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

Report 3, Problem 1

Problem Statement

Find the solution to the following L2-ODE-CC: $y^{″}-10y^{\prime }+25y=r(x)$

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

And the following initial conditions: $y(0)=4,y^{\prime }(0)=-5$

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

$y_h=c_1{e}^{5x}+c_{2}x{e}^{5x}$

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

$y_{p1}=C{e}^{5x}$

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

$y_{p1}=C{x}^2{e}^{5x}$

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

$y_{p^{″}1}-10y_{p^{\prime }1}+25y_{p1}=r_1(x)$

$(C{x}^2{e}^{5x}{)}^{″}-10(C{x}^2{e}^{5x}{)}^{\prime }+25(C{x}^2{e}^{5x})=r_1(x)$

$25C{x}^2{e}^{5x}+10Cx{e}^{5x}+10Cx{e}^{5x}+2C{e}^{5x}-10(5C{x}^2{e}^{5x}+2Cx{e}^{5x})+25C{x}^2{e}^{5x}=7e^{5x}$

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

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

so $C=\frac{7}{2}$ and the first particular solution is,

$y_{p1}=\frac{7}{2}{x}^2{e}^{5x}$

Second Particular Solution

$r_2(x)=-2x^2$,

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

$y_{p2}=a_2{x}^2+a_{1}x+a_0$

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

$y_{p^{″}2}-10y_{p^{\prime }2}+25y_{p2}=r_2(x)$

$(a_2{x}^2+a_{1}x+a_0{)}^{″}-10(a_2{x}^2+a_{1}x+a_0{)}^{\prime }+25(a_2{x}^2+a_{1}x+a_0)=-2x^2$

$2a_2-10(2a_{2}x+a_1)+25(a_2{x}^2+a_{1}x+a_0)=-2x^2$

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

$25a_2{x}^2+(25a_1-20a_2)x+(25a_0-10a_1+2a_2)=-2x^2$

$\left[\begin{array}{ccc}2& -10& 25\\ -20& 25& 0\\ 25& 0& 0\end{array}\right]\left[\begin{array}{c}{a}_2\\ a_1\\ a_0\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ -2\end{array}\right]$

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

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

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

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

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

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

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

Second initial condition $y^{\prime }(0)=-5$

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

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

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

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

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)

File:Plot3 1.jpg

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