

using $$ \; {y}''+a {y}'+b y=0 $$ for the given basis.

Solution
Case: Real double root ODE

where $$ \; \lambda = -a/2 $$

(2.5b.0) and the general solution is $$ \; y= e ^ {-ax} (c_1+ c_2x)  $$

(2.5b.1) So $$   \; e ^ { \lambda x} =  e ^ {- \sqrt {5} x}  $$

Plugging in $$ \; \lambda = -a/2 $$

(2.5b.2) $$  \; e ^ {-ax/2} =  e ^ {- \sqrt {5} x}  $$

Solving for a,

(2.5b.3) $$  \;  -ax/2= - \sqrt {5} x  $$

$$  \; a= 2 \sqrt {5}  $$

(2.5b.4) By plugging a into the determinant $$ \; a^2-4b=0 $$, Solve for b.

(2.5b.5) $$  \; {({2 \sqrt {5}})}^2 -4b=0  $$

(2.5b.6) $$  \; 4(5)=4b $$

$$  \; b=5  $$

By plugging in a and b, the ODE is:

(2.5b.7) $$  \; {y}''+ 2 \sqrt {5}{y}'+ 5y=0 $$

Section 5 Lecture Notes

Section 5 Notes

Author: Neisha Ramnarine Reviewed : Frank Portillo

Problem Statement
Using the spring-dashpot-mass system shown below (from Fig p1.4) and following similar characteristics of (3)p5.5 but using the double real root $$\displaystyle \lambda = -3 $$, find values for the parameters k,c,m.

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$$  \displaystyle {(\lambda -5)}^{2} = {\lambda }^{2}-10\lambda +25 = 0 $$     ((3)p5.5)
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Solution
Rewriting ((3)p5.5) with the double real root $$\displaystyle \lambda = -3 $$ gives the new characteristic equation

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$$  \displaystyle {(\lambda -3)}^{2} = {\lambda }^{2}-6\lambda +9 = 0 $$     (6.0)
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From kinematics:

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$$  \displaystyle y = {y}_{k}+{y}_{c} $$     (6.1)
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From kinetics:

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$$  \displaystyle my''+{f}_{1} = f(t) $$     (6.2)
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$$  \displaystyle {f}_{1} = {f}_{k} = {f}_{c} $$     (6.3)
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Relationships:

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$$  \displaystyle {f}_{k} = k{y}_{k} $$     (6.4)
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$$  \displaystyle {f}_{c} = c{y'}_{c} $$     (6.5)
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From equation (6.1) take the second derivative

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$$  \displaystyle y = {y}_{k} = {y''}_{c} $$     (6.6)
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From equations (6.3), (6.4) & (6.5)

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$$  \displaystyle {f}_{k} = {f}_{c} \Rightarrow k{y}_{k} = c{y'}_{c} \Rightarrow {y'}_{c} = \frac{k}{c}{y}_{k} $$     (6.7)
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From equations (6.6) & (6.7)

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$$  \displaystyle y = {y}_{k}+({y'}_{c})' \Rightarrow {y''}_{k}+(\frac{k}{c}{y}_{k})' $$     (6.8)
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From equations (6.2), (6.3) & (6.8)

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$$  \displaystyle m({y''}_{k}+\frac{k}{c}{y'}_{k})+k{y}_{k} = f(t) $$     (6.9)
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From the characteristic equation (6.0), the corresponding homogeneous L2-ODE-CC is

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$$  \displaystyle y''+6y'+9y = 0 $$     (6.10)
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Now take equation (6.9) and divide through by m

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$$  \displaystyle {y''}_{k}+\frac{k}{cm}{y'}_{k}+\frac{k}{m}{y}_{k} = \frac{f(t)}{m} = 0 $$     (6.11)
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Comparing to equation (6.10) gives

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$$  \displaystyle \frac{k}{cm} = 6 $$     (6.12) &
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$$  \displaystyle \frac{k}{m} = 9 $$     (6.13)
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There are an infinite number of solutions to these two equations, one of these is

$$   \displaystyle k = 9\;\;m = 1\;\;c = \frac{3}{2}$$

Section 1 Lecture Notes

Section 5 Lecture Notes

Section 5 Notes

Author: Steven Rossmeissl Reviewed : Frank Portillo

Problem Statement
Develop the MacLaurin series (Taylor series at t=0) for

$$e^{t} \qquad \sin t \qquad \ cos t\ $$

To develop any MacLaurin series, it is important to follow the following formula:

(7.0) $$ f(x) = f(a) + f'(a)\frac{(x-a)}{1!}^{1}\ + f(a)\frac{(x-a)}{2!}^{2} +f'(a)\frac{(x-a)}{3!}^{3}\ + \cdots f^{n}\frac{(x-a)}{n!}^{n}$$

This equation can be simplified by

(7.1) $$ \sum_{n=0}^{\infty} f^{n}(a)\frac{(x-a)}{n!}^{n}$$

When writing the Maclaurin series for $$ e^{t}$$, using equation 7.0, it looks like

(7.2) $$ 1 + (1)\frac{(x-0)}{1!}^{1} + (1)\frac{(x-0)}{2!}^{2} + (1)\frac{(x-0)}{3!}^{3}$$

which can be simplified to

Solution R2.7a

(7.3) $$ e^{t} = 1 + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}\cdots\frac{x^n}{n!} $$

Similarly, starting with the function

$$ \sin t\ $$

the MacLaurin series looks like:

(7.4) $$ 1\frac{(t-0)}{1!}^{1}+ 0 \frac{(t-0)}{2!}^{2} +1\frac{(t-0)}{3!}^{3} + 0\frac{(t-0)}{4!}^{4} $$

which can be simplified into

Solution R2.7b

(7.5) $$ \sin t = t - \frac{t^3}{3!} + \frac{t^5}{5!} +\cdots - \frac{t^{n-1}}{(t-1)!} + \frac{t^n}{n!} $$

Finally, starting with

(7.6) $$ \cos t\ $$, the MacLaurin series looks like

(7.7) $$ 1 - (0)\frac{(x-0)}{1!}^{1} - (1)\frac{(x-0)}{2!}^{2} + (0)\frac{(x-0)}{3!}^{3} + (1)\frac{(x-0)}{4!}^{4}$$

which can be simplified into

Solution R2.7c

(7.8) $$ \cos t = 1 - \frac{t^2}{2!} + \frac{t^4}{4!} +\cdots - \frac{t^{n-1}}{(t-1)!} + \frac{t^n}{n!} $$

Problem Statement
8. Find the general solution. Check by substitution.

$$ y''+y'+3.25y=0 $$

Solution
The original problem can be re-written as:
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$$  \displaystyle \frac{d^2y}{dx^2}+\frac{dy}{dx}+3.25y=0 $$     (8.0a)
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then let
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$$  \displaystyle \frac{d}{dx}=\lambda $$     (8.1a) By substituting equation 8.1a into 8.0a we get:
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$$  \displaystyle \lambda^2+\lambda+3.25=0 $$     (8.2a) Solve for $$ \; \lambda $$ using the quadratic equation.
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 * }

Quadratic equation:
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$$  \displaystyle \lambda=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$     (8.3a) By using equation 8.3a it is shown that $$ \lambda=\frac{-1\pm i \sqrt{12}}{2}$$
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 *

 * }

Since $$ \; \lambda$$ is a solution to the ODE then
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$$  \displaystyle y=e^{-\frac{1}{2}x}(c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x) $$     (8.4a) is the solution of the ODE with real and distinct roots.
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General Solution $$ y=e^{-\frac{1}{2}x}(c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x) $$

Checking by Substitution To check by substitution, you need to find $$ \; y, y', y''$$ and plug them into the original ODE.

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$$  \displaystyle y'=-\frac{1}{2}e^{-\frac{1}{2}x}(c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x)+e^{-\frac{1}{2}x}(-\sqrt{12}c_1 \sin\sqrt{12}x+\sqrt{12}c_2 \cos\sqrt{12}x) $$     (8.5a)
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$$  \displaystyle y''=\frac{1}{4}e^{-\frac{1}{2}x}(c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x)-\frac{1}{2}e^{-\frac{1}{2}x}(-\sqrt{12}c_1 \sin\sqrt{12}x+\sqrt{12}c_2 \cos\sqrt{12}x)-\frac{1}{2}e^{-\frac{1}{2}x}(-\sqrt{12}c_1 \sin\sqrt{12}x+\sqrt{12}c_2 \cos\sqrt{12}x)$$
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$$+\frac{1}{2}e^{-\frac{1}{2}x}(-12c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x) $$     (8.6a) Since we already know what $$ \; y $$ is equal to from our general solution we are ready to substitute the general solution, equation 8.5a, and 8.6a back into the original equation.
 *

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After substitution we get:
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$$  \displaystyle y''+y'+3.25y=0 $$ So the solution to the ODE is: $$ y=e^{-\frac{1}{2}x}(c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x) $$
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Section 6 Lecture Notes

Author: Ryan Hagberg Reviewed : Frank Portillo

Problem Statement
15. Find a general solution. Check by substitution.

$$  \; {y}''+0.54 {y}'+(0.0729+ \pi )y=0 $$

Solution
The given ODE can be rewritten as:

$$  \; {y}''+0.54 {y}'+(0.0729+ \pi )y= \frac{d^2y}{dx^2}\ + 0.54 \frac{dy}{dx}\ + (0.00729+ \pi )y=0   $$

If  $$ \; \frac{d}{dx}\ = \lambda $$,

The ODE becomes:

(2.8b.0) $$ \; \lambda ^{2}+0.54 \lambda +(0.0729+ \pi)=0 $$

Solving for $$ \; \lambda $$ by using the quadratic formula,

(2.8b.1) $$ \lambda_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\,\!$$

(2.8b.2) $$ \lambda_{1} = -0.27 -{ \sqrt{ \pi } i} $$

(2.8b.3) $$ \lambda_{2} = -0.27 +{ \sqrt{ \pi } i} $$

(2.8b.4) In the standard form: $$ \; \lambda=a + bi $$.

General Solution:

The general homogeneous solution is:

(2.8b.5) $$ y= c_1 e^{a_1 x} \cos b_1 x + c_2 e^{a_2 x} \sin b_2 x\ \!$$

When plugging in the values for a and b, the general solution becomes:

(2.8b.6) $$ y= c_1 e^{-0.27 x} \cos \sqrt{ \pi }x + c_2 e^{-0.27 x} \sin \sqrt{ \pi }x\ \!$$

(2.8b.7) So $$ y=e^{-0.27 x} ( c_1 \cos \sqrt{ \pi }x + c_2 \sin \sqrt{ \pi }x) \ \!$$

Check by Substitution:

Solving for derivatives:

(2.8b.8) $$ y=e^{-0.27 x} ( c_1 \cos \sqrt{ \pi }x + c_2 \sin \sqrt{ \pi }x) \ \!$$

(2.8b.9) $$ \; y'=e^{-0.27 x}(- \sqrt{ \pi } c_1 \sin \sqrt{ \pi }x + \sqrt{ \pi } c_2 \cos \sqrt{ \pi }x)-0.27e^{-0.27x}( c_1 \cos \sqrt { \pi }x + c_2 \sin \sqrt{ \pi }x) \ \!$$

(2.8b.10) $$ \; y''=e^{-0.27 x}(- \pi c_1 \cos \sqrt{ \pi }x - \pi c_2 \sin \sqrt{ \pi }x)+ 0.0729e^{-0.27x}( c_1 \cos \sqrt { \pi }x + c_2 \sin \sqrt{ \pi }x)-0.54e^{-0.27 x}(- \sqrt{ \pi } c_1 \sin \sqrt{ \pi }x + \sqrt{ \pi } c_2 \cos \sqrt{ \pi }x) \ \!$$

Plugging in  $$   \; y, y', y'' $$ into

(2.8b.11) $$  \; {y}''+0.54 {y}'+(0.0729+ \pi )y=0 $$:

(2.8b.11) $$ \; [e^{-0.27 x}(- \pi c_1 \cos \sqrt{ \pi }x - \pi c_2 \sin \sqrt{ \pi }x)+ 0.0729e^{-0.27x}( c_1 \cos \sqrt { \pi }x + c_2 \sin \sqrt{ \pi }x)-0.54e^{-0.27 x}(- \sqrt{ \pi } c_1 \sin \sqrt{ \pi }x + \sqrt{ \pi } c_2 \cos \sqrt{ \pi }x)]$$

(2.8b.12) $$ \; +0.54 [e^{-0.27 x}(- \sqrt{ \pi } c_1 \sin \sqrt{ \pi }x +  \sqrt{ \pi } c_2 \cos \sqrt{ \pi }x)-0.27e^{-0.27x}( c_1 \cos \sqrt { \pi }x + c_2 \sin \sqrt{ \pi }x)]+(0.0729+ \pi )[e^{-0.27 x} ( c_1 \cos \sqrt{ \pi }x + c_2 \sin \sqrt{ \pi }x)]=0 $$

When plugging in, the general solution is confirmed because it equals zero.

Section 5 Notes

Author: Neisha Ramnarine Reviewed : Frank Portillo

Problem Statement
With the given initial conditions, find and plot the solution to

$$ \lambda^2 + 4\lambda + 13 = 0 \!$$

The initial conditions are:

$$ y(0) = 1 \qquad y'(0) = 0 \!$$

It important to remember, there is no excitation in this problem,

$$ i.e. \qquad r(x)=0\ $$

According to Kreyszig, this solution can be found by using the quadratic to find the roots. These roots come out to be

(9.0) $$ \lambda_1 = -2 + 3i \qquad \lambda_2= -2 - 3i$$

We can now write the general solution to the problem as

(9.1) $$ y_h= e^{-2x} [Acos(3x) + Bsin(3x)] \!$$

it is now possible to apply the initial conditions to this solution, since we have two, it is necessary to take the derivative of the general solution.

(9.2) $$ y(0) = e^0 [A\cos(0) + B\sin(0) ]= 1 \color{Red} \rightarrow A=1 $$

Using the second pair of initial conditions yields

(9.3) $$y'(0) = 0 = -2 \color{Red} (1) \color{Black} e^{0} \cos(0) + 3\color{Red} B \color{Black} e^{0} \ cos(0) \color{Red} \rightarrow B = 2/3$$

Solution
$$ y_h= e^{-2x} [cos(3x) + \frac{2}{3}sin(3x)] \!$$

Section 6 Lecture Notes

Author: Frank Portillo Reviewed : Frank Portillo

= Contributing Members =

Table code referenced from Team 1

title: User:Egm4313.s12.team13.r2


text:
User:Egm4313.s12.team13.fsp.r1

=Report 1=

Problem R1

Derive the equation of the motion of a spring dashpot system in parallel with a mass and applied force $$ f(t)\,\!$$

First we examine the figure, and declare constants. $$ y \,\!$$ is the distance the coil expands, and k is the spring constant. Therefore, we are able to define a force $$ F_1\,\!$$ as

(1.0) $$ F_1= -ky\,\!$$.

The force is negative because in this model, the up position is negative and the down position is positive. This is a force opposing the motion of the mass, thus it is negative. Thereafter, we know Newton's second law as

$$ Force= mass* acceleration \,\!$$, or otherwise known as

$$ F=m\,\!$$$$ \frac{dV}{dt}\,\!$$

From calculus, it is also known that $$ \frac{dy}{dx}\,\!$$ can also be written as $$ y',\!$$. Furthermore, acceleration is known as

$$ \frac{dV}{dt}\,\!$$ $$ =\!$$ $$ \frac{d^2y}{dx^2}\,\!$$ $$ = \,\!$$ $$ y'',\!$$ (1.1)

If we sum the forces and equal them to zero we ca write Newtons second law for this system as shown below

$$ \sum{F}\!$$ $$ = \!$$$$ m\,\!$$$$ \frac{dV}{dt}\,\!$$ (1.2)

$$ my''+ky=0 \!$$ (Notice it is important to have all terms on the same side of the)

It is important to note, this force only includes the restoring force of the spring. It is now necessary to include the damper term from the dashpot. The damping constant will be defined as $$ C\!$$. It is also assumed the damping force is proportional to the velocity (Kreyszig, 2011). Therefore, we are able to write

$$ F_2= -c y',\!$$ (1.3)

and by summing forces one more time:

$$ \sum{F}\! = $$ $$ \frac{dV}{dt}\,\!$$ $$ \longrightarrow\!$$ $$F_1 + F_2=my \!$$ $$ \longrightarrow,\!$$$$ my+ cy' +ky =0 \!$$ (1.5)

To get the differential equation in standard form, the only thing required is to divide the equation by the mass.

Solution
$$ y''+ \frac{c}{m}y' +\frac{c}{m}y =0 \!$$

Since this is a parallel system, we are able to say the spring force relation and the damper force relation are equal to each other, which is the reason we don't differ from the $$ y's,\!$$ with subscripts.

Picture obtained from here Section 1 Lecture Notes

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

title: User:Egm4313.s12.team13.fsp.r1


text:
User:Egm4313.s12.team13.fsp.r2

Problem Statement
Find the solution for the general excitation for each of the two systems listed below:

No excitation when $$ r(x)=0\!$$. Solution will then be plotted.

Since the roots of the differential equations have already been given to us, it is simply a matter of transforming it into a general solution from here.

$$ \mathbf{Part\ a}\!$$

$$ \lambda_1=3,\!\qquad\lambda_2= -6\!$$

$$ y(0)=-5,\qquad\! y'(0)=2$$

The roots of the differential equation can be written as

(2.0) $$\color{Red} y(x)=c_1 e^{3x} + c_2 e^{-6x}\!$$

To find the constants, we shall now apply the initial conditions. Since the there are two unknowns in this expression, it will be necessary to take the derivative of equation 2.0 to be able to apply the second initial condition.

(2.1 $$ \frac{d }{dx}\,$$ $$y(x)= \color{Red}y'= 3c_1 e^{3x} - 6c_2 e^{-6x}\!$$

When applying the first initial condition, it results in

(2.2) $$ -5=c_1 e^{0} + c_2 e^{0}\longrightarrow\!\color{Red}-5=c_1+c_2\!$$

Applying the second initial conditions yields

(2.3) $$ 2=3c_1 e^{0} + -6c_2 e^{0}\longrightarrow\!\color{Red}2=3c_1-6c_2\!$$

By solving the system of equations, the constants come out to be

$$\color{Blue}c_1= \frac{-28}{9}\,\qquad c_2=\frac{-17}{9}\,$$

Solution Part a
$$\color{Red}\mathbf{\therefore\! \qquad y(x)=\frac{-28}{9} e^{3x} + \frac{-17}{9}e^{-6x} }$$

$$ \mathbf{Part\ b}\!$$

$$ \lambda_1=-2,\!\qquad\!\lambda_2=+5\!$$

$$ y(0)=1,\!\qquad y'(0)=0$$

Following the exact same procedure in part b as in part a, the general solution is shown below:

(2.4) $$ \color{Red} y(x)=c_1 e^{-2x} + c_2 e^{5x}\!$$

The derivative of equation 2.4 is

(2.3) $$ \frac{d }{dx}\,$$ $$y(x)= \color{Red}y'= -2c_1 e^{-2x} +5c_2 e^{5x}\!$$

Applying the first initial condition yields

(2.5) $$ 1=c_1 e^{0} + c_2 e^{0}\longrightarrow\!\color{Red}1=c_1+c_2\!$$

Applying the second initial condition yields

(2.6) $$ 0=-2c_1 e^{0} + 5c_2 e^{0}\longrightarrow\!\color{Red}0=-2c_1+5c_2\!$$

Solving the system of equations, the constants are figured out to be

$$\color{Blue}c_1= \frac{-5}{2}\,\qquad c_2=\frac{2}{7}\,$$

Solution Part b
$$\color{Red}\mathbf{\therefore\! \qquad y(x)=\frac{-5}{2} e^{-2x} + \frac{2}{7}e^{5x} }$$

Section 3 Notes

title: User:Egm4313.s12.team13.fsp.r2


text: File:Html challenge 8 form style.png

title: File:Html challenge 8 form style.png


text: File:F(x)R2 1.jpg

title: File:F(x)R2 1.jpg


title: File:CORDIC.Background.4.A.20120204.pdf


text:
Bromley by Bow Centre/Study Circle One

The information here is designed to relate to:
 * UEL: IS3501 - 2011-12 Organising People, Resources and Power
 * UEL: IS3503 - 2011-12 Business Models and Client Based Design

title: Bromley by Bow Centre/Study Circle One


text:
Butterfly Software Implementations

Purpose

 * developing FFT Butterfly parallel algorithms using MPI, OpenMP, ....

Background
 FFTW
 * [[Media:BGND.1.A.FFTW.20130202.pdf       | 1.A FFTW]]

 P3DFFT
 * [[Media:BGND.2.A.P3DFFT.20120000.pdf     | 2.A P3DFFT]]

 2DECOMP FFT
 * [[Media:BGND.3.A.2DECOMP_FFT.20120000.pdf | 3.A 2DECOMP&FFT]]

 SPIRAL
 * [[Media:BGND.4.A.SPIRAL.20120000.pdf     | 4.A SPIRAL]]

 MPI
 * See MPI programming in plain view

 Beowulf Cluster

Idea Sketch

 * Idea.1.A ([[Media:Butterfly.SW.1.A.Idea.20120208.2.pdf |pdf]])

Applications
go to [ Electrical_%26_Computer_Engineering_Studies ]
 * geoscience & remote sensing
 * nest
 * esa
 * sdr
 * debian remote sensing

title: Butterfly Software Implementations


text:
Bromley by Bow Centre/Business Models and Client Based Design

Assignments
1) Case study (2000 words) 50%
 * Describe/analyse/compare/contrast two different business models, employed either by the same organisation, or by different organisations, showing evidence of their effectiveness, and suggesting explanations for what you find.
 * Learning Outcomes to be demonstrated:

 * Demonstrate an understanding of business models and related concepts
 * Critically analyse opportunities in specific markets
 * Assess strategic options for a range of service providers and other not-for-profit organisations
 * Examine and explain demand, supply and value issues, and complexities, in specific markets
 * Distinguish between cases, situations, and types of business model
 * Critically evaluate the performance, policies and approaches of organisations in relation to their business models.

2) Proposal (2000 words) 50%
 * Write a proposal for the development or adoption of a new or changed business model by a real-world organisation, showing evidence to support the idea, and providing an argument for its advantages over other options.
 * Learning Outcomes to be demonstrated:

 * Critically analyse opportunities in specific markets
 * Assess strategic options for a range of service providers and other not-for-profit organisations
 * Examine and explain demand, supply and value issues, and complexities, in specific markets
 * Distinguish between cases, situations, and types of business model
 * Diagnose management and other problems in actual and proposed services and other organisational undertakings
 * Produce rationales for the adoption of specific business models
 * Collaborate in the investigation of cases, and articulation of findings and analysis

Business model

 * "Every Good Regulator of a system must be a model of that system" Conant and Ross Ashby
 * "Every Good Key Must Be A Model Of The Lock It Opens" Daniel L. Scholten
 * "The gist of Ashby’s First Law is that any model is better than no model at all, which basically means that even if we can’t figure out the truth of a situation, we are almost always better off just making up any old thing rather than muddling around in the muck of our own ignorance."

Viable Systems model

 * Viable Systems Model by Jon Walker
 * A Viable System Model: Consideration of Knowledge Management by Allenna Leonard

Complexity

 * "Complexity is not the same as variety (Saviotti 1996) . Variety refers to a diversity of types. Complexity exists only when such variety exists within a structured system. In short, complexity in the sense used here is systemically interconnected and interactive variety. By this definition, increasing economic complexity means a growing diversity of interactions between human beings and between people and their technology.
 * "The measure of complexity is called VARIETY. VARIETY is defined as the number of possible states of whatever it is whose complexity we want to measure."

Situations
Leibniz's whole phenomenal theory finally stands on these two very concepts, isomorpohism and situation - which are both taken from geometry.

Wikimedia

 * Business Model


text:
Literature/2004/Osterwalder

Business models

 * "[T]his work tries to dig into the details and define a generic model to describe business models." (p 2)
 * "A ‘model’ is a standard or example for imitation or comparison. It provides a pattern on which to base an artifact." (p. 2)
 * "A quick lookup in the online version of the Cambridge Learner's Dictionary (Cambridge 2003) returns no result for the full combined term but the following definitions for the two separate terms:
 * business: the activity of buying and selling goods and services, or a particular company that does this, or work you do to earn money.
 * model: a representation of something, either as a physical object which is usually smaller than the real object, or as a simple description of the object which might be used in calculations." (p. 14)

 * "In general the purpose of creating a model is to help understand, describe, or predict how things work in the real world by exploring a simplified representation of a particular entity or phenomenon. Thus, in the case of a business model the model (i.e. representation) shall help understand, describe and predict the "activity of buying and selling goods and services" and "earning money" of a particular company. But as the notion buying and selling seems too narrow, I try to extend it. So differently put, the business model is an abstract representation of the business logic of a company. (p. 14)
 * "[T]here is an ongoing discussion on the difference between strategy and business models (Stähler 2002; Seddon and Lewis 2003) . Currently, there are different points of view that differ widely. In this dissertation I will not address this discussion and simply look at a business model as the translation of a company's strategy into a blueprint of the company's logic of earning money. Putting strategy, business models and process models together one can say that they address similar problems (e.g. the one of earning money in a sustainable way) on different business layers." (p. 14)
 * "A business model is a conceptual tool that contains a set of elements and their relationships and allows expressing a company's logic of earning money. It is a description of the value a company offers to one or several segments of customers and the architecture of the firm and its network of partners for creating, marketing and delivering this value and relationship capital, in order to generate profitable and sustainable revenue streams. (p. 15)

Business Model Types

 * Design Busniess Models
 * Finance Busniess Models
 * Implement Busniess Models

Complexity

 * "[O]ne of the major impacts of ICTs has been an increase in the possible business configurations a company can adopt because of the reduced coordination and transaction costs (see Coase 1937 ; Williamson 1975 )" (p 2)
 * "According to Hodgson (2003) capitalism naturally leads to more complexity driven by powerful economic forces. Under this he understands a growing diversity of interactions between human beings and between people and their technology (Hodgson 2003).

Uncertainty

 * "The problem with uncertainty is that it increases the environmental risk that a company faces because the future becomes unpredictable. Referring to this Courtney, Kirkland et al. (1997) speak of four levels of uncertainty that managers face (...). At the first level there is a single view of the future, at the second level one of several futures will occur, at the third level there is a range of possible futures and at the fourth level true ambiguity rules in regard to future. Managing uncertainty is probably one of the most important challenges that managers face today. Providing a specification of the conceptualization of business models could eventually improve scenario approcaches and one day lead to simulation. This would help managers to be better prepared for the future. (p 13)

title: Literature/2004/Osterwalder


text:
User:Egm6322.s12.TEAM1.KIM/HW2

Problem 3: Legendre homogeneous solution
 Solved without assistance

Given
$$\displaystyle {{Q}_{0}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)={{\tanh }^{-1}}x $$ (3.1) $$\displaystyle {{Q}_{1}}(x)=\frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1=x{{\tanh }^{-1}}x-1 $$ (3.2)

Find
Show Eq (3.1) and Eq (3.2)

Solution
Order 0 Legendre equation is that $$\displaystyle(1-{{x}^{2}}){y}''-2x{y}'=0$$ Let z = y' $$\displaystyle {z}'+\frac{2x}{{{x}^{2}}-1}z=0$$ To know the solution, I use integrating factor method. $$\displaystyle h(x) =\exp \left[ \int_ – ^{x}{\frac{2x}{{{x}^{2}}-1}dx} \right] =\exp \left[ \int_ – ^{x}{\frac{1}{1-{{x}^{2}}}d(1-{{x}^{2}})} \right] =\exp [\log (1-{{x}^{2}})] =1-{{x}^{2}}$$  From the previous lecture (PEA1), $$\displaystyle z ={{h}^{-1}}(x)\int_ – ^{x}{h(x)*0dx} =\frac{1-{{x}^{2}}}$$  $$ y =\int{zdx}+{{k}_{1}} ={{k}_{1}}+\int_ – ^{}{\frac{1-{{x}^{2}}}dx} ={{k}_{1}}+\frac{2}\int_ – ^{}{\left( \frac{1}{1+x}+\frac{1}{1-x} \right)dx} ={{k}_{1}}+{{k}_{2}}\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) $$ $$\displaystyle Q_0$$ is one of homogeneous solutions. $$\displaystyle {{Q}_{0}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)$$  Now, We have to find what tanhx is. $$\displaystyle \tanh x = \frac{\sinh x}{\cosh x} = \frac {e^x - e^{-x}} {e^x + e^{-x}} = \frac{e^{2x} - 1} {e^{2x} + 1}$$ $$ x=\frac{1}{2}\log \left( \frac{1+\tanh x}{1-\tanh x} \right)$$ $$ {{\tanh }^{-1}}x=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) $$ These show that $$\displaystyle {{Q}_{0}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)={{\tanh }^{-1}}x$$

Order 1 Legendre equation is that $$\displaystyle (1-{{x}^{2}}){y}''-2x{y}'+2y=0$$ $$\displaystyle {y}''+\frac{2x}{({{x}^{2}}-1)}{y}'+\frac{2}{(1-{{x}^{2}})}y=0$$  using Integrating Factor Method, $$\displaystyle h(x)={{u}_{1}}^{2}(x)\exp \left[ \int_ – ^ – {{{a}_{1}}(x)dx} \right] ={{[{{P}_{1}}(x)]}^{2}}\exp \left[ \int_ – ^ – {\frac{2x}{{{x}^{2}}-1}dx} \right] ={{x}^{2}}\exp \left[ \int_ – ^ – {\frac{1}{1-{{x}^{2}}}d(1-{{x}^{2}})} \right] ={{x}^{2}}(1-{{x}^{2}})$$ $$\displaystyle {{Q}_{1}}(x) ={{u}_{2}}(x) ={{u}_{1}}(x)\int_ – ^ – {\frac{1}{h(x)}dx} =x\int{\frac{1}{{{x}^{2}}(1-{{x}^{2}})}dx} =x\int{\frac{1}+\frac{1}{(1-{{x}^{2}})}dx} =x\int{\frac{1}{(1-{{x}^{2}})}dx+}x\int{\frac{1}dx} =\frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 $$ These show that $$\displaystyle {{Q}_{1}}(x)=\frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1=x{{\tanh }^{-1}}x-1$$

Given
$$Q_n(x) = P_n(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^J \frac {2n-2j+1} {(2n-j+1)j} P_{n-j}(x)$$ $$\displaystyle J = 1+2[\frac {n-1}{2}]$$ $$\displaystyle P_0(x) = 1$$ $$\displaystyle P_1(x) = x$$ $$\displaystyle P_2(x) = \frac {1} {2} (3x^2-1)$$ $$\displaystyle P_3(x) = \frac {1} {2} (5x^3-3x)$$ $$\displaystyle P_4(x) = \frac {1} {8} (35x^4 - 30x^2 + 3)$$

Find
Verify $$\displaystyle Q_0, \,\, Q_1, \,\, Q_2 $$ using above equation.

Solution
When n = 0, $$\displaystyle J = 1+2[\frac {0-1}{2}] = 0$$ $$\displaystyle Q_0(x) = P_0(x) tanh^{-1}(x) = tanh^{-1}(x) $$ When n = 1, $$\displaystyle J = 1+2[\frac {1-1}{2}] = 1$$ $$\displaystyle Q_1(x) = P_1(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^1 \frac {2-2+1} {(2-1+1)1} P_{1-1}(x)$$ $$\displaystyle Q_1(x) = P_1(x) tanh^{-1}(x) - 1 = xtanh^{-1}(x) - 1$$ When n = 2, $$\displaystyle J = 1+2[\frac {2-1}{2}] = 2$$ $$\displaystyle Q_2(x) = P_2(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^2 \frac {2n-2j+1} {(2n-j+1)j} P_{n-j}(x)$$ $$\displaystyle Q_2(x) = (\frac 1 2 (3x^2-1)) tanh^{-1}(x) - 2 \sum_{j=1,3,5,...}^2 \frac {2(2)-2(j)+1} {(2(2)-(j)+1)(j)} P_{2-j}(x)$$ $$\displaystyle Q_2(x) = (\frac 1 2 (3x^2-1)) tanh^{-1}(x) - (\frac 3 2) (x) $$

Given
$$Q_n(x) = P_n(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^J \frac {2n-2j+1} {(2n-j+1)j} P_{n-j}(x)$$ $$\displaystyle J = 1+2[\frac {n-1}{2}]$$

Find
Show that Qn is odd or even depending on the index n.

Solution
$$\displaystyle P_2k(x) $$ is even $$\displaystyle P_{2k+1}(x)$$ is odd

Hence, Let 2k = n = even number k = 1,2,3......  $$Q_2k(x) = P_{2k}(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^J \frac {4k-2j+1} {(4k-j+1)j} P_{2k-j}(x)$$ $$\displaystyle tanh^{-1}(x)$$ is always odd. $$\displaystyle P_{2k-j}(x) $$ is odd.

$$\displaystyle Q_2k(x) = (even)(odd) - (odd) $$ When n = even, Q(x) is odd.

Let 2k+1 = n = odd number k = 1,2,3......  $$Q_{2k+1}(x) = P_{2k+1}(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^J \frac {2(2k+1)-2j+1} {(2(2k+1)-j+1)j} P_{2(2k+1)-j}(x)$$ $$\displaystyle tanh^{-1}(x)$$ is always odd. $$\displaystyle P_{2(2k+1)-j}(x) $$ is even. $$\displaystyle Q_{2k+1}(x) = (odd)(odd) - (even) $$ When n = odd, Q(x) is even.

title: User:Egm6322.s12.TEAM1.KIM/HW2


title: File:Pea2.s12.sec44.a.djvu


text:
Talk:Bromley by Bow Centre/Business Models and Client Based Design

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text:
Talk:Bromley by Bow Centre/Study Circle One

 * Centre/Study Circle One Metrics Jan 2012
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text:
Talk:Bromley by Bow Centre

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title: Talk:Bromley by Bow Centre


text:
Talk:Bromley by Bow Centre/Study Circles on Wikiversity

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text:
User:RojeshShikhrakar/Yamari

Yamari is a food in Newar community.It is composed up of two word Ya means favorite and mari means dish. So, in local language it means favorite dish. it is a white conical shaped dish filed with caramel, made only on special occasions like yamari puni, Birthdays, and so on.

title: User:RojeshShikhrakar/Yamari


text:
User talk:RojeshShikhrakar

 Hello RojeshShikhrakar, 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. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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title: User talk:RojeshShikhrakar


text:
User:Egm4313.s12.team13.akm.r1

Problem R1.2
Derive the equation of motion of the spring-mass-dashpot in fig. 5.3 in K 2011 p 85 with an applied force r(t) on the ball.

In order to derive the equation of motion of the spring-mass-dashpot with an applied force of $$ \displaystyle r(t)$$, the formula of a non-homogenous equation must first be understood.

In the non homogenous equation:
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$$  \displaystyle y'' + p(x)y' + q(x)y = r(x) $$     (2.0) The $$\displaystyle y $$ represents position, $$ \displaystyle y' $$ represents velocity, and $$ \displaystyle y'' $$ represents acceleration.
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Figure 5.3 consist of a mass $$\displaystyle m$$, a spring constant $$\displaystyle c$$, and a damper coefficent $$\displaystyle k$$ These components create 3 internal forces- intertia $$\displaystyle my''$$, damping force $$\displaystyle cy'$$, and restoring force $$\displaystyle ky$$. Along with these 3 forces is also a driving force of $$ \displaystyle r(t)$$.

Solution
Using these 4 components, equation 1 becomes:
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$$\displaystyle my'' + cy' + ky = r(t)$$
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Section 1 Lecture Notes

title: User:Egm4313.s12.team13.akm.r1


text:
User:Egm4313.s12.team13.akm.r2

Problem R2.5a
Problem 16 on pg 59-Find an ODE $$ \displaystyle y'' + ay' +by=0$$ for the given basis of  $$ \displaystyle e^{2.6x}, e^{-4.3x}$$ This is a case with two distinct Real-Roots. Therefore the general equation for the ODE will be

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$$  \displaystyle y=c_1e^{\lambda_1x}+c_2e^{\lambda_2x} $$     (5.0a)
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Where $$\displaystyle {\lambda_1x}=2.6x $$  and   $$ \displaystyle {\lambda_2x} = -4.3x $$ so that

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$$  \displaystyle y=c_1e^{2.6}+c_2e^{-4.3} $$     (5.1a)
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The auxiliary equation is therefore

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$$  \displaystyle ({\lambda}-2.6)({\lambda}+4.6)=0 $$     (5.2a)
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$$  \displaystyle {\lambda^2} + 1.7{\lambda}-11.18=0 $$     (5.3a)
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which results in

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$$  \displaystyle y'' + 1.7y'-11.18y=0 $$     (5.4a)
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Solution

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$$\displaystyle y'' + 1.7y'-11.18y=0 $$
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Section 1 Lecture Notes

title: User:Egm4313.s12.team13.akm.r2


text:
User:Egm4313.s12.team3.pliskow/R2work

=Problem 8: Finding the General Solution of Homogeneous ODE's=

Problem Statement
Find the general solution for each of the ODE's given below. Use substitution to verify the answer.
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$$\displaystyle y''+y'+3.25y=0 $$     (8.1)
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$$\displaystyle y''+0.54y'+(0.0729+\pi)y=0 $$     (8.2)
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Part A

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$$\displaystyle y''+y'+3.25y=0 $$     (8.1) This is a homogeneous, linear, second order ODE with constant coefficients in the standard form of:
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$$\displaystyle y''+ay'+by=0 $$ The corresponding characteristic equation would be:
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$$\displaystyle \lambda^2+\lambda+3.25=0 $$ We can see that the value of the discriminant of the characteristic equation $$\ a^2-4b $$ will be negative. This means that the roots will be complex conjugates.
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$$\displaystyle \lambda_{1,2}=-\frac{1}{2}a \pm i\omega $$ Where
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$$\displaystyle \omega^2=b-\frac{1}{4}a^2 $$ Therefore the roots of the characteristic equation are:
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$$\displaystyle \lambda_{1,2}=-\frac{1}{2} \pm i \sqrt{3} $$ Refer to [[media:iea.s12.sec6.djvu| Loc Vu-Quoc class EGM 4313 notes Sec 6]] for the derivation of the form of the general solution (with no excitation) given below for the case of complex conjugate roots:
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$$\displaystyle y_h(x)=e^{-ax/2}[Acos(wx)+Bsin(wx)] $$ Given the specific characteristic equation, we can apply the determined roots to get the general solution:
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$$\displaystyle y=e^{-x/2}[Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)] $$ Below, we verify that this is indeed a solution to the original ODE.
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$$\displaystyle y''+y'+3.25y=0 $$     (8.1) Before verifying that (1) satisfies the original ODE (8.1), we must calculate the first derivative:
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$$\displaystyle y'=e^{-\frac{x}{2}}[-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)]-\frac{1}{2}e^{-\frac{x}{2}}[Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)] $$    (2)
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$$\displaystyle y'=e^{-\frac{x}{2}}cos(\sqrt{3}x)[\sqrt{3}B-\frac{1}{2}A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-\sqrt{3}A-\frac{1}{2}B] $$    (3) And the second derivative:
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$$\displaystyle y''=e^{-\frac{x}{2}}[-3Acos(\sqrt{3}x)-3Bsin(\sqrt{3}x)]-\frac{1}{2}e^{-\frac{x}{2}}[-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)]-\frac{1}{2}e^{-\frac{x}{2}}[-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)]+\frac{1}{4}e^{-\frac{x}{2}}[Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)] $$    (4)
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$$\displaystyle y''=e^{-\frac{x}{2}}cos(\sqrt{3}x)[-3A-\frac{\sqrt{3}}{2}B-\frac{\sqrt{3}}{2}B+\frac{1}{4}A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-3B+\frac{\sqrt{3}}{2}A+\frac{\sqrt{3}}{2}A+\frac{1}{4}B] $$    (5) With the derivatives calculated, we can check to see if it satisfies the original ODE.
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$$\displaystyle y''+y'+3.25y=e^{-\frac{x}{2}}cos(\sqrt{3}x)[-3A-\frac{\sqrt{3}}{2}B-\frac{\sqrt{3}}{2}B+\frac{1}{4}A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-3B+\frac{\sqrt{3}}{2}A+\frac{\sqrt{3}}{2}A+\frac{1}{4}B]+ e^{-\frac{x}{2}}cos(\sqrt{3}x)[\sqrt{3}B-\frac{1}{2}A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-\sqrt{3}A-\frac{1}{2}B]+3.25e^{-\frac{x}{2}}[Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)] $$    (6)
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$$\displaystyle y''+y'+3.25y=e^{-\frac{x}{2}}cos(\sqrt{3}x)[-3A-\sqrt{3}B+\frac{1}{4}A+\sqrt{3}B-\frac{1}{2}A+3.25A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-3B+\sqrt{3}A+\frac{1}{4}B-\sqrt{3}A-\frac{1}{2}B+3.25B]
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$$    (7)
 *

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$$\displaystyle y''+y'+3.25y=e^{-\frac{x}{2}}cos(\sqrt{3}x)[-3A+\frac{1}{4}A-\frac{1}{2}A+3.25A-\sqrt{3}B+\sqrt{3}B]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[\sqrt{3}A-\sqrt{3}A+\frac{1}{4}B-\frac{1}{2}B+3.25B-3B] $$    (8)
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$$\displaystyle y''+y'+3.25y=e^{-\frac{x}{2}}cos(\sqrt{3}x)[0]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[0] $$    (8)
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$$\displaystyle y''+y'+3.25y=0 $$    (8)
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Since (8) matches the form of the original ODE, we can conclude that the general solution does indeed satisfy the ODE.
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Part B

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$$\displaystyle y''+0.54y'+(0.0729+\pi)y=0 $$     (8.2) This is a homogeneous, linear, second order ODE with constant coefficients in the standard form of:
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 *

 * }

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$$\displaystyle y''+ay'+by=0 $$ The corresponding characteristic equation would be:
 * style="width:95%" |
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 *

 * }

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

$$\displaystyle \lambda^2+0.54\lambda+(0.0729+\pi)=0 $$ We can see that the value of the discriminant of the characteristic equation $$\ a^2-4b $$ will be negative. This means that the roots will be complex conjugates.
 * style="width:95%" |
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 *

 * }

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$$\displaystyle \lambda_{1,2}=-\frac{1}{2}a \pm i\omega $$ Where
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 *

 * }

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$$\displaystyle \omega^2=b-\frac{1}{4}a^2 $$ Therefore the roots of the characteristic equation are:
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 *

 * }

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$$\displaystyle \lambda_{1,2}=-0.27 \pm i \sqrt{\pi} $$ Refer to [[media:iea.s12.sec6.djvu| Loc Vu-Quoc class EGM 4313 notes Sec 6]] for the derivation of the form of the general solution (with no excitation) given below for the case of complex conjugate roots:
 * style="width:95%" |
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 *

 * }

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

$$\displaystyle y_h(x)=e^{-ax/2}[Acos(wx)+Bsin(wx)] $$ Given the specific characteristic equation, we can apply the determined roots to get the general solution:
 * style="width:95%" |
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 *

 * }

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$$\displaystyle y=e^{-0.27x}[Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)] $$
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 *

 * }

Below, we verify that this is indeed a solution to the original ODE.

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

$$\displaystyle y''+0.54y'+(0.0729+\pi)y=0 $$     (8.2) Before verifying that (1) satisfies the original ODE (8.2), we must calculate the first derivative:
 * style="width:95%" |
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 *

 * }

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$$\displaystyle y'=e^{-0.27x}[-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x)]-0.27e^{-0.27x}[Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)] $$    (2)
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 *

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$$\displaystyle y'=e^{-0.27x}cos(\sqrt{\pi}x)[\sqrt{\pi}B-0.27A]+e^{-0.27x}sin(\sqrt{\pi}x)[-\sqrt{\pi}A-0.27B] $$    (3) And the second derivative:
 * style="width:95%" |
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 *

 * }

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$$\displaystyle y''=e^{-0.27x}[-\pi Acos(\sqrt{\pi}x)-\pi Bsin(\sqrt{\pi}x)]-0.27e^{-0.27x}[-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x)]-0.27e^{-0.27x}[-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x)]+(0.27)^{2}e^{-0.27x}[Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)] $$    (4)
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 *

 * }

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$$\displaystyle y''=e^{-0.27x}cos(\sqrt{\pi}x)[-\pi A-0.27\sqrt{\pi}B-0.27\sqrt{\pi}B+(0.27)^{2}A]+e^{-0.27x}sin(\sqrt{\pi}x)[-\pi B+0.27\sqrt{\pi}A+0.27\sqrt{\pi}A+(0.27)^{2}B] $$    (5) With the derivatives calculated, we can check to see if it satisfies the original ODE.
 * style="width:95%" |
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 *

 * }

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$$\displaystyle y''+0.54y'+(0.0729+\pi)y=e^{-0.27x}cos(\sqrt{\pi}x)[-\pi A-0.27\sqrt{\pi}B-0.27\sqrt{\pi}B+(0.27)^{2}A+0.54\sqrt{\pi}B-(0.54)(0.27)A+(0.0729+\pi)]+e^{-0.27x}sin(\sqrt{\pi}x)[-\pi B+0.27\sqrt{\pi}A+0.27\sqrt{\pi}A+(0.27)^{2}B-0.54\sqrt{\pi}A-(0.54)(0.27)B+(0.0729+\pi)B] $$    (6)
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$$\displaystyle y''+0.54y'+(0.0729+\pi)y=e^{-0.27x}cos(\sqrt{\pi}x)[-\pi A-(0.54)(0.27)A+(0.0729+\pi)A+(0.27)^{2}A-0.27\sqrt{\pi}B-0.27\sqrt{\pi}B+0.54\sqrt{\pi}B]+e^{-0.27x}sin(\sqrt{\pi}x)[0.27\sqrt{\pi}A+0.27\sqrt{\pi}A-0.54\sqrt{\pi}A-\pi B+(0.27)^{2}B-(0.54)(0.27)B+(0.0729+\pi)B] $$    (7)
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$$\displaystyle y''+0.54y'+(0.0729+\pi)y=e^{-0.27x}cos(\sqrt{\pi}x)[0A+0B]+e^{-0.27x}sin(\sqrt{\pi}x)[0A+0B] $$    (8)
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$$\displaystyle y''+0.54y'+(0.0729+\pi)y=e^{-0.27x}cos(\sqrt{\pi}x)[0]+e^{-0.27x}sin(\sqrt{\pi}x)[0] $$    (8)
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$$\displaystyle y''+0.54y'+(0.0729+\pi)y=0 $$    (8)
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Since (8) matches the form of the original ODE, we can conclude that the general solution does indeed satisfy the ODE.
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Problem Statement
In this problem we will consider several mathematical representations of physical systems using ODE's. Given the characteristic equation (or its roots), find the general solution for the corresponding ODE. Then, use the initial conditions to determine the particular solution. Plot the solution for each. Part A
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$$\displaystyle \lambda^{2}+4\lambda+13=0 $$     (9.1)
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Part B
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$$\displaystyle \lambda_{1,2}=-3 $$     (9.2)
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Part C
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$$\displaystyle \lambda_1=-2 \;\;,\;\; \lambda_2=5 $$     (9.3) Consider all of the above systems given the following initial conditions.
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Part A

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$$\displaystyle \lambda^{2}+4\lambda+13=0 $$     (9.1)
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This characteristic equation would correspond to the following homogeneous ODE.

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$$\displaystyle {y}''+4{y}'+13y=0 $$ To find the general solution to this ODE, we need to determine the roots of the characteristic equation. This is done below.
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$$  \displaystyle \lambda^2+4\lambda+13=0 $$  $$   \displaystyle \lambda_1=\frac{-4+\sqrt{16-4*13}}2 \; \; \; \; \; \; \; \; \; \; \; \;\;\;\;\;\;\;\; \lambda_2=\frac{-4-\sqrt{16-4*13}}2 $$ $$   \displaystyle \lambda_1=\frac{-4+\sqrt{-36}}2 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\;\;\;\; \lambda_2=\frac{-4-\sqrt{-36}}2 $$ $$   \displaystyle \lambda_1=\frac{-4+6i}{2} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\; \lambda_2=\frac{-4-6i}{2} $$ $$   \displaystyle \lambda_1=-2+3i \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \lambda_2=-2-3i $$ In this case, we have two imaginary roots. We can refer to [[media:iea.s12.sec6.djvu| Loc Vu-Quoc class EGM 4313 notes Sec 6]] for the derivation of the form of the general solution below.
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$$\displaystyle y_h(x)=e^{-ax/2}[Acos(wx)+Bsin(wx)] $$ Given the specific characteristic equation, we can apply the determined roots to get the general solution:
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$$\displaystyle y_h(x)=e^{-2x}[Acos(3x)+Bsin(3x)] $$ Applying the initial condition $$\ y(0)=1 $$ we get:
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$$\displaystyle 1=e^{0}[Acos(0)+Bsin(0)] $$
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$$\displaystyle 1=A(1)+B(0) $$
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$$\displaystyle A=1 $$ We now take the first derivative of the general solution:
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$$\displaystyle {y_h}'(x)=e^{-2x}[-3Asin(3x)+3Bcos(3x)]-2e^{-2x}[Acos(3x)+Bsin(3x)] $$ Applying the initial condition $$\ y'(0)=0 $$:
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$$\displaystyle 0=e^{0}[-3Asin(0)+3Bcos(0)]-2e^{0}[Acos(0)+Bsin(0)] $$
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$$\displaystyle 0=[-3A(0)+3B(1)]-2[A(1)+B(0)] $$
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$$\displaystyle 0=3B-2A $$
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$$\displaystyle 0=3B-2(1) $$
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$$\displaystyle B=\frac{2}{3} $$ We have now fully determined the particular solution for the given initial conditions. This is shown below.
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$$\displaystyle y_h(x)=e^{-2x}[cos(3x)+\frac{2}{3}sin(3x)] $$ The figure below shows a plot of the solution.
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Part B

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$$\displaystyle \lambda_{1,2}=-3 $$     (9.2) In this instance, the roots of the characteristic equation are already provided. Although we technically do not need the corresponding characteristic equation and homogeneous ODE to proceed, they are shown below for the sake of reference. Characteristic Equation:
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$$\displaystyle (\lambda-(-3))^2=0 $$
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$$\displaystyle (\lambda+3)^2=0 $$
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$$\displaystyle \lambda^{2}+6\lambda+9=0 $$ Corresponding Homogeneous ODE:
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$$\displaystyle y''+6y'+9y=0 $$ In this case, we have a double real root. We can refer to [[media:iea.s12.sec5.djvu| Loc Vu-Quoc class EGM 4313 notes Sec 5]] for the derivation of the form of the general solution below.
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$$\displaystyle y_h(x)=c_1e^{\lambda x}+c_2xe^{\lambda x} $$ Applying the root value to the form above, we obtain the general solution:
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$$\displaystyle y_h(x)=c_1e^{-3x}+c_2xe^{-3x} $$ Applying the initial condition $$\ y(0)=1 $$ we get:
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$$\displaystyle 1=c_1e^{0}+c_2(0)e^{0} $$
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$$\displaystyle 1=c_1(1)+0 $$
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$$\displaystyle c_1=1 $$ We now take the first derivative of the general solution:
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$$\displaystyle {y_h}'(x)=-3c_1e^{-3x}+c_2e^{-3x}-3c_2xe^{-3x} $$ Applying the initial condition $$\ y'(0)=0 $$:
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$$\displaystyle 0=-3c_1e^{0}+c_2e^{0}-3c_2(0)e^{0} $$
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$$\displaystyle 0=-3c_1+c_2 $$
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$$\displaystyle 0=-3(1)+c_2 $$
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$$\displaystyle c_2=3 $$ We have now fully determined the particular solution for the given initial conditions. This is shown below.
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$$\displaystyle y_h(x)=e^{-3x}+3xe^{-3x} $$
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Part C

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$$\displaystyle \lambda_1=-2 \;\;,\;\; \lambda_2=5 $$     (9.3) In this instance, the roots of the characteristic equation are already provided. The corresponding characteristic equation and homogeneous ODE were determined in problem 1. They are shown below. Characteristic Equation:
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$$\displaystyle \lambda^{2}-3\lambda+-10=0 $$ Corresponding Homogeneous ODE:
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$$\displaystyle y''-3y'-10y=0 $$ In this case, we have two distinct, real roots. We can refer to [[media:iea.s12.sec3.djvu| Loc Vu-Quoc class EGM 4313 notes Sec 3]] for the form of the general solution below.
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$$\displaystyle y_h(x)=c_1e^{\lambda _1 x}+c_2e^{\lambda_2 x} $$ Applying the distinct root values to the form above, we obtain the general solution:
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$$\displaystyle y_h(x)=c_1e^{-2 x}+c_2e^{5 x} $$ Applying the initial condition $$\ y(0)=1 $$ we get:
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$$\displaystyle 1=c_1e^{0}+c_2e^{0} $$
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$$\displaystyle 1=c_1(1)+c_2 $$
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$$\displaystyle c_1+c_2=1 $$ We now take the first derivative of the general solution:
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$$\displaystyle {y_h}'(x)=-2c_1e^{-2x}+5c_2e^{5x} $$ Applying the initial condition $$\ y'(0)=0 $$:
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$$\displaystyle 0=-2c_1e^{0}+5c_2e^{0} $$
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$$\displaystyle 0=-2c_1+5c_2 $$ We have 2 equations and 2 unknowns. Solving for $$\ c_1 $$:
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$$\displaystyle 0=-2c_1+5(1-c_1) $$
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$$\displaystyle 0=-7c_1+5 $$
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$$\displaystyle c_1=\frac{5}{7} $$ Using this value to determine $$\ c_2 $$:
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$$\displaystyle 1=c_1+c_2 $$
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$$\displaystyle 1=\frac{5}{7}+c_2 $$
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$$\displaystyle c_2=\frac{2}{7} $$
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We have now fully determined the particular solution for the given initial conditions. This is shown below.
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$$\displaystyle y_h(x)=\frac{5}{7}e^{-2x}+\frac{2}{7}e^{5x} $$
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Plot of All Solutions
Below the solutions to Parts A,B,and C have been plotted. For reference, Part A (shown in green) was the case of imaginary roots. Part B (shown in red) was the case of a double real root. Part C (shown in blue) was the case of distinct, real roots.

title: User:Egm4313.s12.team3.pliskow/R2work


text:
User:Egm4313.s12.team13.steinberg.r1

Problem Statement
4.) Find the general solution to the following ODE and check the result by substitution.

$$ y'' + 4y' + (\pi^2 + 4)y = 0\,\!$$

Solution
The homogeneous characteristic equation for linear 2nd order ODE's with constant coefficients is $$ \lambda^2 + a\lambda + b = 0\,\!$$(5.1)

For the ODE in this problem, the characteristic equation becomes

$$ \lambda^2 + 4\lambda + (\pi^2 + 4) = 0\,\!$$

To solve for the solutions, the discriminant to the quadratic equation must first be calculated.

$$ \Delta = a^2 - 4b\,\!$$(5.2)

$$ \Delta = 16 - 4(\pi^2 + 4)\,\!$$ $$ = 16 - 4\pi^2 - 16\,\!$$ $$ = -4\pi^2\,\!$$

Since the discriminant is negative, there will be two imaginary solutions to the ODE. The solutions can be obtained by solving the remainder of the quadratic formula.

$$ \lambda_{1,2} = \frac{-a \pm \sqrt{a^2 - 4b}}{2}\,\!$$(5.3)

$$ \lambda_{1,2} = \frac{-4 \pm \sqrt{-4\pi^2}}{2}\,\!$$ $$ = \frac{-4 \pm 2\pi i}{2}\,\!$$ $$ = -2 \pm \pi i\,\!$$

Therefore,

$$ \lambda_{1} = -2 + \pi i \,\!$$(5.4) $$ \lambda_{2} = -2 - \pi i \,\!$$(5.5)

The two distinct, linearly independent, homogeneous solutions for the case with two imaginary roots are

$$ y_{h,1}(x) = e^{a_1 x} \cos b_1 x\,\!$$(5.6)

$$ y_{h,2}(x) = e^{a_2 x} \sin b_2 x\,\!$$(5.7)

The homogeneous solution is

$$ y_{h} = c_1 y_{h,1} + c_2 y_{h,2}\,\!$$(5.8)

$$ y_{h}(x) = c_1 e^{a_1 x} \cos b_1 x + c_2 e^{a_2 x} \sin b_2 x\,\!$$

$$ y_{h}(x) = c_1 e^{-2 x} \cos \pi x + c_2 e^{-2 x} \sin \pi x\,\!$$

The final general solution is therefore

$$ y_{h}(x) = e^{-2 x}(c_1 \cos \pi x + c_2 \sin \pi x)\,\!$$(5.9)

To check this result using substitution, we must take the first and second derivatives of the general solution to substitute back into the original ODE.

$$ y'(x) = -2e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) + e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)\,\!$$(5.10)

$$ y''(x) = 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - 2e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) - 2e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)\,\!$$(5.11)

$$ = 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - 4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)\,\!$$

Substituting back into the original ODE gives

$$ y'' + 4y' + (\pi^2 + 4)y = 0\,\!$$(5.12)

$$ 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - \cancelto{0}{4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)} + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x) - 8e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) + \cancelto{0}{4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)} + (\pi^{2} + 4)(e^{-2x})(c_1 \cos \pi x + c_2 \sin \pi x) = 0\,\!$$

$$ \cancelto{0}{-4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x)} + \cancelto{0}{e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)} + \cancelto{0}{(\pi^{2} + 4)(e^{-2x})(c_1 \cos \pi x + c_2 \sin \pi x)} = 0\,\!$$

Since all terms cancel to 0, $$ y_(x) = e^{-2 x}(c_1 \cos \pi x + c_2 \sin \pi x)\,\!$$ is a general solution to the original ODE.

Section 2 Lecture Notes

title: User:Egm4313.s12.team13.steinberg.r1


text:
University of Florida/Egm4313/s12.team13.steinberg.r1

Problem Statement
4.) Find the general solution to the following ODE and check the result by substitution.

$$ y'' + 4y' + (\pi^2 + 4)y = 0\,\!$$

Solution
The homogeneous characteristic equation for linear 2nd order ODE's with constant coefficients is $$ \lambda^2 + a\lambda + b = 0\,\!$$(5.1)

For the ODE in this problem, the characteristic equation becomes

$$ \lambda^2 + 4\lambda + (\pi^2 + 4) = 0\,\!$$

To solve for the solutions, the discriminant to the quadratic equation must first be calculated.

$$ \Delta = a^2 - 4b\,\!$$(5.2)

$$ \Delta = 16 - 4(\pi^2 + 4)\,\!$$ $$ = 16 - 4\pi^2 - 16\,\!$$ $$ = -4\pi^2\,\!$$

Since the discriminant is negative, there will be two imaginary solutions to the ODE. The solutions can be obtained by solving the remainder of the quadratic formula.

$$ \lambda_{1,2} = \frac{-a \pm \sqrt{a^2 - 4b}}{2}\,\!$$(5.3)

$$ \lambda_{1,2} = \frac{-4 \pm \sqrt{-4\pi^2}}{2}\,\!$$ $$ = \frac{-4 \pm 2\pi i}{2}\,\!$$ $$ = -2 \pm \pi i\,\!$$

Therefore,

$$ \lambda_{1} = -2 + \pi i \,\!$$(5.4) $$ \lambda_{2} = -2 - \pi i \,\!$$(5.5)

The two distinct, linearly independent, homogeneous solutions for the case with two imaginary roots are

$$ y_{h,1}(x) = e^{a_1 x} \cos b_1 x\,\!$$(5.6)

$$ y_{h,2}(x) = e^{a_2 x} \sin b_2 x\,\!$$(5.7)

The homogeneous solution is

$$ y_{h} = c_1 y_{h,1} + c_2 y_{h,2}\,\!$$(5.8)

$$ y_{h}(x) = c_1 e^{a_1 x} \cos b_1 x + c_2 e^{a_2 x} \sin b_2 x\,\!$$

$$ y_{h}(x) = c_1 e^{-2 x} \cos \pi x + c_2 e^{-2 x} \sin \pi x\,\!$$

The final general solution is therefore

$$ y_{h}(x) = e^{-2 x}(c_1 \cos \pi x + c_2 \sin \pi x)\,\!$$(5.9)

To check this result using substitution, we must take the first and second derivatives of the general solution to substitute back into the original ODE.

$$ y'(x) = -2e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) + e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)\,\!$$(5.10)

$$ y''(x) = 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - 2e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) - 2e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)\,\!$$(5.11)

$$ = 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - 4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)\,\!$$

Substituting back into the original ODE gives

$$ y'' + 4y' + (\pi^2 + 4)y = 0\,\!$$(5.12)

$$ 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - \cancelto{0}{4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)} + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x) - 8e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) + \cancelto{0}{4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)} + (\pi^{2} + 4)(e^{-2x})(c_1 \cos \pi x + c_2 \sin \pi x) = 0\,\!$$

$$ \cancelto{0}{-4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x)} + \cancelto{0}{e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)} + \cancelto{0}{(\pi^{2} + 4)(e^{-2x})(c_1 \cos \pi x + c_2 \sin \pi x)} = 0\,\!$$

Since all terms cancel to 0, $$ y_(x) = e^{-2 x}(c_1 \cos \pi x + c_2 \sin \pi x)\,\!$$ is a general solution to the original ODE.

Section 2 Lecture Notes


text:
User:Egm4313.s12.team13.steinberg.r2

Problem Statement
3.) Find the general solution to the following ODE and check the result by substitution.

$$ y'' + 6y' + 8.96y = 0\,\!$$

Solution
The homogeneous characteristic equation for linear 2nd order ODE's with constant coefficients is $$ \lambda^2 + a\lambda + b = 0\,\!$$(3.1a)

For the ODE in this problem, the characteristic equation becomes

$$ \lambda^2 + 6\lambda + 8.96 = 0\,\!$$

To solve for the solutions, the discriminant to the quadratic equation must first be calculated.

$$ \Delta = a^2 - 4b\,\!$$(3.2a)

$$ \Delta = 36 - 4(8.96) = 0.16\,\!$$

Since the discriminant is positive, there will be two distinct real solutions to the ODE. The solutions can be obtained by solving the remainder of the quadratic formula.

$$ \lambda_{1,2} = \frac{-a \pm \sqrt{a^2 - 4b}}{2}\,\!$$(3.3a)

$$ \lambda_{1,2} = \frac{-6 \pm \sqrt{0.16}}{2}\,\!$$ $$ = \frac{-6 \pm 0.4}{2}\,\!$$ $$ = -3 \pm 0.2\,\!$$

Therefore,

$$ \lambda_{1} = -2.8 \,\!$$(3.4a) $$ \lambda_{2} = -3.2 \,\!$$(3.5a)

The two distinct, linearly independent, homogeneous solutions for the case with two real roots are

$$ y_{h,1}(x) = e^{\lambda_1 x}\,\!$$(3.6a)

$$ y_{h,2}(x) = e^{\lambda_2 x}\,\!$$(3.7a)

The homogeneous solution is

$$ y_{h} = c_1 y_{h,1} + c_2 y_{h,2}\,\!$$(3.8a)

$$ y_{h}(x) = c_1 e^{\lambda_1 x} + c_2 e^{\lambda_2 x}\,\!$$

The final general solution is therefore

$$ y_{h}(x) = c_1 e^{-2.8 x} + c_2 e^{-3.2 x}\,\!$$(3.9a)

To check this result using substitution, we must take the first and second derivatives of the general solution to substitute back into the original ODE.

$$ y'(x) = -2.8 c_1 e^{-2.8x} - 3.2 c_2 e^{-3.2x}\,\!$$(3.10a)

$$ y''(x) = 7.84 c_1 e^{-2.8x} + 10.24 c_2 e^{-3.2x}\,\!$$(3.11a)

Substituting back into the original ODE gives

$$ y'' + 6y' + 8.96y = 0\,\!$$(3.12a)

$$ 7.84 c_1 e^{-2.8x} + 10.24 c_2 e^{-3.2x} + 6(-2.8 c_1 e^{-2.8x} - 3.2 c_2 e^{-3.2x}) + 8.96(c_1 e^{-2.8 x} + c_2 e^{-3.2 x}) = 0\,\!$$

$$ \cancelto{0}{(7.84 - 16.8 + 8.96)c_1 e^{-2.8x}} + \cancelto{0}{(10.24 - 19.2 + 8.96)c_2 e^{-3.2x}} = 0\,\!$$

Since all terms cancel to 0, $$ y_(x) = c_1 e^{-2.8 x} + c_2 e^{-3.2 x}\,\!$$ is a general solution to the original ODE.

Section 3 Lecture Notes

Section 5 Lecture Notes

Problem Statement
4.) Find the general solution to the following ODE and check the result by substitution.

$$ y'' + 4y' + (\pi^2 + 4)y = 0\,\!$$

Solution
The homogeneous characteristic equation for linear 2nd order ODE's with constant coefficients is $$ \lambda^2 + a\lambda + b = 0\,\!$$(3.1b)

For the ODE in this problem, the characteristic equation becomes

$$ \lambda^2 + 4\lambda + (\pi^2 + 4) = 0\,\!$$

To solve for the solutions, the discriminant to the quadratic equation must first be calculated.

$$ \Delta = a^2 - 4b\,\!$$(3.2b)

$$ \Delta = 16 - 4(\pi^2 + 4)\,\!$$ $$ = 16 - 4\pi^2 - 16\,\!$$ $$ = -4\pi^2\,\!$$

Since the discriminant is negative, there will be two imaginary solutions to the ODE. The solutions can be obtained by solving the remainder of the quadratic formula.

$$ \lambda_{1,2} = \frac{-a \pm \sqrt{a^2 - 4b}}{2}\,\!$$(3.3b)

$$ \lambda_{1,2} = \frac{-4 \pm \sqrt{-4\pi^2}}{2}\,\!$$ $$ = \frac{-4 \pm 2\pi i}{2}\,\!$$ $$ = -2 \pm \pi i\,\!$$

Therefore,

$$ \lambda_{1} = -2 + \pi i \,\!$$(3.4b) $$ \lambda_{2} = -2 - \pi i \,\!$$
(3.5b)

The two distinct, linearly independent, homogeneous solutions for the case with two imaginary roots are

$$ y_{h,1}(x) = e^{a_1 x} \cos b_1 x\,\!$$
(3.6b)

$$ y_{h,2}(x) = e^{a_2 x} \sin b_2 x\,\!$$
(3.7b)

The homogeneous solution is

$$ y_{h} = c_1 y_{h,1} + c_2 y_{h,2}\,\!$$
(3.8b)

$$ y_{h}(x) = c_1 e^{a_1 x} \cos b_1 x + c_2 e^{a_2 x} \sin b_2 x\,\!$$

$$ y_{h}(x) = c_1 e^{-2 x} \cos \pi x + c_2 e^{-2 x} \sin \pi x\,\!$$

The final general solution is therefore

$$ y_{h}(x) = e^{-2 x}(c_1 \cos \pi x + c_2 \sin \pi x)\,\!$$
(3.9b)

To check this result using substitution, we must take the first and second derivatives of the general solution to substitute back into the original ODE.

$$ y'(x) = -2e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) + e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)\,\!$$
(3.10b)

$$ y''(x) = 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - 2e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) - 2e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)\,\!$$
(3.11b)

$$ = 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - 4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)\,\!$$

Substituting back into the original ODE gives

$$ y'' + 4y' + (\pi^2 + 4)y = 0\,\!$$
(3.12b)

$$ 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - \cancelto{0}{4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)} + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x) - 8e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) + \cancelto{0}{4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)} + (\pi^{2} + 4)(e^{-2x})(c_1 \cos \pi x + c_2 \sin \pi x) = 0\,\!$$

$$ \cancelto{0}{-4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x)} + \cancelto{0}{e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)} + \cancelto{0}{(\pi^{2} + 4)(e^{-2x})(c_1 \cos \pi x + c_2 \sin \pi x)} = 0\,\!$$

Since all terms cancel to 0, $$ y_(x) = e^{-2 x}(c_1 \cos \pi x + c_2 \sin \pi x)\,\!$$ is a general solution to the original ODE.

Section 5 Lecture Notes

Section 6 Lecture Notes

title: User:Egm4313.s12.team13.steinberg.r2


text:
User:Egm6322.s12.team2.shieldsdraft

= R 13.3 - Relative convergence of power series and continued fraction expansions=

Given
Power series expansion of $$\tan x$$:
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$$  \displaystyle \tan x = \sum_{n=1}^\infty \frac{(-1)^{n-1}2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}x^{2n-1}, x \in ]-\pi/2,\pi/2[ $$     (13.3.1)
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in which $$B_n$$ are the Bernoulli numbers:

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$$  \displaystyle \begin{align} B_1 &= -\frac{1}{2}, B_{2n+1} = 0, n = 1,2,\dots\\ B_0 &= 1, B_2 = \frac{1}{6}, B_4 = -\frac{1}{30}, B_6 = \frac{1}{42}, B_8 = -\frac{1}{30}, \dots \end{align} $$     (13.3.2)
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Continued fraction expansion of $$\tan x$$:
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$$  \displaystyle \tan x = \frac{x}{1-}\frac{x^2}{3-}\frac{x^2}{5-}\frac{x^2}{7-}\frac{x^2}{\ddots} $$     (13.3.3)
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Find
Take $$x=\pi/4$$ and compare the term by term convergence of the power series (Eq. 13.3.1) and the continued fraction (Eq. 13.3.3). Provide a plot to visualize the convergence.

Solution
The power series expansion of a function is a standard approximation technique although it does not necessarily show the fastest convergence. The expression for the expansion of $$\tan x$$ is given in Eqn 13.3.1 and is seen to be a function of the Bernoulli numbers. These values are provided in Eq. 13.3.2 and can also be found on Wolfram Alpha. The term-by-term expansion of the function can be created in Matlab very easily by looping over the number of desired terms and summing the values generated by Eq. 13.3.1. This can be seen in the Matlab script provided below.

Creating the continued fraction expansion of $$\tan x$$ is slightly more complicated as it involves a new loop for a given number of terms. Thus, in the Matlab script provided below, the j index corresponds to the desired number of terms and the i index computes the necessary terms for that expansion. It is also necessary to provide an if statement to distinguish between the first term of Eq. 13.3.3, in which the numerator is $$x$$ instead of $$x^2$$.

The convergence for $$x = \pi/4$$ is shown below in the left figure. It can be seen that both techniques have the same initial value $$\tan x = x = \pi/4$$, but the continued fraction expansion converges more rapidly than the power series expansion. The second term of the continued fraction expansion is within the 5\% bounds of the exact solution while the power series is not quite as accurate. Within 4 terms, both solutions are quite accurate.

A potential explanation for the rapid convergence of both methods in the problem is the simple nature of the x-value, as $$\tan(pi/4)=1$$. To investigate this, a different x-value of $$x=\pi/4 + \sqrt(pi/11)$$ was selected; the squared term on the right hand side is small enough that the value still fits within the bounds of $$x\in ]-\pi/2,\pi/2[$$, but does not have a simple solution for $$\tan x $$. No other parameters in the code need to be changed, and the result is seen in the figure on the right. Again, both methods have the same initial condition. The continued fraction expansion requires only 3 terms to converge within the 5\% bounds and additional terms indicate no discernible difference from the exact solution; however, the power series expansion requires 3 times as many terms to converge to the 5\% bounds and still does not appear to be completely converged after 10 terms. This indicates that the continued fraction expansion is more robust for a wide variety of input conditions.

Matt Shields 18:03,6 April 2012 (UTC)

title: User:Egm6322.s12.team2.shieldsdraft


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Video journalism/Develop a current affairs story

Writing treatments – current affairs research/planning exercise

Objectives
By the end of this exercise:
 * 1) You should have an ability to put together a research plan and treatment, on the word processor, for a TV current affairs feature.
 * 2) You should be writing with visual emphasis while always considering sound.
 * 3) You should then be able to take your story on to developing it into a finished work in that it could be given to an editor who will be able to interpret the edit instructions and edit accordingly.

Keep objective three in mind, there is always the possibility of your work being screened or even re-worked for inclusion in other programmes. This means that the product being designed here may have commercial value. Any footage made available to you, or any material that you shoot, will have to be cleared to you, you will have to own the rights or obtain the material from Creative Commons or the like.

Such permission should be applied for in writing and signed by all parties; in terms of chain-of-title, copyright and other encumbrances (legal or moral) that come with filming someone or using their material.

Before you start, you might read "The last of punchy current affairs: Profile interview: Mark Davis,Dateline, Special Broadcasting Service"

http://ro.uow.edu.au/apme/vol1/iss15/17

Assignment - Part A
Consider your story possibilities, scenarios, or briefs that you chose in Assignment 1 research. Focus on the final one as chosen with the view of developing a current affairs story that could run at around seven minutes screening time.

As researcher/scriptwriter you are asked to research the story possibility very briefly. Using the internet or some equally fast and efficient search technique, choose extra material for further development, while working with the scenario you chose. If this is unworkable, find a new subject area and go through the same newsworthy testing as the case in the first assignment.

Research and write a plan and treatment for a TV current affairs story of 7 minutes. If the original idea is still viable, from Assignment 1, then try to develop this into a Current Affairs story that will run onwards into Part B in this section. This Assignment at this stage is not a script, unlike Assignment 1, but it is a plan for a script. It should show an established angle, story synopsis, research sources, possible locations and possible visually stimulating events for shooting. The treatment should show possible quirky angles, possible irresistible aspects and possible talent (interviewees and subjects). There should also be a paragraph or so on possible library, archival and overlay footage and so on — where relevant.

Although this assignment document is required by the producer (COS) at a length of only four pages, your first response is overwhelming panic, followed by a calm as you decide to use the Internet for finding further material and backgrounding.

Any Internet research should include web addresses. Your story structure and its basis, and, any possible story telling footage from a library (archival, Vimeo or Creative Commons) should all be clearly identified.

You should always note and be on the look-out for possible shooting locations for original material that can go into the story. These should be briefly described in terms of the event set for filming, including potential interviewees who have expressed that they would be willing to consent to be filmed.

A treatment should include a clear angle and direction, it should provide research data, addresses and descriptive comment. Treatments with technical and data based information should be accurate.

While the treatment and all the documents that follow will contain elements of your intellectual property and copyright, they will also carry some moral and legal responsibilities.

The assignment should be presented in the accepted treatment structure and form that applies to your region, location and niche to which you hope to target your developing work. If you don't know how this formatting might be undertaken, try to obtain some examples so that you are writing within an accepted format, even genre.

Please refer to resources online, where examples of scripts, treatments and research plans are provided for you to use as ideas and templates. Always choose the examples and scenarios that best suit you study plan. This should be based on what you want from this Study Schedule, your career aspirations in television, and ultimately, your study interest.

There is no absolutely perfect or correct way to perform any of these tasks, as there is no perfect film or current affairs piece. You should use other examples and resources to guide you and find the best path through the exercise, while keeping in mind that the better treatments are those that have a strong visual impact emanating from the words. They will have a strong story line and a simple and clear structure.

Throughout this process of thinking and writing you should visualise the organisational and research work that must have gone into other film stories. Try to mentally deconstruct them so you can inform your own process in constructing your piece. In other words work within the genre. Read the section on genre in the Topics. In some instances the scripts are provided in examples you may find online, showing how a work changed from script to broadcast work. There might even be discussion about the process and how legal issues or aesthetic considerations brought about change to the original script.

Try to make your thinking, decision making and conclusions transparent and apparent in a Rationale section of your TV treatments.

Assignment - Part B
Synopsis, treatment and possible shooting plan (Shooting Script) 15 minutes.

Continue to work with your chosen scenario from Part A, or if it has lost freshness, apply the same assignment brief and parameters to a fresh news / current affairs scenario that relates to your working project or geographical region and interest.

Apply the following assignment criteria to write a longer (15 minute) current affairs synopsis, treatment, possible shooting plan and draft shooting script.

The Brief
The Executive Producer of the current affairs programme in which youre a journalist is asking that you shoot, cut and voice a 15 minute current affairs (mainstream TV) story based on the news story that you chose in the first assignment.

In the first instance you are to provide a written synopsis, treatment and possible shooting plan (shooting script). You dont have much time, but there is enough time to run a search on the web for alternative data. You can shoot a limited amount of development material while scripting and you're aware that you have access to existing library footage. You're told that all material is of broadcast quality.

For this assignment you should present a treatment and early script. The assignment objective is for you to develop an overview of how to structure, plan, shoot and cut such a story. You must consider balance, editing grammar and rhythm while using the TV codes of mainstream current affairs. You are not required to submit a budget.

Always weigh up your priorities with what is realistically achievable and document all research considerations. If your story requires filming in foreign countries, you will need a fixer, an on location organiser and you should provide their address and brief bio. You will need to consider visas, travel requirements and filming permission in addition to the standard TV network requirements of equipment, crews and legalities.

While working as ethically and fairly as possible, and knowing your copyright parameters; write the plan, synopsis and draft (rough) shooting script. You should write so that angle, sequencing and list of possible vision is clearly defined. Try to make clear in notes or asides, the dilemmas, double standards or possible conflicts of interest you encountered or might encounter in filming.

Base your structure on the templates provided in the Resources. You will need to have a realistic script that is stimulating, visually oriented and runs and reads, while being read aloud, at 15 minutes. You would need to take 20 pages to achieve a satisfactory result.


text:
User:Egm4313.s12.team13.Neisha.R1

Problem Statement
5. Find a general solution. Check by substitution.

$$  \; {y}''+2\pi {y}'+\pi ^{2}y=0 $$

Solution
$$  \; {y}''+2\pi {y}'+\pi ^{2}y= \frac{d^2y}{dx^2}\ + 2 \pi \frac{dy}{dt}\ + \pi ^{2}y=0   $$ (5.0b)

If $$ \; \frac{d}{dx}\ = \lambda $$,

The homogeneous linear ODE with constant coefficients becomes:

$$ \; \lambda ^{2}+2\pi \lambda +\pi ^{2}=0 $$ (5.1b)

By using the quadratic formula, or in this case factoring, we find the solutions of the equation:

Solving for $$ \; \lambda $$,

$$ \; (\lambda +\pi) (\lambda +\pi)=0 $$ (5.2b)

$$ \; \lambda =-\pi, -\pi $$

General Solution:

General solution in the case of distinct real roots:

$$  \; y=c_1e ^ { \lambda _1 x} + c_2e ^ { \lambda _2 x}  $$ (5.3b)

So the general solution becomes:

$$  \; y=c_1e ^ {-\pi x} + c_2xe ^ {-\pi x}= e ^ {-\pi x} (c_1+c_2x) $$ (5.4b)

Check by Substitution:

To check by substitution, you find $$  \; y, y', y'' $$ and plug into the original ODE.

$$  \; y= e ^ {-\pi x} (c_1+c_2x) $$

$$  \; y'= -\pi c_1e ^ {-\pi x} + c_2 (e ^ {-\pi x} -\pi xe ^ {-\pi x})= - \pi e ^{- \pi x} (c_1 + c_2x) + c_2 e^ {- \pi x} $$(5.5b)

$$  \; y''= \pi ^{2} c_1e ^ {- \pi x} - \pi c_2e ^ {-\pi x} -\pi c_2(e ^ {- \pi x} - \pi xe^ {- \pi x} )= \pi ^ {2} e^ {- \pi x} (c_1+ c_2x) -2 \pi c_2 e^ {- \pi x} $$ (5.6b)

Plugging in  $$   \; y, y', y $$ into $$   \; {y}+2\pi {y}'+\pi ^{2}y=0 $$ :

$$  \; [ \pi ^ {2} e^ {- \pi x} (c_1 + c_2x) - 2 \pi c_2e^ {- \pi x}] + 2 \pi [ - \pi e^ {- \pi x} (c_1 + c_2x) + c_2e^ {- \pi x} ] + \pi ^ {2} [ e^ {- \pi x} (c_1+c_2x)]=0 $$

When plugging in, the general solution is confirmed because it equals zero.

Section 2 Notes

title: User:Egm4313.s12.team13.Neisha.R1


text:
User:Egm4313.s12.team13.rlh.r1

Problem Statement
For the spring-dashpot-mass system on p.1-4 (found in notes) draw the FBDs and derive the equation of motion.

A free body diagram of the spring-mass-dashpot found in Sec 1 of the notes must be broken up into three different parts. The spring, the dashpot, and the mass.

The positive y direction equals $$ \overrightarrow{y+} \! $$

The spring's free body diagram is: $$ F_k \longleftarrow\! $$ $$ \longrightarrow\! F_k $$

The dashpot's free body diagram is: $$ F_k \longleftarrow\! $$ $$ \longrightarrow\! F_c $$

The free body diagram of the mass is: $$ F_c \longleftarrow\! $$ $$ \longrightarrow\! F(t) $$

Since there are only two opposing forces on the dashpot those two forces must be equal. :{| style="width:100%" border="0" $$  \displaystyle F_c = F_k = F_i $$     (3.0)
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By definition
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$$  \displaystyle ma=my'' $$     (3.1)
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Using Newtons second law :
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$$  \displaystyle \sum F = ma $$ (3.2)
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using substitution and equations (3.2), (3.1), and (3.0) we get
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$$  \displaystyle \sum F_y = F_(t) - F_i = my'' $$     (3.3)
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Rearranging equation (3.3) gives us the solution

$$ F(t) = F_i + my''$$

title: User:Egm4313.s12.team13.rlh.r1