Question

Find a particular solution to the differential equation y′′+2y′+y=2t2−t+3e−2t. y′′+2y′+y=2t2−t+3e−2t.

Answer 1

$y''+2y'+y=2t^2-t+3e^{-2t}$

The characteristic equation is

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

so the characteristic solution is

$y_c=C_1e^{-t}+C_2te^{-t}$

For the particular solution, we can try looking for a solution of the form

$y_p=at^2+bt+c+de^{-2t}$
$\implies{y_p}'=2at+b-2de^{-2t}$
$\implies{y_p}''=2a+4de^{-2t}$

Substituting into the ODE, we have

$(2a+4de^{-2t})+2(2at+b-2de^{-2t})+(at^2+bt+c+de^{-2t})=2t^2-t+3e^{-2t}$
$at^2+(4a+b)t+(2a+2b+c)+de^{-2t}=2t^2-t+3e^{-2t}$
$\implies\begin{cases}a=2\\4a+b=-1\\2a+2b+c=0\\d=3\end{cases}\implies a=2,b=-9,c=14,d=3$

So the general solution to the ODE is

$y=y_c+y_p$
$y=C_1e^{-t}+C_2te^{-t}+2t^2-9t+14+3e^{-2t}$