Finding the general solution of the given differential equation

[bigu01]

y"+2y'+y=2e^-t
I tried to find the solution for this nonhomogenous diff. Equation but i could not. First i took a function Y(t)=Ae^-t but i was getting 0=2e^-t.
To get rid of that i took another y'+y=2e^-t and found the solution y=2te^-t + ce^-t. Noticed that first part of this finding is solution of my nonhomogenous diff equation so i took another function Y(t)=Ate^-t but then again i am finding the same answer, i need some tips on how to continue. The general solution for this equation as homogenous equation has repeated roots.

 

[ShayanJ]

When the characteristic equation has repeated roots,the answer is $Ae^{qt}+Bte^{qt}$.
For finding the particular solution,take $y_p=f(t)e^{-t}$ and try finding f(t).

 

[CAF123]

... Noticed that first part of this finding is solution of my nonhomogenous diff equation so i took another function Y(t)=Ate^-t but then again i am finding the same answer...

The homogeneous differential equation is second order and so the general solution is composed of two linearly independent pieces. One of those pieces is precisely of the form Y(t) = Ate^-t, so subbing this in as a particular solution is redundant. You have to look for another form.

 

[bigu01]

I multiplied my function by t^2, since by multiplying only by t was not working, and it worked.