text:
User:Egm4313.s12.team9/R2

= R2.1: Non-Homogeneous L2-ODE-CC =

This problem is presented in three parts and will thus be solved in three parts: Part 1 can be found on [[media:Egm4313.s12.team9.3-7.jpg|lecture slide 3-7]] Part 2 can be found on [[media:Egm4313.s12.team9.3-7.jpg|lecture slide 3-7]] Part 3 can be found on [[media:Egm4313.s12.team9.3-9.jpg|lecture slide 3-9]]

Given
Two roots:

$$\lambda_1=-2, \ \lambda_2=-2$$

Initial Conditions:

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

 * $$\displaystyle y(0)=1 $$
 * $$\displaystyle(Eq.1)$$
 * }
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle y'(0)=0 $$
 * $$\displaystyle (Eq.2)$$
 * }

Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation $$r(x) \ $$.

Methods
From the characteristic equation:

$$(\lambda-\lambda_1)(\lambda-\lambda_2)=0 \ $$

We Have:

$$(\lambda+2)(\lambda-5)=0 \ $$

Expanding to the standard form:

$$\lambda^{2}-3\lambda-10 \ $$

Thus, the non-homogeneous L2-ODE-CC is:

$$y''-3y'-10y=r(x) \ $$

Its corresponding homogeneous solution is:

$$y(x)=c_1e^{5x}+c_2e^{-2x} \ $$

The overall solution $$y(x) \ $$is:

$$y(x)=c_1e^{5x}+c_2e^{-2x}+y_p(x) \ $$

and its derivative $$y'(x) \ $$ is:

$$y'(x)=5c_1e^{5x}-2c_2e^{-2x}+y'_p(x) \ $$

Satisfying the initial conditions (Eq.1 and Eq.2):

$$y(0)=c_1+c_2+y_p(0)=1 \ $$

$$y'(0)=5c_1-2c_2+y'p(0) \ $$

Multiplying (Eq.1) by 2 and adding it to (Eq.2) yields:

$$2=7c_1+0c_2+2y_p(0)+y'_p(0) \ $$

Solving for $$c_1 \ $$:

$$7c_1=2-2y_p(0)-y'_p(0) \ $$

$$c_1=\frac{1}{7}(2-2y_p(0)-y'_p(0)) \ $$

Substituting $$c_1 \ $$ into (Eq.1) yields:

$$1=\frac{1}{7}(2-2y_p(0)-y'_p(0))+c_2+y_p(0) \ $$

Rearranging to solve for $$c_2 \ $$:

$$7=2-2y_p(0)-y'_p(0)+7c_2+7y_p(0) \ $$

$$7c_2=7-2+2y_p(0)-7y_p(0)+y'_p(0) \ $$

$$c_2=\frac{1}{7}(5-5y_p(0)+y'_p(0)) \ $$

Solution
Plugging $$c_1,c_2 \ $$ into $$y(x) \ $$:

$$y(x)=\frac{1}{7}(2-2y_p(0)-y'_p(0))e^{5x}+\frac{1}{7}(5-5y_p(0)+y'_p(0))e^{-2x}+y_p(x) \ $$

Given
Consider the particular case of the non-homogeneous L2-ODE-CC solved above with no excitation:

$$r(x)=0 \ $$

and plot the solution.

Methods
Since $$r(x)=0 \ $$, thus $$y_p(x)=0 \ $$ as well as its corresponding 1st and 2nd derivatives.

We can thus solve for the real values of $$c_1, c_2 \ $$:

$$c_1=\frac{2}{7}, \ c_2=\frac{5}{7}$$

Using MATLAB to plot $$y(x) \ $$ we use the following code:

Solution
Plugging in $$c_1, c_2 \ $$ to $$y(x) \ $$ we get:

$$y(x)=\frac{2}{7}e^{5x}+\frac{5}{7}e^{-2x}$$

The corresponding plot for $$y(x) \ $$ is:

Given
Part 3 asks us to generate 3 non-standard & non-homogeneous L2-ODE-CC that still admit the two initial values:

$$\lambda_1=-2, \ \lambda_2=5$$

found in figure 3a on [[media:Lecture_Slide_3-7.png|Lecture Slide 3-7]] as the two roots in the corresponding characteristic equation.

Methods
The standard form can be found from Eq.1 on [[media:Lecture_Slide_3-8.png|Lecture Slide 3-8]] when $$M=1 \ $$. Non-standard forms can be found using any integer $$M\neq0,1 \ $$.

Thus, three non-standard corresponding equations can be found by using the values:

$$M=2,3,5 \ $$

and the characteristic equation:

$$M(\lambda-\lambda_1)(\lambda-\lambda_2)=0 \ $$.

Plugging in these values of $$M \ $$ we get:

$$M=2 \rightarrow 2\lambda^{2}-6\lambda-20 \ $$

$$M=3 \rightarrow 3\lambda^{2}-9\lambda-30 \ $$

$$M=5 \rightarrow 5\lambda^{2}-15\lambda-50 \ $$

Solution
The corresponding non-standard and non-homogeneous L2-ODE-CC equations with matching roots are:

$$M=2 \rightarrow r(x)=2y''-6y'-20y \ $$

$$M=3 \rightarrow r(x)=3y''-9y'-30y \ $$

$$M=5 \rightarrow r(x)=5y''-15y'-50y \ $$

Author
Solved and typed by Egm4313.s10.team9.delp 04:24, 7 February 2012 (UTC)

Reviewed by Egm4313.s10.team9.lebo 13:25, 8 February 2012 (UTC)

Edited by Egm4313.s10.team9.lebo 13:25, 8 February 2012 (UTC)

= R2.2: Solution for Linear 2nd Order ODE (CC) with Double Root = This problem can be found on [[media:Egm4313.s12.team9.5-6.jpg|lecture slide 5-6]].

Given
Given the following linear, 2nd-order ODE with constant coefficients,

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

$$ The initial conditions:
 * $$\displaystyle {y}''-10{y}'+25y=r(x) \! $$
 * $$\displaystyle (Eq.1)
 * $$\displaystyle (Eq.1)
 * }
 * }

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

$$ And no excitation:
 * $$\displaystyle y(0)=1\text{, }{y}'(0)=0 \! $$
 * $$\displaystyle (Eq.1b)
 * $$\displaystyle (Eq.1b)
 * }
 * }

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

$$ Find and plot the solution.
 * $$\displaystyle r(x)=0 \! $$
 * $$\displaystyle (Eq.1c)
 * $$\displaystyle (Eq.1c)
 * }
 * }

Methods
Let $$\lambda =\frac{dy}{dx}$$ in Equation (1).

The characteristic equation of Equation (1) is

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

$$
 * $$\displaystyle {{\lambda }^{2}}-10\lambda +25=0 \! $$
 * $$\displaystyle (Eq.2)
 * $$\displaystyle (Eq.2)
 * }
 * }

The case of the equation can be determined by $${{a}^{2}}-4b \! $$ where $$a=-10 \! $$ and $$b=25 \! $$ as shown below:

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

 * $$\displaystyle {{a}^{2}}-4b={{(-10)}^{2}}-4(25)=100-100=0 \! $$
 * }
 * }
 * }

Since this is equal to 0, the equation is a Case II equation with a real double root.

Solution
Equation (2) can be factored as shown below to determine the root:

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

 * $$\displaystyle {{(\lambda -5)}^{2}}=0 \! $$
 * }
 * {| style="width:100%" border="0" align="left"
 * }
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle \text{double root: }\lambda =5 \! $$
 * }
 * }
 * }

From [[media:Egm4313.s12.team9.5-4.jpg|lecture slide 5-4]] and from Kreyszig 2011 pg.55, the general homogenous solution for the equation is:

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

 * $$\displaystyle {{y}_{h}}={{c}_{1}}{{e}^{\lambda x}}+{{c}_{2}}x{{e}^{\lambda x}} \! $$
 * }
 * }
 * }

Therefore, the general solution for this example is:

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

$$
 * $$\displaystyle {{y}_{h}}(x)={{c}_{1}}{{e}^{5x}}+{{c}_{2}}x{{e}^{5x}} \! $$
 * $$\displaystyle (Eq.3)
 * $$\displaystyle (Eq.3)
 * }
 * }

The initial conditions are used to solve for $${{c}_{1}}\text{ and }{{c}_{2}} \! $$, as shown below.

From Equations (1b) and (3),

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

 * $$\displaystyle y(0)={{c}_{1}}{{e}^{0}}+(0){{c}_{2}}{{e}^{0}}={{c}_{1}}=1 \! $$
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle {y}'(0)=5{{c}_{1}}{{e}^{5x}}+{{c}_{2}}{{e}^{5x}}+x(5{{c}_{2}}{{e}^{5x}})={{e}^{5x}}(5{{c}_{1}}+{{c}_{2}}+5x{{c}_{2}})=5{{c}_{1}}+{{c}_{2}}+5(0){{c}_{2}}=5(1)+{{c}_{2}}=0 \! $$
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle \therefore {{c}_{2}}=-5 \! $$
 * }
 * }
 * }

Substituting the solutions for $${{c}_{1}}\text{ and }{{c}_{2}} \! $$ into Equation (3), the final solution is:

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

$$\displaystyle y(x)={{e}^{5x}}-5x{{e}^{5x}} \!$$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle $$
 * }

Using MATLAB to plot $$y(x) \ $$ we use the following code:

The corresponding plot for $$y(x) \ $$ is:

Author
Solved and typed by --Egm4313.s12.team9.wright 05:53, 8 February 2012 (UTC)

Reviewed by --Egm4313.s12.team9.ehlers.km 15:08, 8 February 2012 (UTC)

Edited by

=R2.3: Find a general solution for problems 3 and 4 (K2011 pg. 59)= This problem can be found on [[media:Egm4313.s12.team9.5-6.jpg|lecture slide 5-6]].

Given
$$ y'' + 6y' + 8.96y = 0 \! $$ $$ \lambda^{2} + 6\lambda + 8.96 = 0 \! $$

Methods
Using the Quadratic Equation: $$ a=1, b=6, c=8.96 \! $$  $$ \frac{-b\frac{+}{-}\sqrt{b^{2} - 4ac}}{2a} \! $$ $$ \frac{-6\frac{+}{-}\sqrt{6^{2} - 35.84}}{2} \! $$ $$ \frac{-6 + 0.4}{2} \! $$ $$ \frac{-6 - 0.4}{2} \! $$ Root 1: $$ -2.8 \! $$ Root 2: $$ -3.2 \! $$

Solution
General Solution:

Given
From Kreyszig 2011 pg. 59 problem 4,

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

{y}''+4{y}'+({{\pi }^{2}}+4)y=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq.1)
 * }
 * }

Find a general solution for Equation (1) and check the answer by substitution.

Methods
Let $$\lambda =\frac{dy}{dx}$$ in Equations (1). The characteristic equation of Equation (1) is

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

{{\lambda }^{2}}+4\lambda +({{\pi }^{2}}+4)=0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

The case of the equation can be determined by $$\displaystyle {{a}^{2}}-4b$$, where $$\displaystyle a=4$$ and $$\displaystyle b={{\pi }^{2}}+4$$ , as shown below:

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

{{a}^{2}}-4b={{4}^{2}}-4({{\pi }^{2}}+4)=16-4{{\pi }^{2}}-16=-4{{\pi }^{2}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Since this is below 0, the equation is a Case III equation with complex conjugate roots.

Solution
The roots of the characteristic equation is given by:

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

& \lambda =\frac{1}{2}\left( -a\pm \sqrt{{{a}^{2}}-4b} \right) \\ & \lambda =\frac{1}{2}\left( -4\pm \sqrt{-4{{\pi }^{2}}} \right) \\ & \lambda =-2\pm \pi i \\ \end{align}$$
 * $$\displaystyle\begin{align}
 * $$\displaystyle\begin{align}
 * }
 * }

Therefore, in this problem,
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle \omega =\pi$$
 * }
 * }
 * }

From page 57, a real general solution in Case III is

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

y={{e}^{-ax/2}}(A\cos \omega x+B\sin \omega x)$$ where A and B are arbitrary constant coefficients. In this example, the general solution is
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * {| style="width:0%" border="0" align="left"

$$\displaystyle y={{e}^{-2x}}(A\cos \pi x+B\sin \pi x) $$
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

To check by substitution, the first and second derivatives of the general solution are taken and substituted into Equation (1).

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

& {y}'=-2{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)+{{e}^{-2x}}\left( -A\pi \sin \pi x+B\pi \cos \pi x \right) \\ & {y}''=4{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)-2{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)-2{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)+{{e}^{-2x}}(-A{{\pi }^{2}}\cos \pi x-B{{\pi }^{2}}\sin \pi x) \\ & {y}''=4{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)-4{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)+{{e}^{-2x}}(-A{{\pi }^{2}}\cos \pi x-B{{\pi }^{2}}\sin \pi x) \\ \end{align}$$
 * $$\displaystyle\begin{align}
 * $$\displaystyle\begin{align}
 * }
 * }

Substituting into Equation (1),

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

4{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)-4{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)+{{e}^{-2x}}(-A{{\pi }^{2}}\cos \pi x-B{{\pi }^{2}}\sin \pi x)+4\left( -2{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)+{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x) \right)+({{\pi }^{2}}+4)\left( {{e}^{-2x}}(A\cos \pi x+B\sin \pi x) \right)=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Multiplying and rearranging terms,
 * {| style="width:100%" border="0" align="left"

4{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)-8{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)+4{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)-4{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)+4{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)+{{e}^{-2x}}(-A{{\pi }^{2}}\cos \pi x-B{{\pi }^{2}}\sin \pi x)+{{\pi }^{2}}{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

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

0=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Since all terms cancel, the solution is
 * {| style="width:100%" border="0" align="left"

y={{e}^{-2x}}(A\cos \pi x+B\sin \pi x)$$.
 * $$\displaystyle
 * }
 * }

Author
Solved and typed by --Egm4313.s12.team9.tjl 18:17, 8 February 2012 (UTC)

Reviewed by --Egm4313.s12.team9.wright 05:53, 8 February 2012 (UTC)

Edited by --Egm4313.s12.team9.wright 05:53, 8 February 2012 (UTC)

= R2.4: Find a general solution for problems 5 and 6 (K2011 pg. 59) =

Problem 5
This problem can be found on [[media:Egm4313.s12.team9.5-6.jpg|lecture slide 5-6]] and from Kreyszig 2011 pg. 59 problem 5.

Given
Problem Statement: Find a general solution and check by substitution.

Methods
Becomes:

Let $$\lambda =\frac{dy}{dx}$$ in Equation (1).

Equation (1) becomes:

Solving for $$\lambda \! $$ through factoring,

Root: $$\lambda =-\pi \! $$

Solution
General Solution:

So the general solution becomes:

$$ \; y = c_1 e^{-\pi x} + c_2 x e^{-\pi x} $$

Check by Substitution:

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

$$ \; y= e ^ {-\pi x} (c_1 + c_2 x) $$

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

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

Substitute the values of $$ \; y, y', y $$ into $$ \; {y}+2\pi {y}'+\pi ^{2}y=0 $$ :

$$ \; [ \pi ^ {2} e^ {- \pi x} (c_1 + c_2 x) - 2 \pi c_2e^ {- \pi x}] + 2 \pi [ - \pi e^ {- \pi x} (c_1 + c_2 x) + c_2e^ {- \pi x} ] + \pi ^ {2} [ e^ {- \pi x} (c_1 + c_2 x)]=0 $$

After substitution, it proves the general solution because the substitution equals zero.

Problem 6
This problem can be found on [[media:Egm4313.s12.team9.5-6.jpg|lecture slide 5-6]] and from Kreyszig 2011 pg. 59 problem 6.

Given
Problem Statement: Find a general solution and check by substitution.

Methods
Equation (1) can be rewritten as:

Let $$\lambda =\frac{dy}{dx}$$ in Equation (2).

The characteristic equation can then be written as:

Solve for $$ \; \lambda $$ by using the quadratic equation.

$$ \; \lambda=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$

$$ \; \lambda = \frac{8}{5} $$

Solution
General solution:

The general solution then becomes:

Check by Substitution:

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

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

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

$$ \; y'' = \frac{64}{25} e^{\frac{8}{5}x}(c_1 + c_2 x) + \frac{16}{5}c_2 e^{\frac{8}{5}x} $$

Substitute the values of $$ \; y, y', y $$ into $$ \; 10 y - 32 y' + 25.6 y = 0 $$ :

$$ \; \frac{64}{25} e^{\frac{8}{5}x}(c_1 + c_2 x) - \frac{64}{25} e^{\frac{8}{5}x}(c_1 + c_2 x) + 32 c_2 e^{\frac{8}{5}x} - 32 c_2 e^{\frac{8}{5}x} = 0 $$

After substitution, it proves the general solution because the substitution equals zero.

Author
Solved and typed by --Egm4313.s12.team9.ehlers.km 15:03, 8 February 2012 (UTC)

Reviewed by --Egm4313.s12.team9.wright 05:53, 8 February 2012 (UTC)

Edited by --Egm4313.s12.team9.wright 05:53, 8 February 2012 (UTC)

= R2.5: Find an ODE for problems 16 and 17 (K2011 pg. 59) = This problem can be found on [[media:Egm4313.s12.team9.5-6.jpg|lecture slide 5-6]].

Given
The DE given is:

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

$$
 * $$\displaystyle {y}''+a{y}'+by = 0 $$
 * $$\displaystyle (Eq.1)
 * $$\displaystyle (Eq.1)
 * }
 * }

We are being asked to find an ODE for the given basis solutions shown below.

For problem 16:

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

 * $$\displaystyle e^{2.6x}\,,\,e^{-4.3x} $$
 * }
 * }
 * }

Methods
From Kreyszig 2011 pg. 58:

In the case of two real distinct roots, the following general solution should be used:

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

$$
 * $$\displaystyle y = c_{1}e^{\lambda x} + c_{2}e^{\lambda x} $$
 *
$$\displaystyle (Eq.2)
 *
$$\displaystyle (Eq.2)
 * }
 * }

Solution
Let

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

 * $$\displaystyle y_{1} = e^{2.6x}\,,\,y_{2} = e^{-4.3x} $$
 * }
 * }
 * }

Since $$ y_{1} \!$$ and $$ y_{2} \!$$ are defined for all real values, the general solution of the ODE would look like this

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

$$
 * $$\displaystyle y = c_{1}e^{2.6x} + c_{2}e^{-4.3x} $$
 *
$$\displaystyle (Eq.3)
 *
$$\displaystyle (Eq.3)
 * }
 * }

Comparing the general solution (3) to the "model" general solution (2), we find that

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

 * $$\displaystyle \lambda_{1}=2.6\,,\,\lambda_{2}=-4.3 $$
 * }
 * }
 * }

Which are distinct real values; therefore, the characteristic equation can be obtained by

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

 * $$\displaystyle (\lambda-2.6)\,[\lambda-(-4.3)]=0 $$
 * }
 * }
 * }

Distributing,

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

 * $$\displaystyle \lambda^{2}-2.6\lambda+4.3\lambda-11.18=0 $$
 * }
 * }
 * }

So, the ODE with the given basis solution that we are looking for is:

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

$$\displaystyle {y}''+1.7{y}'-11.18y = 0 $$
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

Given
The DE given is:

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

 * $$\displaystyle {y}''+a{y}'+by = 0 $$
 * }
 * }
 * }

We are being asked to find an ODE for the given basis solutions shown below.

For problem 17:

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

 * $$\displaystyle e^{-\sqrt{5}x}\,,\,xe^{-\sqrt{5}x} $$
 * }
 * }
 * }

Methods
From Kreyszig 2011 pg. 58:

In the case of a real double root, the following general solution should be used:

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

$$
 * $$\displaystyle y = (c_{1}+xc_{2})e^{\lambda x} $$
 *
$$\displaystyle (Eq.4)
 *
$$\displaystyle (Eq.4)
 * }
 * }

Solution
Let

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

 * $$\displaystyle y_{1} = e^{-\sqrt{5}x}\,,\,y_{2} = xe^{-\sqrt{5}x} $$
 * }
 * }
 * }

Since $$ y_{1} \!$$ and $$ y_{2} \! $$ are defined for all real values, the general solution of the ODE is written as

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

$$
 * $$\displaystyle y = (c_{1}+c_{2}x)e^{-\sqrt{5}x} $$
 *
$$\displaystyle (Eq.5)
 *
$$\displaystyle (Eq.5)
 * }
 * }

Comparing the general solution (5) to the "model" general solution (4), we obtain

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

 * $$\displaystyle \lambda=-\sqrt{5} $$
 * }
 * }
 * }

Which is a real repeated root; therefore, the characteristic equation can be obtained by

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

 * $$\displaystyle [\lambda-(-\sqrt{5})]\,[\lambda-(-\sqrt{5})]=0 $$
 * }
 * }
 * }

Distributing,

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

 * $$\displaystyle \lambda^{2}+2\sqrt{5}\lambda+5=0 $$
 * }
 * }
 * }

So, the ODE with the given basis solution that we are looking for is:

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

$$\displaystyle {y}''+2\sqrt{5}{y}'+5y = 0 $$
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

Author
Solved and typed by E.Beuses 16:15, 8 February 2012 (UTC)

Reviewed by --Egm4313.s12.team9.wright 08:58, 8 February 2012 (UTC)

Edited by

= R2.6 = This problem can be found on [[media:Egm4313.s12.team9.5-6.jpg|lecture slide 5-6]].

Given
We are given an equation from [[media:Lecture_Slide_5-5.png|lecture slide 5-5]]:

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

$$
 * $$\displaystyle (\lambda - 5)^2 = (\lambda)^2 - 10(\lambda) + 25 $$
 *
$$\displaystyle (Eq.1)
 *
$$\displaystyle (Eq.1)
 * }
 * }

and told to use:

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

$$ as our double real root.
 * $$\displaystyle \lambda = -3 $$
 *
$$\displaystyle (Eq.2)
 *
$$\displaystyle (Eq.2)
 * }
 * }

We were asked to find the variables k, c, and m, which correlate to the spring-dashpot-mass system in series shown in [[media:Lecture_Slide_1-4.png|lecture slide 1-4]]:

Methods
So we sub Eq.2 into Eq.1 to get:
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle (\lambda + 3)^2 = 0 $$
 *
$$\displaystyle (Eq.3)
 *
$$\displaystyle (Eq.3)
 * }
 * }

Now, we expand Eq.3:
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle {y}'' + 6{y}' + 9 = 0 $$
 *
$$\displaystyle (Eq.4)
 *
$$\displaystyle (Eq.4)
 * }
 * }

Solution
With the expansion of the equation, we see (from [[media:Egm4313.s12.team9.1-5.jpg|lecture slide 1-5]], equation 3) how the coefficients relate to the variables we want to find. So the coefficient in front of $$ y'' \ $$ relates to m, the coefficient in front of $$ y' \ $$ relates to $$\frac{mk}{c}$$, and the coefficient in front of $$ y \ $$ relates to k:

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

 * $$\displaystyle m=1 $$
 * }
 * }
 * }

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

 * $$\displaystyle c=\frac{3}{2} $$
 * }
 * }
 * }

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

 * $$\displaystyle k=9 $$
 * }
 * }
 * }

Author
Solved and typed by --Egm4313.s12.team9.lebo 18:23, 8 February 2012 (UTC)

Reviewed by --Egm4313.s10.team9.delp 19:56, 8 February 2012 (UTC)

Edited by

=R2.7: Develop the MacLaurin Series for $$ e^{t}, cos(t), sin(t) \! $$= This problem can be found on [[media:Egm4313.s12.team9.6-6.jpg|lecture slide 6-6]].

$$ P_n(t) = f(a) + f'(a)(t-a) + \frac{f''(a)}{2!}(t-a)^{2} +...+\frac{f^{n}(a)}{n!}(t-a)^{n} \! $$

MacLaurin Series(Taylor Series at t=0) for $$ e^{t} \! $$
$$ f(t) = e^{t}.....f(0)=1 \! $$ $$ f'(t) = e^{t}.....f'(0)=1 \! $$ $$ f(t) = e^{t}.....f(0)=1 \! $$ $$ f(t) = e^{t}.....f(0)=1 \! $$ $$ f'(t) = e^{t}.....f'(0)=1 \! $$ $$ P_5(t) = 1 + t + \frac{t^{2}}{2!} + \frac{t^{3}}{3!} + \frac{t^{4}}{4!} \! $$

MacLaurin Series(Taylor Series at t=0) for $$ cos(t) \! $$
$$ f(t) = \cos{t}.....f(0)=1 \! $$ $$ f'(t) = -\sin{t}.....f'(0)=0 \! $$ $$ f(t) = -\cos{t}.....f(0)=-1 \! $$ $$ f(t) = \sin{t}.....f(0)=0 \! $$ $$ f'(t) = \cos{t}.....f'(0)=1 \! $$ $$ P_5(t) = 1 - \frac{t^{2}}{2!} + \frac{t^{4}}{4!} \! $$

MacLaurin Series(Taylor Series at t=0) for $$ sin(t) \! $$
$$ f(t) = \sin{t}.....f(0)=0 \! $$ $$ f'(t) = \cos{t}.....f'(0)=1 \! $$ $$ f(t) = -\sin{t}.....f(0)=0 \! $$ $$ f(t) = -\cos{t}.....f(0)=-1 \! $$ $$ f'(t) = \sin{t}.....f'(0)=0 \! $$ $$ P_5(t) = t - \frac{t^{3}}{3!} \! $$

Author
Solved and typed by --Egm4313.s12.team9.tjl 18:17, 8 February 2012 (UTC) Reviewed by --Egm4313.s12.team9.murray 09:50, 8 February 2012 (UTC)

Edited by

= R2.8: General Solution for Linear 2nd Order ODEs (CC) with Complex Conjugate Roots = This problem can be found on [[media:Egm4313.s12.team9.6-6.jpg|lecture slide 6-6]].

Given
From Kreyszig pg. 59 problem 8,
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle {y}''+{y}'+3.25y=0 \ $$
 *
$$\displaystyle (Eq.1)
 *
$$\displaystyle (Eq.1)
 * }
 * }

Find a general solution for Equation (1) and check the answer by substitution.

Methods
Let $$\lambda =\frac{dy}{dx}$$ in Equation (1).

The characteristic equation of Equation (1) is
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle {{\lambda }^{2}}+\lambda +3.25=0 \ $$
 *
$$\displaystyle (Eq.2)
 *
$$\displaystyle (Eq.2)
 * }
 * }

The case of the equation can be determined by $${{a}^{2}}-4b \!$$ where $$a=1 \!$$ and $$b=3.25 \!$$, as shown below:
 * {| style="width:100%" border="0" align="left"

Since this is below 0, the equation is a Case III equation with complex conjugate roots.
 * $$\displaystyle {{a}^{2}}-4b={{(1)}^{2}}-4(3.25)=1-13=-12 \ $$
 * }
 * }
 * }

Solution
The roots of the characteristic equation are given by:

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

& \lambda =\frac{1}{2}\left( -a\pm \sqrt{{{a}^{2}}-4b} \right) \\ & \lambda =\frac{1}{2}\left( -1\pm \sqrt{-12} \right)=\frac{1}{2}\left( -1\pm \left( 2\sqrt{3} \right)i \right) \\ & \lambda =-\frac{1}{2}\pm \left( \sqrt{3} \right)i \\ \end{align}$$
 * $$\displaystyle \begin{align}
 * $$\displaystyle \begin{align}
 * }
 * }

Therefore, in this problem,
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle \omega =\sqrt{3} $$
 * }
 * }
 * }

From [[media:Egm4313.s12.team9.6-5.jpg|lecture slide 6-5]] and from Kreyszig page 57, a general homogeneous solution for a Case III equation is
 * {| style="width:100%" border="0" align="left"

{{y}_{h}}={{e}^{-ax/2}}(A\cos \omega x+B\sin \omega x) \ $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

where A and B are arbitrary constant coefficients. In this example, the general solution is
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y={{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right) \!$$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle $$
 * }

To check by substitution, the first and second derivatives of the general solution are taken and substituted into Equation (1).
 * {| style="width:100%" border="0" align="left"

{y}'=-0.5{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)+{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

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

{y}''=0.25{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)-0.5{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)-0.5{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)+{{e}^{-0.5x}}\left( -3A\cos (\sqrt{3})x-3B\sin (\sqrt{3})x \right)$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

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

{y}''=0.25{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)-{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)+{{e}^{-0.5x}}\left( -3A\cos (\sqrt{3})x-3B\sin (\sqrt{3})x \right)$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Substituting into Equation (1),
 * {| style="width:100%" border="0" align="left"

0.25{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)-{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)+{{e}^{-0.5x}}\left( -3A\cos (\sqrt{3})x-3B\sin (\sqrt{3})x \right)-0.5{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)+{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)+3.25{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Multiplying and rearranging terms,

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

0.25{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)-0.5{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)-3{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)+3.25{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)-{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)+{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

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

 * $$0=0 \!$$
 * }
 * }
 * }

Since all terms cancel, the solution is
 * {| style="width:100%" border="0" align="left"

y={{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Given
From Kreyszig pg. 59 problem 15,
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle {y}''+0.54{y}'+(0.0729+\pi )y=0 \! $$
 *
$$\displaystyle (Eq.3)
 *
$$\displaystyle (Eq.3)
 * }
 * }

Find a general solution for Equation (3) and check the answer by substitution.

Methods
Let $$\lambda =\frac{dy}{dx}$$ in Equation (3).

Using the same method, the characteristic equation of Equation (3) is
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle {{\lambda }^{2}}+0.54\lambda +(0.0729+\pi )=0 \ $$
 *
$$\displaystyle (Eq.4)
 *
$$\displaystyle (Eq.4)
 * }
 * }

The case of the equation can be determined by $${{a}^{2}}-4b \! $$ where $$a=0.54 \! $$ and $$b=0.0729+\pi \! $$, as shown below:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle {{a}^{2}}-4b={{(0.54)}^{2}}-4(0.0729+\pi )=0.2916-0.2916-4\pi =-4\pi \ $$
 * }
 * }
 * }

Since this is below 0, the equation is also a Case III equation with complex conjugate roots.

Solution
The roots of the characteristic equation are given by:

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

& \lambda =\frac{1}{2}\left( -a\pm \sqrt{{{a}^{2}}-4b} \right) \\ & \lambda =\frac{1}{2}\left( -0.54\pm \sqrt{-4\pi } \right)=\frac{1}{2}\left( -0.54\pm \left( 2\sqrt{\pi } \right)i \right) \\ & \lambda =-0.27\pm \left( \sqrt{\pi } \right)i \\ \end{align}$$
 * $$\displaystyle \begin{align}
 * $$\displaystyle \begin{align}
 * }
 * }

Therefore, in this problem,

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

 * $$\omega =\sqrt{\pi }$$
 * }
 * }
 * }

From [[media:Egm4313.s12.team9.6-5.jpg|lecture slide 6-5]] and from Kreyszig page 57, a general homogeneous solution for a Case III equation is
 * {| style="width:100%" border="0" align="left"

 * $${{y}_{h}}={{e}^{-ax/2}}(A\cos \omega x+B\sin \omega x) \! $$
 * }
 * }
 * }

where A and B are arbitrary constant coefficients. In this example, the general solution is
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y={{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right) \! $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle $$
 * }

To check by substitution, the first and second derivatives of the general solution are taken and substituted into Equation (2).
 * {| style="width:100%" border="0" align="left"

 * $${y}'=-0.27{{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right)+{{e}^{-0.27x}}\left( -\sqrt{\pi }A\sin (\sqrt{\pi }x)+\sqrt{\pi }B\cos (\sqrt{\pi }x) \right)$$
 * }
 * }
 * }

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

 * $${y}''=0.25{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)-0.5{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)-0.5{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)+{{e}^{-0.5x}}\left( -3A\cos (\sqrt{3})x-3B\sin (\sqrt{3})x \right)$$
 * }
 * }
 * }

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

 * $${y}''=0.25{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)-{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)+{{e}^{-0.5x}}\left( -3A\cos (\sqrt{3})x-3B\sin (\sqrt{3})x \right)$$
 * }
 * }
 * }

Substituting into Equation (2),
 * {| style="width:100%" border="0" align="left"

 * $$0.0729{{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right)-0.54{{e}^{-0.27x}}\left( -\sqrt{\pi }A\sin (\sqrt{\pi }x)+\sqrt{\pi }B\cos (\sqrt{\pi }x) \right)+{{e}^{-0.27x}}\left( -\pi A\cos (\sqrt{\pi }x)-\pi B\sin (\sqrt{\pi }x) \right)+0.54\left[ -0.27{{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right)+{{e}^{-0.27x}}\left( -\sqrt{\pi }A\sin (\sqrt{\pi }x)+\sqrt{\pi }B\cos (\sqrt{\pi }x) \right) \right]+\left( 0.0729+\pi \right)\left[ {{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right) \right]=0$$
 * }
 * }
 * }

Multiplying and rearranging terms,
 * {| style="width:100%" border="0" align="left"

 * $$0.0729{{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right)+0.0729{{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right)-0.1458{{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right)+\pi {{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right)-\pi {{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right)-0.54{{e}^{-0.27x}}\left( -\sqrt{\pi }A\sin (\sqrt{\pi }x)+\sqrt{\pi }B\cos (\sqrt{\pi }x) \right)+0.54{{e}^{-0.27x}}\left( -\sqrt{\pi }A\sin (\sqrt{\pi }x)+\sqrt{\pi }B\cos (\sqrt{\pi }x) \right)=0$$
 * }
 * }
 * }

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

 * $$0=0 \!$$
 * }
 * }
 * }

Since all terms cancel, the solution is
 * {| style="width:100%" border="0" align="left"

 * $$y={{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right)$$
 * }
 * }
 * }

Author
Solved and typed by --Egm4313.s12.team9.wright 08:18, 8 February 2012 (UTC)

Reviewed by Egm4313.s10.team9.delp 16:16, 8 February 2012 (UTC)

Edited by

= R2.9 =

Given:
$$\lambda^2 + 4\lambda +13 = 0 \! $$ Characteristic equation from the ODE: $$y''+4y'+13y=r(x) \! $$ With the following initial conditions: $$ y(0)=1,\, y'(0)=0,\,  r(x)=0 \! $$

Solution I:
First, we solve the Characteristic Equation for $$\lambda$$ using the Quadratic Formula: $$\lambda = \frac{-b \pm \sqrt{b^2-4(a)(c)}}{2(a)} \! $$ $$\lambda = \frac{-4 \pm \sqrt{4^2-4(1)(13)}}{2(1)} \! $$ $$\lambda = \frac{-4 \pm 6i}{2} \! $$ $$\lambda = {-2 \pm 3i} \! $$ Where $$ a=-2 \,\! $$ and $$ \,\omega = 3 \! $$ The solution form for a Linear Second Order Differential Equation with Constant Coefficients is: $$ y(x) = e^{-ax}(C_1\cos({\omega x}) + C_2\sin({\omega x})) \! $$ Now we take the derivative of $$ y(x) $$ $$ y(x) = e^{-2x}(C_1\cos({3x}) + C_2\sin({3x})) \! $$ $$ y(x)' = e^{-2x}(-3C_1\sin({3x}) + 3C_2\cos({3x})) - 2e^{-2x}(C_1\cos({3x}) + C_2\sin({3x})) \! $$ Now, we use the initial conditions listed above to determine the coefficients $$C_1$$ and $$C_2$$. $$y(0)= 1 = e^{-2(0)}(C_1\cos({3(0)}) + C_2\sin({3(0)})) \! $$ $$ y(0)' = 0 = e^{-2(0)}(-3C_1\sin({3(0)}) + 3C_2\cos({3(0)})) - 2e^{-2(0)}(C_1\cos({3(0)}) + C_2\sin({3(0)})) \! $$ $$C_1=1 \! $$ $$C_2=\frac{2}{3} \! $$ Therefore, the final solution will be,

Solution II:
We are then asked to create a superposed plot with solutions from R2.1 and R2.6 using the following equations:

Author
Solved and typed by --Egm4313.s12.team9.murray 09:47, 8 February 2012 (UTC)

Reviewed by E.Beuses 17:15, 8 February 2012 (UTC)

Edited by

= References =

=Contributing Team Members=

--Egm4313.s12.team9.wright 09:28, 8 February 2012 (UTC)

--Egm4313.s10.team9.delp 16:20, 8 February 2012 (UTC)

--Egm4313.s12.team9.murray 18:53, 8 February 2012 (UTC)

--Egm4313.s12.team9.ehlers.km 15:14, 8 February 2012 (UTC)

--Egm4313.s12.team9.beuses 17:23, 8 February 2012 (UTC)

--Egm4313.s12.team9.tjl 18:23, 8 February 2012 (UTC)

--Egm4313.s12.team9.lebo 18:53, 8 February 2012 (UTC)

Table code referenced from Team 4- R1

title: User:Egm4313.s12.team9/R2


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

title: File:Sampling.2.B.BandPass.20120202.pdf


text:
User:FairyTeacher

FairyTeacher is a Lecturer in Information Systems since 2005 and an active researcher in Teaching and Learning in Higher Education, specialising in flexible learning and the learning environment.

title: User:FairyTeacher


text:
User talk:FairyTeacher

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title: User talk:FairyTeacher


text:
Helicopter/Howard Community College/Fall 2011/501 Helicopter

This is a link to my team page Drones

Give instructor top 3 project choice list
For my third project, I chose my top three preferences in the exact order. 1. Helicopter 2. Illuminate Clothing 3. Cavitation

Write problem statement
I chose to stay with the same project once again. This time i am continuing on working with the Arduino Uno and a DC Motor. I will try to retrace the professor's stimulation of an Arduino Duemilanove with a ESC computer and a motor.

Assign Task 1
I will continue working with the motor shield and try to solve my problem with why the DC Motor is not working with the code i am trying given provided by ladyada.net

Compare actual work done to Task 1
After the demonstration, I wanted to incline more on working with a motor and an ESC with the Arduino just like the professor showed in class as a demonstration. So this week I stopped working with the motor shield and started to work on a motor that i had to put together and connect an ESC to it. Well working with the motor brought up some frustration since i have never before worked with a motor yet alone put one together.

Week 1 Narrative
So in my first week, i was given a brand new Hacker Brushless Motor A10 13L and a brand new ES Thunderbird 9. So as i started working with the motor, i noticed that the manual did not teach you by steps how to put the motor together. So i researched online under this motor and i found a few websites, the most helpful was [] were i found the actyal motor online and what it includes incase i might be missing something. Also, i found a really helpful website with an actual image of my ESC that im using with a motor and how it is setup [] In addition, I also found on the web another guy who is using a very similar motor with the same ESC and I plan on using this image sort of as a guide for what my connections should look like [].

I also ran across an issue. I have three wires from the ESC; a white, red, and black wire. On my motor I also have three wires but none of them are labeled nor colored to show which wires I should connect together. I looked online but was not able to find anything and the closest I got to finding an answer is the link #2 where it shows the wiring of the ESC and a motor. In addtion to this, I want to bring up in engineering class of how I should come along with mounting this motor on a board. The reason being is that as you can see by the picture, it looks like what I will have to do is create a hole on a piece of wood that way the bottom of the motor will basically be inside the wood and only the propellers will be sticking out and visible. Maybe hearing what my classmates have to say I might be able to come up with a better solution.

Week 1 Peer Review
I added good reviews on my classmates wiki pages and i am also up to date with what they are doing as far as our new project on this team.

Assign Task 2
For my next task over the weekend I plan on testing the wires to see if i have connected the wires to each other correctly. I will also mount these parts to a 2"by4" piece of wood that way i wont loose anything and will make my work easier.

Compare actual work done to Task 2
For my second week, I proposed my problem to the class during a presentation. That problem was that I did not not if it mattered what 3 wires I would connect from the motor which has 3 wires and the ESC which also has 3 wires. So in class, the professor told me it did not matter because the ESC programs itself to know which wire is which when it is connected and powered. So using but splices, I connected the wires together and mounted this onto my board. But at first I mounted the motor onto the board, and as you read from week one I would have to make a hole through the board in order to get my motor on. So I did that, I drilled a whole through a board in order for my motor to spin freely on it without being disrupted or contacted. I didnt use any screws to keep the motor in place, what I used was epoxy, which I mixed carefully and applied to the motor, and I let it set for a few hours before it hardened. Now the motor will not come off at all, which is what I wanted from the beginning. Next I had to mount the ESC to the board, and to do this i would have to drill 2 small holes and strap the ESC to the board, the only problem is that I do not have a drill at home so I will drill these holes in the class room. In addition, im also adding the Arduino to the board and I will strap it onto the board, not glue it. Rationalize differences

Week 2 Narrative
After letting the epoxy set, I connected the motor and ESC, as you can see in the pictures. I only have a small problem but nothing major, I will have to take the but splice off the red cable because it doesnt give that much room for the motor's wire to go in. That I will take care of in class since it is nothing major and only takes a few seconds. I also took a picture of the Arduino on the board just to have the sense of exactly how my project is going to look like and how my wiring is going to be set up. I now also need to strap the Arduino onto the board so that is more holes that I now need to make on the board and again I will strap the Arduino just as i will with the ESC.

Week2 Peer Review
I looked at both of my team mates wiki pages on Sunday and I found nothing under their second week. I added on their discussion pages that I didnt see anything and saved that on their page.

Assign Task 3
For my next task, I will plan on getting some current over to the ESC to see if it will power the ESC. The ESC should make a noise if the wires are connected to each other correctly. I will also strap the Arduino onto the board and proceed with my attempt to mock the professor's project.

Compare actual work done to Task 3
Rationalize differences

Week 3 Narrative
Upload photos to wikimedia. Upload video to youtube. Convert all spreadsheets, documents, and presentations to wiki format and create wiki subpage of your page for each of them. Link to them here in a narrative that tells a story. High light the problems (engineering problems) you had and how you solved them.

Week 3 Peer Review
Go to each team mates weekly summary wiki page. Read their activities and narrative. Then go the associated discussion page. Say something positive on this page. Try some constructive criticism. Add your name and a signature ~ by entering four tildes in a row to create a time and date stamp. Create a new category if another team mate has already commented here.

Assign Task 4
Record what you are planning on doing for the team during the weekend between week 3 and week 4 of the project here.

Compare actual work done to Task 4
Rationalize differences

Week 4 Narrative
Upload photos to wikimedia. Upload video to youtube. Convert all spreadsheets, documents, and presentations to wiki format and create wiki subpage of your page for each of them. Link to them here in a narrative that tells a story. High light the problems (engineering problems) you had and how you solved them.

Week 4 Peer Review
Go to each team mates weekly summary wiki page. Read their activities and narrative. Then go the associated discussion page. Say something positive on this page. Try some constructive criticism. Add your name and a signature ~ by entering four tildes in a row to create a time and date stamp. Create a new category if another team mate has already commented here.

Complete Project Page
Follow the "project done" format.

Start Next Project Week 0 activities
During this week you will perform the Week 0 activities of the next project.


text:
Engine/Howard Community College/Fall 2011/501 Engine

Link to Team HEMI Page

Write problem statement
The Stirling engine is an external combustion engine. In our project, we will find a way to make it generate enough power to move a set of wheels allowing independent movement.

 * Example of Stirling Engine Motor

Compare actual work done to Task 1
In week 1, we tinkered around with the model engines that were already available to us to see if we could get it to run as is. We were able to get one to run with much assistance eventually. We thought that steam would be a better gas to use in the combustion process but we have no way of trapping steam in the cylinder for use. Air seems to be a better gas to use and placing water inside the cylinder to boil is not a good idea, nor allowed.

Week 1 Narrative

 * Putting water inside the cylinder is not a good idea.
 * We were not very successful with getting it to run just off of air.
 * I believe this is because the cylinder is not airtight so most of the energy that is produced is lost before it can start a cycle.
 * I suggest we find a way to make it airtight. That should solve our problem.

 * Another hypothesis is that we are not generating enough heat to heat the air but the videos that have portrayed a Stirling engine successfully have used flames from candles.
 * We even used the blow torch to no discernible difference or help.

Assign Task 2
I plan to look into other successfully running Stirling engines to see what aspects they used that we could incorporate. Mainly, the types of gases used like air vs. nitrogen. Also, to find the actual numbers that proves air is a good gas to use. I believe it has to do with the specific heat capacity of gasses.

Compare actual work done to Task 2
We worked on the engine more to make it work better with just air in the pressure chamber. The metal of the chamber had seem to expand and meld to the metal around it. This helped to block some of the air from escaping, so it worked to our favor in a way. We were able to get more movement out of the engine and even able to start movement without our having to move it first.

Week 2 Narrative
Model homemade Stirling engine Motor

Week 2 Peer Review
Go to each team mates weekly summary wiki page. Read their activities and narrative. Then go the associated discussion page. Say something positive on this page. Try some constructive criticism. Add your name and 1sfoerster 13:32, 28 August 2011 (UTC) to create a time and date stamp. Create a new category if another team mate has already commented here.

Assign Task 3
Since we don't seem to be making any significant progress in our project. We have been tasked with all building our own versions of stirling engine's at home and then compare our designs in class. We use the best out of all of them to build a functioning engine.

Compare actual work done to Task 3
I am working on my own design of a homemade engine using materials that I have at my disposal.

Week 3 Narrative
Initially, I am finding it hard to make the pieces I need because I do not have proper tools to cut through metal smoothly. Once I do get past this problem, I have looked at several different designs on youtube of people who have successfully made homemade engines before. It seems that the easiest idea would be to make 2 seperate chambers for the pistons instead of filling it into one space.

Week 3 Peer Review
Go to each team mates weekly summary wiki page. Read their activities and narrative. Then go the associated discussion page. Say something positive on this page. Try some constructive criticism. Add your name and 1sfoerster 13:33, 28 August 2011 (UTC) to create a time and date stamp. Create a new category if another team mate has already commented here.

Assign Task 4
Record what you are planning on doing for the team during the weekend between week 3 and week 4 of the project here.

Compare actual work done to Task 4
Rationalize differences

Week 4 Narrative
Upload photos to wikimedia. Upload video to youtube. Convert all spreadsheets, documents, and presentations to wiki format and create wiki subpage of your page for each of them. Link to them here in a narrative that tells a story. High light the problems (engineering problems) you had and how you solved them.

Week 4 Peer Review
Go to each team mates weekly summary wiki page. Read their activities and narrative. Then go the associated discussion page. Say something positive on this page. Try some constructive criticism. Add your name and 1sfoerster 13:33, 28 August 2011 (UTC) to create a time and date stamp. Create a new category if another team mate has already commented here.

Complete Project Page
Follow the "project done" format.

Start Next Project Week 0 activities
During this week you will perform the week 0 activities of the next project.


text:
User talk:OpenScientist/Open grant writing/Wissenswert 2011/Documentation

Beautiful Soup update
Docs update: http://www.crummy.com/software/BeautifulSoup/doc/ Sumanah 19:33, 2 February 2012 (UTC)


text:
User:Egm4313.s12.team6.mcpherson

Team 6

title: User:Egm4313.s12.team6.mcpherson


title: User:Egm4313.s12.team6.nembhard


text:
User:Mwnuk

User
Matthew Wnuk

Vocation
US Navy, SONAR technician

Origin
Milwaukee, Wisconsin

Age
30

Goal
Student for life

Classes

 * Engineering Design
 * Digital Circuit Lab
 * Electrical Circuits

title: User:Mwnuk


text:
User:Egm4313.s12.team19.R2

R2.1
Contributed by Vladimir Horwitz.

Problem Statement:
a) Find a nonhomogeneous L2ODECC for the roots and initial conditions listed here
 * {| style = "width: 100%"

!  $$ \lambda_1 = - 2, \ \lambda_2 = 5 $$ !!  (Eq. 1, a and b) The L2ODECC should be of standard form and be written for the general excitation r(x).
 *  $$ \ y (0) = 1, \ y' (0) = 0 $$
 *  (Eq. 2, a and b)
 * }
 * }

b) Plot the solution for $$ \ r (x) = 0 $$.

c) Write 3 L2ODECCs that are not standard and not homogeneous that will accept (eq. 1, a and b) and (eq. 2, a and b).

Solution:
a) The characteristic equation for the given roots and initial conditions is of the form $$ \ (\lambda - \lambda_1) * (\lambda - \lambda_2) = 0 $$.

Therefore,

 * {| class = wikitable style="width: 100%; text-align:center"

!  $$ \ y'' - 3 y' - 10 y = r (x) $$  (Eq. 3)
 * }

The solutions should be of the form $$ \ y (x) = y_p (x) + y_h (x) $$.

Where according to (eq. 1, a and b) and (eq. 2, a and b)

 * {| style = "width: 100%"

!  $$ \ y_h (0) = c_1 + c_2 $$ !!  (Eq. 4)
 *  $$ \ y'_h (0) = c_1 \lambda_1 + c_2 \lambda_2 $$
 *  (Eq. 5)
 * and
 *  $$ \ c_1 = {5 \over 7} = .71429 $$
 *  (Eq. 6)
 *  $$ \ c_2 = {2 \over 7} = .28571 $$
 *  (Eq. 7)
 * Thus,
 *  $$ \ y(x) = \overbrace {.71429 e^{(-2 x)} + .28571 e^{(5 x)}}^{y_h (x)} + y_p (x) $$
 *  (Eq. 8)
 * }
 * Thus,
 *
 $$ \ y(x) = \overbrace {.71429 e^{(-2 x)} + .28571 e^{(5 x)}}^{y_h (x)} + y_p (x) $$
 *
 (Eq. 8)
 * }
 *
 (Eq. 8)
 * }

b) We can now plot the homogeneous form of (eq. 8) $$ \underbrace {y_h (x) = .71429 e^{(-2 x)} + .28571 e^{(5 x)}}_{\downarrow} $$.

c) Three more equations can be attained for the given (eq. 1, a and b) and (eq. 2, a and b), because of the linearity.

 * {| class = wikitable style="width: 100%; text-align:center"

!
 $$ \ 3 y'' - 9 y' - 30 y = r (x) $$
 (Eq. 9)
 * }

 * {| class = wikitable style="width: 100%; text-align:center"

!
 $$ \ 10 y'' - 30 y' - 100 y = r (x) $$
 (Eq. 10)
 * }

 * {| class = wikitable style="width: 100%; text-align:center"

!
 $$ \ 5 y'' - 15 y' - 50 y = r (x) $$
 (Eq. 11)
 * }

R2.2
Contributed by Gabriel Arab.

Problem Statement:
Find and plot the solution for the L2-ODE-CC,

 * {| style = "width: 100%"

!
 $$ \ y'' - 10 y' + 25 y = r (x) $$ !!
 (Eq. 1)
 * Initial conditions.
 *
 $$ \ y (0) = 1, \ y' (0) = 0 $$
 *
 (Eq. 2)
 * No excitation.
 *
 $$ \ r (x) = 0 $$
 *
 (Eq. 3)
 * }
 *
 $$ \ r (x) = 0 $$
 *
 (Eq. 3)
 * }
 * }

Solution:
Converting original ODE into λ notation and factoring,

 * {| style = "width: 100%"

!
 $$ \ \lambda^2 - 10 \lambda + 25 = r (x) $$ !!
 (Eq. 4)
 *
 $$ \ (\lambda - 5)^2 = r (x) = 0 $$
 *
 (Eq. 5)
 * We find that
 *
 $$ \ \lambda = 5 $$ is a double real root.
 *
 (Eq. 6)
 * }
 *
 $$ \ \lambda = 5 $$ is a double real root.
 *
 (Eq. 6)
 * }

Implementing the solution equation for real double roots and deriving,

 * {| style = "width: 100%"

!
 $$ \ y_h (x) = c_1 e^{5 x} + c_2 x e^{5 x} $$ !!
 (Eq. 7)
 *
 $$ \ y_h' (x) = 5 c_1 e^{5 x} + 5 c_2 x e^{5 x} + c_2 e^{5 x} $$
 *
 (Eq. 8)
 * }
 * }

Plugging in initial conditions,

 * {| style = "width: 100%"

!
 $$ \ y_h (0) = c_1 e^{5 (0)} + c_2 (0) e^{5 (0)} = 1 $$ !!
 (Eq. 9)
 *
 $$ \ 1 = c_1 $$
 *
 (Eq. 10)
 * and
 *
 $$ \ y_h' (0) = 5 c_1 e^{5 (0)} + 5 c_2 (0) e^{5 (0)} + c_2 e^{5 (0)} = 0 $$
 *
 (Eq. 11)
 *
 $$ \ 5 c_1 + c_2 = 0 $$
 *
 (Eq. 12)
 * Thus,
 *
 $$ \ c_2 = -5 $$
 *
 (Eq. 13)
 * }
 * Thus,
 *
 $$ \ c_2 = -5 $$
 *
 (Eq. 13)
 * }
 *
 (Eq. 13)
 * }

Writing final solution equation and plotting on WolframAlpha engine,

 * {| class = wikitable style="width: 100%; text-align:center"

!
 $$ \ y_h (x) = e^{5 x} - 5 x e^{5 x} $$
 (Eq. 14)
 * }

R2.3
Contributed by Adam Burton.

Given:

Problem Statement:
From Kreyszig 2011 page 59, problems 3 and 4.

Find a general solution. Check your answers by substitution.

Solution:
3) :{| style = "width: 100%" !
 $$ \ y'' + 6y' + 8.96y = 0  $$ !!
 (Eq. 1)
 * }

Substituting

!
 $$ \ h = d/dx  $$ !!
 (Eq. 2)
 * }

 * {| style = "width: 100%"

!
 $$ \ h^2 + 6h + 8.96 = 0 $$ !!
 (Eq. 3)
 * }

To solve this, we will use the quadratic formula:

 * {| style = "width: 100%"

!
 $$ \ h = \frac{-b \plusmn\! \sqrt{b^2-4ac}}{2a} $$ !!
 (Eq. 4)
 * }

 * {| style = "width: 100%"

!
 $$ \ h = \frac{-6 \plusmn\! \sqrt{6^2-4(1)(8.96)}}{2(1)} $$ !!
 (Eq. 4a)
 * }

Through calculation, we see that:

 * {| style = "width: 100%"

!
 $$ \ h_{1,2} = -2.8, -3.2$$ !!
 (Eq. 5)
 * }

Therefore, the general solution is:

 * {| class = wikitable style="width: 100%; text-align:center"

!
 $$ \ y = c_1e^{-2.8x} + c_2e^{-3.2x} $$
 (Eq. 6)
 * }

Check:

 * {| style = "width: 100%"

!
 $$ \ y = c_1e^{-2.8x} + c_2e^{-3.2x} $$ !!
 (Eq. 6)
 * }

 * {| style = "width: 100%"

!
 $$ \ y' = -2.8c_1e^{-2.8x} - 3.2c_2e^{-3.2x} $$ !!
 (Eq. 7)
 * }

 * {| style = "width: 100%"

!
 $$ \ y'' = 7.84c_1e^{-2.8x} + 10.24c_2e^{-3.2x} $$ !!
 (Eq. 8)
 * }

Substituting these 3 equations in to equation 1 we get:

 * {| style = "width: 100%"

!
 $$ \ 7.84c_1e^{-2.8x} + 10.24c_2e^{-3.2x} + 6(-2.8c_1e^{-2.8x} - 3.2c_2e^{-3.2x}) + 8.96(c_1e^{-2.8x} + c_2e^{-3.2x}) = 0 $$ !!
 (Eq. 9)
 * }

 * {| style = "width: 100%"

!
 $$ \ 0 = 0 $$ !!
 (Eq. 10)
 * }

Through substitution, we can see that the solution above is correct!

4)
 * {| style = "width: 100%"

!
 $$ \ y'' + 4y' + (\pi^2 + 4)y = 0 $$ !!
 (Eq. 1)
 * }

Substituting

!
 $$ \ h = d/dx $$ !!
 (Eq. 2)
 * }

 * {| style = "width: 100%"

!
 $$ \ h^2 + 4h + (\pi^2+4) = 0 $$ !!
 (Eq. 3)
 * }

To solve this, we will use the quadratic formula:

 * {| style = "width: 100%"

!
 $$ \ h = \frac{-b \plusmn\! \sqrt{b^2-4ac}}{2a} $$ !!
 (Eq. 4)
 * }

 * {| style = "width: 100%"

!
 $$ \ h = \frac{-4 \plusmn\! \sqrt{4^2-4(1)(\pi^2+4)}}{2(1)} $$ !!
 (Eq. 4a)
 * }

 * {| style = "width: 100%"

!
 $$ \ h = \frac{-4 \plusmn\! \sqrt{2\pi}}{2(1)} $$ !!
 (Eq. 4b)
 * }

Through calculation, we see that:

 * {| style = "width: 100%"

!
 $$ \ h_{1,2} = -2 \plusmn\ i\pi$$ !!
 (Eq. 5)
 * }

Therefore, the general solution is:

 * {| class = wikitable style="width: 100%; text-align:center"

!
 $$ \ y = e^{-2x}(c_1\cos\pi x + c_2\sin\pi x) $$
 (Eq. 6)
 * }

Check:

 * {| style = "width: 100%"

!
 $$ \  y = e^{-2x}(c_1\cos\pi x + c_2\sin\pi x)  $$ !!
 (Eq. 6)
 * }

 * {| style = "width: 100%"

!
 $$ \  y' = e^{-2x}(-\pi c_1\sin\pi x + \pi c_2\cos\pi x) - 2e^{-2x}(c_1\cos\pi x + c_2\sin\pi x) $$ !!
 (Eq. 7)
 * }

 * {| style = "width: 100%"

!
 $$ \  y'' = 9e^{-2x}(c_1\cos\pi x + c_2\sin\pi x) + e^{-2x}(-\pi c_1\cos\pi x + \pi c_2\sin\pi x) - 6e^{-2x}(-\pi c_1\sin\pi x + \pi c_2\cos\pi x) $$ !!
 (Eq. 8)
 * }

Substituting these 3 equations in to equation 1 we get:

 * {| style = "width: 100%"

!
 $$ \  9e^{-2x}(c_1\cos\pi x + c_2\sin\pi x) + e^{-2x}(-\pi c_1\cos\pi x + \pi c_2\sin\pi x) - 6e^{-2x}(-\pi c_1\sin\pi x + \pi c_2\cos\pi x) \ + \ \downarrow $$
 *
 $$ \uparrow + \ 4(e^{-2x}(-\pi c_1\sin\pi x + \pi c_2\cos\pi x) - 2e^{-2x}(c_1\cos\pi x + c_2\sin\pi x)) + (\pi^2 + 4)(e^{-2x}(c_1\cos\pi x + c_2\sin\pi x)) = 0 $$
 *
 (Eq. 9)
 * }
 * }

 * {| style = "width: 100%"

!
 $$ \  0 = 0 $$ !!
 (Eq. 10)
 * }

Through substitution, we can see that the solution above is correct!

R2.4
Contributed by Joshua Campbell.

Problem 5

Given:

 * {| style = "width: 100%"

!
 $$ \ y'' + 2\pi y' + \pi^{2} y =0 $$ !!
 (Eq. 1)
 * }

Problem Statement:
Find a general solution. Check your answer by substitution.

Solution:
The characteristic equation is given by:

 * {| style = "width: 100%"

!
 $$ \ r^{2} +2\pi +\pi^{2}= 0 $$ !!
 (Eq. 2)
 * }

or

 * {| style = "width: 100%"

!
 $$ \ (r + \pi)(r+ \pi) = 0 $$ !!
 (Eq. 3)
 * }

Since this is a repeated root, the general solution is of the form:

 * {| style = "width: 100%"

!
 $$ \ y(x) = c_1 e^{x} + c_2 x e^{x} $$ !!
 (Eq. 4)
 * }

A general solution is then

 * {| class = wikitable style="width: 100%; text-align:center"

!
 $$ \ y(x) = (c_1 + c_2 x) e^{-\pi x}  $$
 (Eq. 5)
 * }

Problem 6

Given:

 * {| style = "width: 100%"

!
 $$ \ 10y'' - 32 y' + 25.6 y =0 $$ !!
 (Eq. 6)
 * }

Problem Statement:
Find a general solution. Check your answer by substitution.

Solution:
The characteristic equation is given by:

 * {| style = "width: 100%"

!
 $$ \ 10 r^{2} -32 r + 25.6= 0 $$ !!
 (Eq. 7)
 * }

the expression can be factored using the quadratic formula since it is of the form:

 * {| style = "width: 100%"

!
 $$ \ a x^{2} + b x + c = 0$$ !!
 (Eq. 8)
 * }

the quadratic formula states that the roots of the characteristic equation, x, are:

 * {| style = "width: 100%"

!
 $$ \ x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}$$ !!
 (Eq. 9)
 * }

We see from the result of the calculation that the solution to the characteristic equation is a repeated root of

 * {| style = "width: 100%"

!
 $$ \ x = 1.6 $$ !!
 (Eq. 10) Since this is again a repeated root, the general solution is of the form:
 * }

 * {| style = "width: 100%"

!
 $$ \ y(x) = c_1 e^{x} + c_2 x e^{x} $$ !!
 (Eq. 11)
 * }

A general solution is then

 * {| class = wikitable style="width: 100%; text-align:center"

!
 $$ \ y(x) = (c_1 + c_2 x) e^{-1.6 x}  $$
 (Eq. 12)
 * }

R2.5
Contributed by Charles Chiamchittrong.

Problem Statement:
Kreyszig 2011 p.59 pbs. 16-17

We are asked to find a second-order homogeneous linear ODE of the form

 * {| style = "width: 100%"

!
 $$ \ y'' + ay' + by= 0 $$ !!
 (Eq. 1) for a given basis of solutions.
 * }

Problem 16
 * {| style = "width: 100%"

!
 $$ \ e^{2.6x} \, \ e^{-4.3x} $$ !!

 * }

Problem 17
 * {| style = "width: 100%"

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

 * }

Solution:
Problem 16

Examining the given basis, we can see that this is a case of two distinct real roots.

The general solution in this case is:
 * {| style = "width: 100%"

!
 $$ \ y = c_1e^{\lambda_1x} + c_2e^{\lambda_2x} $$ !!
 (Eq. 2)
 * }

Substituting the basis of solutions, we obtain:
 * {| style = "width: 100%"

!
 $$ \ y = c_1e^{2.6x} + c_2e^{-4.3x} $$ !!
 (Eq. 3)
 * }

Comparing Eq. 2 and Eq. 3 we can determine the roots $$\lambda_1$$ and $$\lambda_2$$:

 * {| style = "width: 100%"

!
 $$ \ \lambda_1 = 2.6 \, \ \lambda_2 = -4.3 $$ !!
 (Eq. 4a, 4b)
 * }

Now we can write the characteristic equation:

 * {| style = "width: 100%"

!
 $$ \ (\lambda - \lambda_1)(\lambda - \lambda_2) = 0 $$ !!
 (Eq. 5)
 * }

 * {| style = "width: 100%"

!
 $$ \ (\lambda - 2.6)(\lambda + 4.3) = 0 $$ !!
 (Eq. 6)
 * }

 * {| style = "width: 100%"

!
 $$ \ \lambda^2 + 1.7\lambda - 11.18 = 0 $$ !!
 (Eq. 7)
 * }

Finally, we can convert the characteristic equation into the corresponding homogeneous ODE:

 * {| class = wikitable style="width: 100%; text-align:center"

!
 $$ \ y'' + 1.7y' - 11.18y = 0 $$
 (Eq. 8)
 * }

Problem 17

Examining the given basis, we can see that this is a case of a real double root.

The general solution in this case is:
 * {| style = "width: 100%"

!
 $$ \ y = (c_1 + c_2x)e^{\lambda x} $$ !!
 (Eq. 9)
 * }

Substituting the basis of solutions, we obtain:
 * {| style = "width: 100%"

!
 $$ \ y = (c_1 + c_2x)e^{-\sqrt{5} x} $$ !!
 (Eq. 10)
 * }

Comparing Eq. 8 and Eq. 9 we can determine the root $$\lambda_1$$:

 * {| style = "width: 100%"

!
 $$ \ \lambda_1 = -\sqrt{5} $$ !!
 (Eq. 11)
 * }

Now we can write the characteristic equation:

 * {| style = "width: 100%"

!
 $$ \ (\lambda - \lambda_1)^2 = 0 $$ !!
 (Eq. 12)
 * }

 * {| style = "width: 100%"

!
 $$ \ (\lambda + \sqrt{5})^2 = 0 $$ !!
 (Eq. 13)
 * }

 * {| style = "width: 100%"

!
 $$ \ \lambda^2 + 2\sqrt{5}\lambda + 5 = 0 $$ !!
 (Eq. 14)
 * }

Finally, we can convert the characteristic equation into the corresponding homogeneous ODE:

 * {| class = wikitable style="width: 100%; text-align:center"

!
 $$ \ y'' + 2\sqrt{5}y' + 5y = 0 $$
 (Eq. 15)
 * }

R2.6
Contributed by Santiago Marin.

Problem Statement:
Realize a spring-dashpot-mass systems in series like the one shown in the figure below.

Using the real double root λ = -3, find the values for the parameters k,c,m.

Solution:
Using equation (3) p.5-5 and putting into standard form, the motion of a spring-dashpot-mass system is:

!
 $$ y'' + \frac{k}{cm}y' + \frac{k}{m}y = f(t)\!$$ !!
 (Eq. 1)
 * }

Using the real double root λ= -3, the characteristic equation is:
 * {| style = "width: 100%"

!
 $$ y'' + 6y' + 9 = 0 \!$$ !!
 (Eq. 2)
 * }

Matching the coefficients of equation (Eq. 1) and (Eq. 2) a solution is:

$$ c = \frac{3}{2} \!$$

$$ k = 9 \!$$

$$ m = 1 \!$$

R2.7
Contributed by Grant (Kyle) Uppercue.

Problem statement:
Develop the MacLaurin series (Taylor series at t = 0) for:
 * {| style = "width: 100%"

!
 $$ \ e^t, cost , sint $$ !!
 (Eq. 1)
 * }

Solutions:
General Taylor Series Expansion at point a:
 * {| style = "width: 100%"

!
 $$ f(a) + \frac{f'(a)} {1!} (x-a) + \frac{f(a)} {2!} (x-a)^2 + \frac{f'(a)} {3!} (x-a)^3 + ... $$ !!
 (Eq. 2)
 * }

$$ \ e^t $$, where t = 0:
 * {| style = "width: 100%"

!
 $$ e^0 + \frac{e^0} {1!} (x) + \frac{e^0} {2!} (x)^2 + \frac{e^0} {3!} (x)^3 + ... $$ !!
 (Eq. 3)
 * }

 * {| class = wikitable style="width: 100%; text-align:center"

!
 $$ = 1 + x + \frac{x^2} {2} + \frac{x^3} {6} + ... $$
 (Eq. 4)
 * }

$$ \ \cos(t) $$, where t = 0:

 * {| style = "width: 100%"

!
 $$ \cos(0) + \frac{-\sin(0)} {1!} (x) + \frac{-\cos(0)} {2!} (x)^2 + \frac{sin(0)} {3!} (x)^3 + \frac{cos(0)} {4!} (x)^4 + ... $$ !!
 (Eq. 5)
 * }

 * {| class = wikitable style="width: 100%; text-align:center"

!
 $$ = 1 - \frac{x^2} {2} + \frac{x^4} {24} + ... $$
 (Eq. 6)
 * }

$$ \ \sin(t) $$, where t = 0:

 * {| style = "width: 100%"

!
 $$ \sin(0) + \frac{cos(0)} {1!} (x) + \frac{-\sin(0)} {2!} (x)^2 + \frac{-cos(0)} {3!} (x)^3 + \frac{\sin(x)} {4!} (x)^4 + \frac{\cos(0)} {5!} (x)^5 + ... $$ !!
 (Eq. 7)
 * }

 * {| class = wikitable style="width: 100%; text-align:center"

!
 $$ = x - \frac{x^3} {3!} + \frac{x^5} {5!} + ... $$
 (Eq. 8)
 * }

R2.8
Contributed by Vladimir Horwitz.

Problem Statement:
Answer questions 8 and 15 from Kreyszing.

Solution:
Prob. 8)

 * {| style = "width: 100%"

!
 $$ \ y'' + y' + 3.25 y = 0 $$ !!
 (Eq. 1)
 *
 $$ \ \lambda^2 + \lambda + 3.25 = 0 $$
 *
 (Eq. 2)
 * }
 * }

The characteristic equation has these roots

 * {| style = "width: 100%"

 *
 $$ \lambda = -1/2 + \sqrt {3} i, \bar \lambda = -1/2 - \sqrt {3} i $$
 *
 (Eq. 3, a and b)
 * }
 * }

Due to the fact that these roots are complex, the following equations from Kreyszig 2011, page 57 case 3, can be applied.

 * {| class = wikitable style="width: 100%; text-align:center"

!
 $$ \ y = e^{-1/2} * (A \cos (\sqrt 3 x) + B \sin (\sqrt 3 x)) $$
 (Eq. 4)
 * }

And following this, we can say

 * {| style = "width: 100%"

 *
 $$ y' = -.5 e^{-.5} \ (A \cos (\sqrt 3 x) + B \sin (\sqrt 3 x)) \ + \ e^{-.5} \ (-A \sqrt 3 \sin (\sqrt 3 x) + B \sqrt 3 \cos (\sqrt 3 x)) $$
 *
  (Eq. 5)
 * and
 *
 $$ y'' = .25 e^{-.5} \ (A \cos \sqrt 3 x \ + B \sin \sqrt 3 x) \ + \ e^{-.5} \ (-3 A \cos \sqrt 3 x \ - 3 B \sin \sqrt 3 x) + \downarrow $$
 *
 $$ \uparrow -.5 e^{-.5} \ (-A \sqrt 3 \sin \sqrt 3 x \ + B \sqrt 3 \cos \sqrt 3 x) \ - \ .5 e^{-.5} \ (-A \sqrt 3 \sin \sqrt 3 x \ + B \sqrt 3 \cos \sqrt 3 x) $$
 *
 (Eq. 6)
 * }
 *
 $$ \uparrow -.5 e^{-.5} \ (-A \sqrt 3 \sin \sqrt 3 x \ + B \sqrt 3 \cos \sqrt 3 x) \ - \ .5 e^{-.5} \ (-A \sqrt 3 \sin \sqrt 3 x \ + B \sqrt 3 \cos \sqrt 3 x) $$
 *
 (Eq. 6)
 * }

which simplifies to

 * {| style = "width: 100%"

 *
 $$ 0 = (3.25 - .5 + .25 - 3) * e^{-.5} * (A \cos \sqrt 3 x \ + B \sin \sqrt 3 x) $$
 *
 (Eq. 7)
 * }

Prob. 15)

 * {| style = "width: 100%"

!
 $$ \ y'' + .54 y' + (.0729 + \pi) y = 0 $$ !!
 (Eq. 1)
 *
 $$ \ \lambda^2 + .54 \lambda + .0729 + \pi = 0 $$
 *
 (Eq. 2)
 * }
 * }

The characteristic equation has these roots

 * {| style = "width: 100%"

 *
 $$ \lambda = -.27 + \sqrt {\pi} i, \bar \lambda = -.27 - \sqrt {\pi} i $$
 *
 (Eq. 3, a and b)
 * }

Due to the fact that this roots are complex, the following equations from Kreyszig 2011, page 57 case 3, can be applied.

 * {| class = wikitable style="width: 100%; text-align:center"

!
 $$ \ y = e^{-.27} * (A \cos (\sqrt \pi x) + B \sin (\sqrt \pi x)) $$
 (Eq. 4)
 * }

And then we can show

 * {| style = "width: 100%"

 *
 $$ y' = -.27 e^{-.27} \ (A \cos (\sqrt \pi x) + B \sin (\sqrt \pi x)) \ + \ e^{-.27} \ (-A \sqrt \pi \sin (\sqrt \pi x) + B \sqrt \pi \cos (\sqrt \pi x)) $$
 *
 (Eq. 5)
 * and
 *
 $$ y'' = .0729 e^{-.27} \ (A \cos \sqrt \pi x \ + B \sin \sqrt \pi x) \ + \ e^{-.27} \ (- \pi A \cos \sqrt \pi x \ - \pi B \sin \sqrt \pi x) + \downarrow $$
 *
 $$ \uparrow -.27 e^{-.27} \ (-A \sqrt \pi \sin \sqrt \pi x \ + B \sqrt \pi \cos \sqrt \pi x) \ - \ .27 e^{-.27} \ (-A \sqrt \pi \sin \sqrt \pi x \ + B \sqrt \pi \cos \sqrt \pi x) $$
 *
 (Eq. 6)
 * Then, after major calculator work, we find that
 *
 $$ 0 = (0) * e^{-.27} * (A \cos \sqrt \pi x \ + B \sin \sqrt \pi x) $$
 *
 (Eq. 7)
 * }
 * Then, after major calculator work, we find that
 *
 $$ 0 = (0) * e^{-.27} * (A \cos \sqrt \pi x \ + B \sin \sqrt \pi x) $$
 *
 (Eq. 7)
 * }
 * }

R2.9
Contributed by Gabriel Arab.

Problem Statement:
Find and plot the solution for the L2-ODE-CC,

 * {| style = "width: 100%"

!
 $$ \ \lambda^2 + 4 \lambda + 13 = 0 $$ !!
 (Eq. 1)
 * Initial conditions,
 *
 $$ \ y (0) = 1, \ y' (0) = 0 $$
 *
 (Eq. 2)
 * No excitation,
 *
 $$ \ r (x) = 0 $$
 *
 (Eq. 3)
 * }
 *
 $$ \ r (x) = 0 $$
 *
 (Eq. 3)
 * }
 * }

Solution:
Solving for roots using Quadratic Equation with a=1, b=4 and c=13

 * {| style = "width: 100%"

!
 $$ \ \lambda = {-b \pm \sqrt {b^2 - 4 a c} \over 2 a} $$ !!
 (Eq. 4)
 *
 $$ \ \lambda = {-4 \pm \sqrt {4^2 - 4 * 1 * 13} \over 2 * 1} $$
 *
 (Eq. 5)
 * which is complex
 *
 $$ \ \lambda = -2 \pm 3 i $$
 *
 (Eq. 6)
 * }
 *
 $$ \ \lambda = -2 \pm 3 i $$
 *
 (Eq. 6)
 * }

Implementing the solution equation for complex roots and deriving

 * {| style = "width: 100%"

!
 $$ \ y_h (x) = e^{-2 x} (A \cos(3 x) + B \sin(3 x)) $$ !!
 (Eq. 7)
 *
 $$ \ y_h' (x) = -2 e^{-2 x} [A \cos(3 x) + B \sin(3 x)] + e^{-2 x} [-3 A \sin(3 x) + 3 B \cos(3 x)] $$
 *
 (Eq. 8)
 * }
 * }

Plugging in initial conditions into (7) and (8),

 * {| style = "width: 100%"

!
 $$ \ y_h (0) = e^{-2 (0)} [A \cos(3(0)) + B \sin(3(0))] = 1 $$ !!
 (Eq. 9)
 *
 $$ \ A = 1 $$
 *
 (Eq. 10)
 * and
 *
 $$ \ y_h' (0) = -2 e^{-2 (0)} [A \cos(3 (0)) + B \sin(3 (0))] + e^{-2 (0)} [-3 A \sin(3 (0)) + 3 B \cos(3 (0))] = 0 $$
 *
 (Eq. 11)
 *
 $$ \ 2 A + 3 B = 0 $$
 *
 (Eq. 12)
 * Thus,
 *
 $$ \ B = 2 / 3 $$
 *
 (Eq. 13)
 * }
 * Thus,
 *
 $$ \ B = 2 / 3 $$
 *
 (Eq. 13)
 * }
 *
 (Eq. 13)
 * }

Writing final solution equation and plotting on WolframAlpha engine,

 * {| class = wikitable style="width: 100%; text-align:center"

!
 $$ \ y_h (x) = e^{-2 x} (\cos(3 x) + {2 \over 3} \sin(3 x)) $$
 (Eq. 14)
 * }

Page information
Page designed by Vladimir Horwitz on 02:31, 3 February 2012 (UTC).

Last edited by Egm4313.s12.team19.Horwitz on 19:51, 8 February 2012 (UTC).

title: User:Egm4313.s12.team19.R2


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Crank Shaft
A co-worker of mine had some thoughts on the crank. he suggested circles instead of arms. This would be more sturdy and help with balance.

B

xC
B Fb

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Psychology (A-Level)

This is a brief overview of the A-Level course for Psychology.

Unit 1 - Cognitive Psychology, Developmental Psychology and Research Methods

 * Memory
 * Attachments
 * Research Methods

Unit 2 - Biological Psychology, Social Psychology and Individual Differences

 * Stress
 * Conformity and Group Influence
 * Abnormalities

Unit 3 - Topics in Psychology
You will learn three of the following subjects for Unit 3.
 * Biological Rhythms and Sleep
 * Perception
 * Relationships
 * Agression
 * Eating Behaviour
 * Gender
 * Intelligence and Learning
 * Cognition and Development

Unit 4 - Psychopathology, Psychology in Action and Research Methods

 * Psychopathology (One of the following)
 * Schizophrenia
 * Depression
 * Phobic Disorders
 * Obsessive Compulsive Disorder
 * Psychology in Action
 * Media Psychology
 * The Psychology of Addictive Behaviour
 * Anomalistic Psychology
 * Research Methods

title: Psychology (A-Level)