$$y'' + 2y' – 8y = e^{2x} $$
How do I find the particular solution?
I tried setting: $y = Ae^{2x} => y' = 2Ae^{2x} => y''= 4Ae^{2x}$
If I substitute I get: $4Ae^{2x} + 4Ae^{2x} – 8Ae^{2x} = e^{2x} => 0 = e^{2x}$
What am I doing wrong?
[Math] Finding particular solution to $y” + 2y’ – 8y = e^{2x} $
ordinary differential equations
Best Answer
That is because $e^{2x}$ is a solution of the homogeneous equation. You can see this from solving the characteristic equation:
$$r^2+2r-8=0$$
It gives you two solutions $-4$ and $2$, which means $e^{2x}$ is a solution to the homogeneous equation.
So for the particular solution, try $Axe^{2x}$ instead. If it fails, try $Axe^{2x}+Bx^2e^{2x}$, and so on.