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)`