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I'm given the problem $$y'' + 4y' = t$$ and asked to solve for y. I compute the general solution (using the characteristic equation) to be $$c_1 + c_2e^{-4t}\ ,$$ which I am pretty sure is correct.

Next I compute the particular solution to be $$At + B\ ,$$ as $t$ is a polynomial of degree one. This leaves me with $$0 + 4A = t$$ by substitution back into the initial equation, as $(At + B)'' = 0$ and $(At + B)' = A$. I rearrange and find that $$A = t/4\ ,$$ which doesn't seem like the right answer, and leaves me no apparent way to solve for $B$.

I'm not sure whether I did this wrong or just don't know the next step, but any help would be appreciated.

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You should note that your homogeneous solution $$c_1+c_2e^{-4t}$$ and your attempt at a particular solution $$y_p=At+B$$ have a term in common, namely the constant. So you should multiply by $t$ and try $$y_p=At^2+Bt$$ instead.

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  • $\begingroup$ And suddenly it makes sense. Blind spot, I guess. $\endgroup$ – SquarerootSquirrel May 29 '15 at 2:49
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    $\begingroup$ But it obviously doesn't make sense to some people - I wonder why the downvote? $\endgroup$ – David May 29 '15 at 2:51

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