Question

Find the general solution to the equation y" – y' – 2y=-et

Answer 1

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

The corresponding homogeneous ODE is

$y''-y'-2y=0$

with characteristic equation

$r^2-r-2=(r-2)(r+1)=0$

with roots at $r=2$ and $r=-1$, so the characteristic solution is

$y_c=C_1e^{2t}+C_2e^{-t}$

For the non-homogeneous ODE, assume a particular solution of the form

$y_p=ae^t$

$\implies{y_p}'=ae^t$

$\implies{y_p}''=ae^t$

Substituting $y_p$ and its derivatives into the ODE gives

$ae^t-ae^t-2ae^t=-e^t$

$-2ae^t=-e^t$

$\implies-2a=-1\implies a=\dfrac12$

Then the ODE has the general solution

$\boxed{y(t)=C_1e^{2t}++C_2e^{-t}+\dfrac12e^t}$