Problem source

Given that 
\[
\frac{\mathrm{d}x}{\mathrm{d}t}=4(x-y)\qquad\mbox{ and }\qquad\frac{\mathrm{d}y}{\mathrm{d}t}=x-12(\mathrm{e}^{2t}+\mathrm{e}^{-2t}),
\]
obtain a differential equation for $x$ which does not contain $y$.
Hence, or otherwise, find $x$ and $y$ in terms of $t$ given that $x=y=0$ when $t=0$.

Solution source

\begin{align*}
&& \frac{\d x}{\d t} &= 4(x-y) \\
&& \frac{\d y}{\d t} &= x - 12(e^{2t}+e^{-2t}) \\
\Rightarrow && \frac{\d^2 x}{\d t^2} &= 4 \frac{\d x}{\d t}-4\frac{\d y}{\d t} \\
&&&=  4 \frac{\d x}{\d t}-4 \left ( x - 12(e^{2t}+e^{-2t}) \right) \\ 
\Rightarrow && \frac{\d^2 x}{\d t^2} -   4 \frac{\d x}{\d t}+4x &= 48 (e^{2t}+e^{-2t})
\end{align*}

This differential equation has characteristic polynomial $\lambda^2 - 4\lambda + 4 = (\lambda-2)^2$. Therefore we should expect a general solution of $(At+B)e^{2t}$. 

For particular integrals we should try $ke^{-2t}$ and $Ct^2 e^{2t}$.

For the former, we have:

\begin{align*}
&& 48 &= 4k+8k+k \\ 
\Rightarrow && k &= \frac{48}{13}
\end{align*}
For the latter we have:

\begin{align*}
&&4Ct^2e^{2t} -4C(2te^{2t}+2t^2e^{2t})+2C((1+2t)e^{2t}+2t^2e^{2t}) &= 48e^{2t} \\
\Rightarrow && 2C &= 48 \\
\Rightarrow && C &= 24
\end{align*}

Therefore the solution should be:

\begin{align*}
x = (At+B)e^{2t} + \frac{48}{13}e^{-2t} + 24t^2 e^{2t} \\
x(0) = B + \frac{48}{13} \\
x'(0) = 2B+A-\frac{96}{13} \\
x =\frac{48}{13}((4t-1)e^{2t}+e^{-2t})+24t^2e^{2t} \\
y = x - \frac{1}{4} \frac{\d x}{\d t} 
\end{align*}