Science:Math Exam Resources/Courses/MATH215/December 2011/Question 06 (c)/Solution 1

As there are constants and exponentials in this first-order linear ODE, we will use undetermined coefficients, postulating that the particular solution is

$x_p(t)=\left(\begin{array}{c}{B}_1\\ B_2\end{array}\right)e^{2t}+\left(\begin{array}{c}{C}_1\\ C_2\end{array}\right).$

so that

${x_p}^{\prime }(t)=\left(\begin{array}{c}2B_1\\ 2B_2\end{array}\right)e^{2t}.$

Substituting our solution into the system we get,

$\begin{array}{rl}& Ax+\left(\begin{array}{c}-2e^{2t}\\ 1\end{array}\right)\\ & =\left(\begin{array}{cc}3& 1\\ -4& -1\end{array}\right)\left(\begin{array}{c}{B}_1{e}^{2t}+C_1\\ B_2{e}^{2t}+C_2\end{array}\right)+\left(\begin{array}{c}-2e^{2t}\\ 1\end{array}\right)\\ & =\left(\begin{array}{c}3B_1+B_2-2\\ -4B_1-B_2\end{array}\right)e^{2t}+\left(\begin{array}{c}3C_1+C_2\\ -4C_1-C_2+1\end{array}\right).\end{array}$

This must equal ${\displaystyle }{x_p}^{\prime }$ and so we have

$\begin{array}{rl}\left(\begin{array}{c}3B_1+B_2-2\\ -4B_1-B_2\end{array}\right)e^{2t}+\left(\begin{array}{c}3C_1+C_2\\ -4C_1-C_2+1\end{array}\right)& =\left(\begin{array}{c}2B_1\\ 2B_2\end{array}\right)e^{2t}\\ \left(\begin{array}{c}{B}_1+B_2-2\\ -4B_1-3B_2\end{array}\right)e^{2t}+\left(\begin{array}{c}3C_1+C_2\\ -4C_1-C_2+1\end{array}\right)& =\mathbf{𝟎}\end{array}$

This gives us the system of equations

$\begin{array}{rl}{B}_1+B_2-2& =0\\ -4B_1-3B_2& =0\end{array}$

(based on the $e^{2t}$ terms)

and

$\begin{array}{rl}3C_1+C_2& =0\\ -4C_1-C_2& =-1\end{array}$

(based on comparing the constant vectors).

Looking at the first equation of the first system of equations, we have $B_1=2-B_2$ so that in the second equation,

$\begin{array}{rl}-4(2-B_2)-3B_2& =0\\ B_2& =8.\end{array}$

Using the first equation again, we get

${\displaystyle }{B}_1=-6.$

Looking at the first equation of the second system of equations, we have $C_2=-3C_1$. From the second equation we get

$\begin{array}{rl}-4C_1-(-3C_1)& =-1\\ C_1& =1.\end{array}$

and hence

${\displaystyle }{C}_2=-3$

after substituting back into the first equation.

The general solution to the ODE system will be $u_p+u_h$ where $u_h$ is our homogeneous solution from part (b).

Thus,

$x(t)=\left(\begin{array}{c}1\\ -3\end{array}\right)+\left(\begin{array}{c}-6\\ 8\end{array}\right)e^{2t}+\left(\begin{array}{c}ct{e}^t+b{e}^t\\ -2ct{e}^t+(c-2b)e^t\end{array}\right).$

If

${\displaystyle }x(0)=(1,-5{)}^T$

then

$\left(\begin{array}{c}1\\ -5\end{array}\right)=\left(\begin{array}{c}1-6+b\\ -3+8+(c-2b))\end{array}\right).$

The first component tells us that $b=6$ while using this value of b in the second component gives $c=2$.

Therefore, the solution is

$x(t)=\left(\begin{array}{c}1\\ -3\end{array}\right)+\left(\begin{array}{c}-6\\ 8\end{array}\right)e^{2t}+\left(\begin{array}{c}2t{e}^t+6e^t\\ -4t{e}^t-10e^t\end{array}\right).$