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I'm trying to solve the differential equation $x'' + 4x = 4t$, endpoint value $x(0) = x(1) = 0$, and ask me how to derive the solution $x(t) = t - \sin2t/\sin2$.

The answer says that since the particular solution $t$ is easily found, the general solution is $A\cos2t + B\sin2t + t$.

But I think that due to the characteristic equation being $r^2 - 4 = 0$ the general solution should be $Ae^{2t} + Be^{-2t} + t$.

Why can this be written in terms of trigonometric functions?

Thanks in advance.

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up vote 2 down vote accepted

Here, the characteristic equation is $r^2 + 4=0$.

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Oh, how stupid am I. Thx! – Liang-Yu Pan Jan 15 '13 at 11:08

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