\(\begin{cases}{\mathrm D}^2x-{\mathrm D}y=e^{2t}-6t\\{\mathrm D}x+{\mathrm D}y=y-x+3e^{2t}\end{cases}\)
Adding the two ODEs gives
\({\mathrm D}^2x+{\mathrm D}x=y-x+4e^{2t}-6t\)
Differentiating once gives
\({\mathrm D}^3x+{\mathrm D}^2x={\mathrm D}y-{\mathrm D}x+8e^{2t}-6\)
\({\mathrm D}^3x+{\mathrm D}^2x-{\mathrm D}y+{\mathrm D}x=8e^{2t}-6\)
The first ODE lets you simplify this a bit:
\({\mathrm D}^3x+e^{2t}-6t+{\mathrm D}x=8e^{2t}-6\)
\({\mathrm D}^3x+{\mathrm D}x=7e^{2t}+6t-6\)
Let \(z={\mathrm D}x\), so that \({\mathrm D}^2z={\mathrm D}^3x\), and reduce the order of the ODE to get
\({\mathrm D}^2z+z=7e^{2t}+6t-6\)
Solving this ODE for \(z\) shouldn't be a problem for you; you would find that
\(z=C_1\cos t+C_2\sin t+\dfrac75e^{2t}+6t-6\)
Since \(z={\mathrm D}x\), integrating once with respect to \(t\) gives the general solution for \(x\):
\(x=C_1\sin t-C_2\cos t+\dfrac7{10}e^{2t}+3t^2-6t+C_3\)
Now plug in the second derivative of this solution into the first ODE to find another ODE in \(y\) alone.
\({\mathrm D}^2x=-C_1\sin t+C_2\cos t+\dfrac{14}5e^{2t}+6\)
\(-C_1\sin t+C_2\cos t+\dfrac{14}5e^{2t}+6-{\mathrm D}y=e^{2t}-6t\)
\({\mathrm D}y=\dfrac95e^{2t}+6t+6-C_1\sin t+C_2\cos t\)
Integrate with respect to \(t\) and you get
\(y=\dfrac9{10}e^{2t}+3t^2+6t+C_1\cos t+C_2\sin t+C_3\)
