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

Question

asked Aug 11, 2020 209k views

1 Answer

Maxisme · Answer 1 · 2020-08-16T13:36:31+0000

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

$\implies-2a=-1\implies a=\frac12$

Then the ODE has the general solution

$\boxed{y(t)=C_1e^(2t)++C_2e^(-t)+\frac12e^t}$