University of Florida/Egm4313/s12 Report 4, Problem 4.3

Problem 4.3

Problem 4.3 Part 2

Problem Statement

Given: $y^{″}-3y^{\prime }+2y=r(x)$

where $r(x)=\log (1+x)$

With initial conditions: $y(-\frac{3}{4})=1,y^{\prime }(-\frac{3}{4})=0$

Find the overall solution $y_n(x)$ for $n=4,7,11$ and plot these solutions on the interval from $[-\frac{3}{4},3]$

Solution

First we find the homogeneous solution to the ODE:
The characteristic equation is:
${\lambda }^2-3\lambda +2=0$
$(\lambda -2)(\lambda -1)=0$
Then, $\lambda =1,2$
Therefore the homogeneous solution is:
$y_h=C_1{e}^{(}2x)+c_2{e}^x$

Now to find the particulate solution
For n=4

$r(x)=\sum _0^n-\frac{(-1{)}^n{x}^n}{n\ln (10)}$

$r(x)=\frac{x}{\ln (10)}-\frac{x^2}{2\ln (10)}+\frac{x^3}{3\ln (10)}-\frac{x^4}{4\ln (10)}$

We can then use a matrix to organize the known coefficients:

$\left[\begin{array}{ccccc}2& -3& 2& 0& 0\\ 0& 2& -6& 6& 0\\ 0& 0& 2& -9& 12\\ 0& 0& 0& 2& -12\\ 0& 0& 0& 0& 2\end{array}\right]\left[\begin{array}{c}{K}_0\\ K_1\\ K_2\\ K_3\\ K_4\end{array}\right]=\left[\begin{array}{c}0\\ \frac{1}{ln(10)}\\ \frac{1}{2ln(10)}\\ \frac{1}{3ln(10)}\\ \frac{1}{4ln(10)}\end{array}\right]$

Then, using MATLAB and the backlash operator we can solve for these unknowns:
Therefore
$y_{p4}=4.0444+3.7458x+1.5743x^2+0.3981x^3+0.0543x^4$

Superposing the homogeneous and particulate solution we get
$y_n=4.0444+3.7458x+1.5743x^2+0.3981x^3+0.0543x^4+C_1{e}^{2x}+C_2{e}^x$

Differentiating:
$y{\text{'}}_n=3.7458+3.1486x+1.1943x^2+0.2172x^3+2C_1{e}^{2x}+C_2{e}^x$ Evaluating at the initial conditions:
$y(-0.75)=0.9698261719+0.231301601C_{!}+0.4723665527C_2=1$
$y^{\prime }(-0.75)=1.9645125+0.4462603203C_1+0.4723665527C_2$

We obtain:
$C_1=-4.46$
$C_2=0.055$

Finally we have:
$y_n=4.0444+3.7458x+1.5743x^2+0.3981x^3+0.0543x^4-4.46e^{2x}+0.055e^x$

For n=7

$r(x)=\sum _0^n-\frac{(-1{)}^n{x}^n}{n\ln (10)}$

$r(x)=\frac{x}{\ln (10)}-\frac{x^2}{2\ln (10)}+\frac{x^3}{3\ln (10)}-\frac{x^4}{4\ln (10)}+\frac{x^5}{5\ln (10)}-\frac{x^6}{6\ln (10)}+\frac{x^7}{7\ln (10)}$

We can then use a matrix to organize the known coefficients:

$\left[\begin{array}{cccccccc}2& -3& 2& 0& 0& 0& 0& 0\\ 0& 2& -6& 6& 0& 0& 0& 0\\ 0& 0& 2& -9& 12& 0& 0& 0\\ 0& 0& 0& 2& -12& 20& 0& 0\\ 0& 0& 0& 0& 2& -15& 30& 0\\ 0& 0& 0& 0& 0& 2& -18& 42\\ 0& 0& 0& 0& 0& 0& 2& -21\\ 0& 0& 0& 0& 0& 0& 0& 2\end{array}\right]\left[\begin{array}{c}{K}_0\\ K_1\\ K_2\\ K_3\\ K_4\\ K_5\\ K_6\\ K_7\end{array}\right]=\left[\begin{array}{c}0\\ \frac{1}{ln(10)}\\ \frac{1}{2ln(10)}\\ \frac{1}{3ln(10)}\\ \frac{1}{4ln(10)}\\ \frac{1}{5ln(10)}\\ \frac{1}{6ln(10)}\\ \frac{1}{7ln(10)}\end{array}\right]$

Then, using MATLAB and the backlash operator we can solve for these unknowns:
Therefore
$y_{p7}=377.4833+375.3933x+185.6066x^2+60.5479x^3+14.4946x^4+2.6492x^5+0.3619x^6+0.0310x^7$

Superposing the homogeneous and particulate solution we get
$y_n=377.4833+375.3933x+185.6066x^2+60.5479x^3+14.4946x^4+2.6492x^5+0.3619x^6+0.0310x^7+C_1{e}^{(}2x)+c_2{e}^x$

Differentiating:
$y{\text{'}}_n=375.3933+371.213x+181.644x^2+57.9784x^3+13.46x^4+2.1714x^5+0.214x^6+2C_1{e}^{(}2x)+C_2{e}^x$ Evaluating at the initial conditions:
$y(-0.75)=178.816+0.2231301601C_{!}+0.4723665527C_2=1$
$y^{\prime }(-0.75)=178.413+0.4462603203C_1+0.4723665527C_2$

We obtain:
$C_1=-2.6757$
$C_2=-375.173$