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

$x_{p}'(t)=\left({\begin{array}{c}2B_{1}\\2B_{2}\end{array}}\right)e^{2t}.$

Substituting our solution into the system we get,

${\begin{aligned}&Ax+{\begin{pmatrix}-2e^{2t}\\1\end{pmatrix}}\\&={\begin{pmatrix}3&1\\-4&-1\end{pmatrix}}{\begin{pmatrix}B_{1}e^{2t}+C_{1}\\B_{2}e^{2t}+C_{2}\end{pmatrix}}+{\begin{pmatrix}-2e^{2t}\\1\end{pmatrix}}\\&={\begin{pmatrix}3B_{1}+B_{2}-2\\-4B_{1}-B_{2}\end{pmatrix}}e^{2t}+{\begin{pmatrix}3C_{1}+C_{2}\\-4C_{1}-C_{2}+1\end{pmatrix}}.\end{aligned}}$

This must equal $\displaystyle {}x_{p}'$ and so we have

${\begin{aligned}{\begin{pmatrix}3B_{1}+B_{2}-2\\-4B_{1}-B_{2}\end{pmatrix}}e^{2t}+{\begin{pmatrix}3C_{1}+C_{2}\\-4C_{1}-C_{2}+1\end{pmatrix}}&={\begin{pmatrix}2B_{1}\\2B_{2}\end{pmatrix}}e^{2t}\\{\begin{pmatrix}B_{1}+B_{2}-2\\-4B_{1}-3B_{2}\end{pmatrix}}e^{2t}+{\begin{pmatrix}3C_{1}+C_{2}\\-4C_{1}-C_{2}+1\end{pmatrix}}&=\mathbf {0} \end{aligned}}$

This gives us the system of equations

${\begin{aligned}B_{1}+B_{2}-2&=0\\-4B_{1}-3B_{2}&=0\end{aligned}}$

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

and

${\begin{aligned}3C_{1}+C_{2}&=0\\-4C_{1}-C_{2}&=-1\end{aligned}}$

(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,

${\begin{aligned}-4(2-B_{2})-3B_{2}&=0\\B_{2}&=8.\end{aligned}}$

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

${\begin{aligned}-4C_{1}-(-3C_{1})&=-1\\C_{1}&=1.\end{aligned}}$

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,

$x(t)={\begin{pmatrix}1\\-3\end{pmatrix}}+{\begin{pmatrix}-6\\8\end{pmatrix}}e^{2t}+{\begin{pmatrix}cte^{t}+be^{t}\\-2cte^{t}+(c-2b)e^{t}\end{pmatrix}}.$

If

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

then

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

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

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