University of Florida/Egm4313/s12.team11.imponenti/R3.1

Problem Statement
Find the solution to the following L2-ODE-CC: $$y''-10y'+25y=r(x)\!$$

With the following excitation: $$r(x)=7e^{5x}-2x^2\!$$

And the following initial conditions: $$y(0)=4, y'(0)=-5\!$$

Plot this solution and the solution in the example on p.7-3

Homogeneous Solution
To find the homogeneous solution we need to find the roots of our equation

$$\lambda^2-10\lambda+25=0\!$$ $$(\lambda-5)(\lambda-5)=0\!$$ $$\lambda=5\!$$

We know the homogeneous solution for the case of a real double root with $$\lambda=5\!$$ to be

$$y_h=c_1e^{5x}+c_2xe^{5x}\!$$

Particular Solution
For the given excitation we must use the Sum Rule to the particular solution as follows

$$y_p=y_{p1}+y_{p2}\!$$ where $$y_{p1}\!$$ and $$y_{p2}\!$$ are the solutions to $$r_1(x)=7e^{5x}\!$$ and $$r_2(x)=-2x^2\!$$, respectively

First Particular Solution
$$r_1(x)=7e^{5x}\!$$,

from table 2.1, K 2011, pg. 82 we have

$$y_{p1}=Ce^{5x}\!$$

but this corresponds to one of our homogeneous solutions so we must use the modification rule to get

$$y_{p1}=Cx^2e^{5x}\!$$

Plugging this into the original L2-ODE-CC then substituting;

$$y_{p1}''-10y_{p1}'+25y_{p1}=r_1(x)\!$$

$$(Cx^2e^{5x})''-10(Cx^2e^{5x})'+25(Cx^2e^{5x})=r_1(x)\!$$

$$25Cx^2e^{5x}+10Cxe^{5x}+10Cxe^{5x}+2Ce^{5x}-10(5Cx^2e^{5x}+2Cxe^{5x})+25Cx^2e^{5x}=7e^{5x}\!$$

$$e^{5x}[25Cx^2+10Cx+10Cx+2C-50Cx^2-20Cx+25Cx^2]=7e^{5x}\!$$

$$e^{5x}2C=7e^{5x}\!$$

so $$C=\frac{7}{2}\!$$ and the first particular solution is,
 * $$y_{p1}=\frac{7}{2}x^2e^{5x}\!$$

Second Particular Solution
$$r_2(x)=-2x^2\!$$,

from table 2.1, K 2011, pg. 82 we have

$$y_{p2}=a_2x^2+a_1x+a_0\!$$

Plugging this into the original L2-ODE-CC then substituting;

$$y_{p2}''-10y_{p2}'+25y_{p2}=r_2(x)\!$$

$$(a_2x^2+a_1x+a_0)''-10(a_2x^2+a_1x+a_0)'+25(a_2x^2+a_1x+a_0)=-2x^2\!$$

$$2a_2-10(2a_2x+a_1)+25(a_2x^2+a_1x+a_0)=-2x^2\!$$

grouping like terms we get three equations to solve for the three unknowns, these are written in matrix form

$$25a_2x^2+(25a_1-20a_2)x+(25a_0-10a_1+2a_2)=-2x^2\!$$


 * $$ \begin{bmatrix}

2 & -10 & 25\\ -20 & 25 & 0\\  25 & 0 & 0\end{bmatrix}

\begin{bmatrix} a_2\\ a_1\\ a_0\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ -2\end{bmatrix}\!$$

solving by back subsitution leads to $$a_2=-\frac{2}{25}, a_1=\frac{8}{125}, a_0=\frac{4}{125}\!$$

so the second particular solution is,
 * $$y_{p2}=-\frac{2}{25}x^2+\frac{8}{125}x+\frac{4}{125}\!$$

General Solution
The general solution is the summation of the homogeneous and particular solutions

$$y=y_h+y_{p1}+y_{p2}\!$$

$$y=c_1e^{5x}+c_2xe^{5x}+\frac{7}{2}x^2e^{5x}-\frac{2}{25}x^2+\frac{8}{125}x+\frac{4}{125}\!$$

$$y=e^{5x}(c_1+c_2x+\frac{7}{2}x^2)-\frac{2}{25}x^2+\frac{8}{125}x+\frac{4}{125}\!$$

Applying the first initial condition $$y(0)=4\!$$

$$y(0)=c_1+\frac{4}{125}=4\!$$

$$c_1=\frac{496}{125}\!$$

Second initial condition $$y'(0)=-5\!$$

$$y'=\frac{d}{dx}y=e^{5x}[5(c_1+c_2x+\frac{7}{2}x^2)+c_2+7x]-\frac{4}{25}x+\frac{8}{125}\!$$

$$y'=e^{5x}[\frac{35}{2}x^2+(5c_2+7)x+5c_1+c_2]-\frac{4}{25}x+\frac{8}{125}\!$$

$$y'(0)=5c_1+c_2+\frac{8}{125}=-5\!$$

$$c_2=-\frac{3113}{125}\!$$

The general solution to the differential equation is therefore

$$y(x)=e^{5x}(\frac{496}{125}-\frac{3113}{125}x+\frac{7}{2}x^2)-\frac{2}{25}x^2+\frac{8}{125}x+\frac{4}{125}\!$$

Plot
Below is a plot of this solution and the solution to in the example on p.7-3

our solution $$y(x)=e^{5x}(\frac{496}{125}-\frac{3113}{125}x+\frac{7}{2}x^2)-\frac{2}{25}x^2+\frac{8}{125}x+\frac{4}{125}\!$$ (shown in red)

example on p.7-3 $$y(x)=e^{5x}(4-25*x+*\frac{7}{2}x^2)\!$$ (shown in blue)



Egm4313.s12.team11.imponenti 22:31, 20 February 2012 (UTC)