2nd Order Non-Homogeneous Initial Value Equation with real repeated roots $x''-2x'+x=e^{2t}$

Question

So, I was trying to the question

$$x''-2x'+x=e^{2t} , x(0) = 0 , x'(0) = 1$$

I worked the roots out as +1 repeated and found the general solution to be

$x(t)= (c1)e^{t} + (c2)te^{t}$

The particular solution I tried was

$x_p = e^{at}$

$x'_p = ae^{at}$

$x''_p = (a^2)e^{at}$

I subbed these into the initial equation to get

$(a^2)e^{at} - 2ae^{at} + e^{at} = e^{2t}$

The problem is that I don't know how to carry on. I missed a lot of my college year due to being ill and I am currently trying to self teach the topic for my upcoming august exams since i missed the summer ones.

Any help would be appreciated! ( Also, this is my first post so, apologies for the formatting errors ! )

Answer 1

This is another method for additional purposes: $$x''-2x'+x=e^{2t}$$ $$(e^{-t}x''-e^{-t}x')-(e^{-t}x'-e^{-t}x)=e^{t}$$ $$(e^{-t}x')'-(e^{-t}x)'=e^{t}$$ $$e^{-t}x'-e^{-t}x=e^{t}+C_1$$ $$(e^{-t}x)'=e^{t}+C_1$$ $$e^{-t}x=e^{t}+C_1t+C_2$$ $$x=e^{2t}+C_1te^{t}+C_2e^{t}$$