Find $x(t)$ and $y(t)$ which satisfy the following differential equations

Question

Find $x(t)$ and $y(t)$ which satisfy

$3\dot x + \dot y +5x-y=2e^{-t}+4e^{-3t}$,

$\dot x + 4\dot y -2x+7y=-3e^{-t}+5e^{-3t}$,

subject to $x=y=0$ at $t=0$.

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This is how I tried it:

If we multiply the first equation by $4$ and then substract the second we get the nicer

$\dot x + 2x=e^{-3t}+e^{-t}+y$

Similarly, if we multiply the second equation by $3$ and substract the first, we get

$\dot y + 2y=e^{-3t}-e^{-t}+x$.

If now we try to solve the first one in the usual 'particular solution+general solution' way and substitude in for $\dot y$ from the second, we end up with a new equation for $x$ but it still involves $\dot x$. So it seems like we are just going in circles.

Answer 1

derivative of $$y=x'+2x-e^{-3t}-e^{-t}$$ is

$$y'=x''+2x'+3e^{-3t}+e^{-t}$$

substitute in the second equation to get

$$x''+2x'+3e^{-3t}+e^{-t}+2(x'+2x-e^{-3t}-e^{-t})=e^{-3t}-e^{-t}+x$$ $$x''+4x'+3x=0$$ $$m^2+4m+3=0$$ $$m_1=-1$$ $$m_2=-3$$ so $$x=C_1e^{-t}+C_2e^{-3t}$$ $$x'=-C_1e^{-t}-3C_2e^{-3t}$$

so $$y=-C_1e^{-t}-3C_2e^{-3t}+2(C_1e^{-t}+C_2e^{-3t})-e^{-3t}-e^{-t}$$ $$y=C_1e^{-t}-C_2e^{-3t}-e^{-3t}-e^{-t}$$

Answer 2

Put them in matrix form: let $X(t)=\left( \begin{array} {c} x(t) \\ y(t) \end{array}\right)$; then $\left( \begin{array} {cc} 3 & 1 \\ 1 & 4 \end{array}\right) \dot X (t)+ \left( \begin{array} {cc} 5 & -1 \\ -2 & 7 \end{array}\right) X (t) = \left( \begin{array} {cc} 2 & 4 \\ -3 & 5 \end{array}\right) \left( \begin{array} {c} \mathbb{e}^{-t} \\ \mathbb{e}^{-3t} \end{array}\right)$.