Science:Math Exam Resources/Courses/MATH215/December 2013/Question 05 (b)/Solution 1

We need a particular solution. To obtain it, we will use the method of undetermined coefficients.

If ${\overrightarrow{x}}_p=\left[\begin{array}{c}at{e}^{3t}+b{e}^{3t}\\ ct{e}^{3t}+d{e}^{3t}\end{array}\right]$ then ${\overrightarrow{x_p}}^{\prime }=\left[\begin{array}{c}(a+3b)e^{3t}+3at{e}^{3t}\\ (c+3d)e^{3t}+3ct{e}^{3t}\end{array}\right]$ and

$A{x}_p+\left[\begin{array}{c}25e^{3t}\\ 0\end{array}\right]=\left[\begin{array}{c}3ct{e}^{3t}+(3d+25)e^{3t}\\ (2a+c)t{e}^{3t}+(2b+d)e^{3t}\end{array}\right]$

Therefore if ${\overrightarrow{x_p}}^{\prime }=A\overrightarrow{x_p}+\left[\begin{array}{c}25e^{3t}\\ 0\end{array}\right]$ we get

$\left[\begin{array}{c}(a+3b)e^{3t}+3at{e}^{3t}\\ (c+3d)e^{3t}+3ct{e}^{3t}\end{array}\right]=\left[\begin{array}{c}3ct{e}^{3t}+(3d+25)e^{3t}\\ (2a+c)t{e}^{3t}+(2b+d)e^{3t}\end{array}\right]$.

In comparing the coefficients of $e^{3t}$ from row 1, the coefficients of $t{e}^{3t}$ from row 1, the coefficients of $e^{3t}$ from row 2, and the coefficients of $t{e}^{3t}$ from row 2, the following equations must hold:

$\begin{array}{rl}a+3b& =3d+25\\ 3a& =3c\\ c+3d& =2b+d\\ 3c& =2a+c\end{array}$

We can turn this into a matrix equation and row-reduce:

$\left[\begin{array}{cccc}1& 3& 0& -3\\ 1& 0& -1& 0\\ 0& 2& -1& -2\\ 1& 0& -1& 0\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]=\left[\begin{array}{c}25\\ 0\\ 0\\ 0\end{array}\right]$

Subtracting row 2 from row 4 and row 1 from row 2 yields:

$\left[\begin{array}{cccc}1& 1& 1& -1\\ 0& -3& -1& 3\\ 0& 2& -1& -2\\ 0& 0& 0& 0\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]=\left[\begin{array}{c}25\\ -25\\ 0\\ 0\end{array}\right]$

We next multiply the third row by 3 and then add the second row twice:

$\left[\begin{array}{cccc}1& 1& 1& -1\\ 0& -3& -1& 3\\ 0& 0& -5& 0\\ 0& 0& 0& 0\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]=\left[\begin{array}{c}25\\ -25\\ -50\\ 0\end{array}\right]$

Next, divide the third row by -5 and then add the result to the second row, and subtract it from the first row:

$\left[\begin{array}{cccc}1& 1& 0& -1\\ 0& -3& 0& 3\\ 0& 0& 1& 0\\ 0& 0& 0& 0\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]=\left[\begin{array}{c}15\\ -15\\ 10\\ 0\end{array}\right]$

In our last step we divide the second row by -3 and then subtract the result from the first row:

$\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& -1\\ 0& 0& 1& 0\\ 0& 0& 0& 0\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]=\left[\begin{array}{c}10\\ 5\\ 10\\ 0\end{array}\right]$

We see that d is a free parameter, and since we only need one particular we choose d=0 for simplicity. Then we read off a = 10, b - d = 5, c = 10. Putting this together we find a particular solution

$\overrightarrow{x_p}=\left[\begin{array}{c}10t{e}^{3t}+5e^{3t}\\ 10t{e}^{3t}\end{array}\right]$.

The general solution is the homogeneous solution (which we found in part (a)) plus a particular solution:

$x(t)=C_1\left[\begin{array}{c}1\\ 1\end{array}\right]e^{3t}+C_2\left[\begin{array}{c}3\\ -2\end{array}\right]e^{-2t}+\left[\begin{array}{c}10t{e}^{3t}+5e^{3t}\\ 10t{e}^{3t}\end{array}\right]$.