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 ${\vec {x}}_{p}={\begin{bmatrix}ate^{3t}+be^{3t}\\cte^{3t}+de^{3t}\end{bmatrix}}$ then ${\vec {x_{p}}}'={\begin{bmatrix}(a+3b)e^{3t}+3ate^{3t}\\(c+3d)e^{3t}+3cte^{3t}\end{bmatrix}}$ and

$Ax_{p}+{\begin{bmatrix}25e^{3t}\\0\end{bmatrix}}={\begin{bmatrix}3cte^{3t}+(3d+25)e^{3t}\\(2a+c)te^{3t}+(2b+d)e^{3t}\end{bmatrix}}$

Therefore if ${\vec {x_{p}}}'=A{\vec {x_{p}}}+{\begin{bmatrix}25e^{3t}\\0\end{bmatrix}}$ we get

${\begin{bmatrix}(a+3b)e^{3t}+3ate^{3t}\\(c+3d)e^{3t}+3cte^{3t}\end{bmatrix}}={\begin{bmatrix}3cte^{3t}+(3d+25)e^{3t}\\(2a+c)te^{3t}+(2b+d)e^{3t}\end{bmatrix}}$.

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

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

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

${\begin{bmatrix}1&3&0&-3\\1&0&-1&0\\0&2&-1&-2\\1&0&-1&0\end{bmatrix}}{\begin{bmatrix}a\\b\\c\\d\end{bmatrix}}={\begin{bmatrix}25\\0\\0\\0\end{bmatrix}}$

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

${\begin{bmatrix}1&1&1&-1\\0&-3&-1&3\\0&2&-1&-2\\0&0&0&0\end{bmatrix}}{\begin{bmatrix}a\\b\\c\\d\end{bmatrix}}={\begin{bmatrix}25\\-25\\0\\0\end{bmatrix}}$

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

${\begin{bmatrix}1&1&1&-1\\0&-3&-1&3\\0&0&-5&0\\0&0&0&0\end{bmatrix}}{\begin{bmatrix}a\\b\\c\\d\end{bmatrix}}={\begin{bmatrix}25\\-25\\-50\\0\end{bmatrix}}$

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

${\begin{bmatrix}1&1&0&-1\\0&-3&0&3\\0&0&1&0\\0&0&0&0\end{bmatrix}}{\begin{bmatrix}a\\b\\c\\d\end{bmatrix}}={\begin{bmatrix}15\\-15\\10\\0\end{bmatrix}}$

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

${\begin{bmatrix}1&0&0&0\\0&1&0&-1\\0&0&1&0\\0&0&0&0\end{bmatrix}}{\begin{bmatrix}a\\b\\c\\d\end{bmatrix}}={\begin{bmatrix}10\\5\\10\\0\end{bmatrix}}$

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

${\vec {x_{p}}}={\begin{bmatrix}10te^{3t}+5e^{3t}\\10te^{3t}\end{bmatrix}}$.

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

$x(t)=C_{1}{\begin{bmatrix}1\\1\end{bmatrix}}e^{3t}+C_{2}{\begin{bmatrix}3\\-2\end{bmatrix}}e^{-2t}+{\begin{bmatrix}10te^{3t}+5e^{3t}\\10te^{3t}\end{bmatrix}}$.