Solve autonomous equation

Question

$ y''' - y = e^{2t} $

Actually I need only homogeneous solution of $ y''' - y = 0. $

I can clearly see that it is $ e^x, $ but how to prove it?

I tried with $ \frac{dy}{dt} = v, $ but could conclude nothing.

Answer 1

the roots are $\frac{-1\pm i\sqrt{3}}{2}$. From the first one $-\frac{1}{2}+i\frac{\sqrt{3}}{2}$

you have also the functions $e^{-x/2}\cos \frac{\sqrt{3}x}{2}$ and $e^{-x/2}\sin\frac{\sqrt{3}x}{2}$

Therefore the solution to the homogeneous equation is $y_h=A e^x +Be^{-x/2}\cos \frac{\sqrt{3}x}{2}+Ce^{-x/2}\sin\frac{\sqrt{3}x}{2} $