Question

Solve this system of ODEs by eliminating y.

Answer 1

$\begin{cases}{\mathrm D}^2x-{\mathrm D}y=e^{2t}-6t\\{\mathrm D}x+{\mathrm D}y=y-x+3e^{2t}\end{cases}$

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

$y=\dfrac9{10}e^{2t}+3t^2+6t+C_1\cos t+C_2\sin t+C_3$