# Find a particular solution of $y''-4y = 2e^{2x}$

If I wanted to use the method of undetermined coefficients, why would a guess of $y_p = Ae^{2x}$ be incorrect? In general, how can I avoid choosing incorrect guesses which initially seem to be correct?

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Your $y_p$ is the solution to the homogeneous equation $y''-4y=0$. –  Joe Johnson 126 Nov 4 '12 at 23:30
So how can I change my guess to work around this? –  user1038665 Nov 4 '12 at 23:32
Standard here is $Axe^{2x}$. –  André Nicolas Nov 4 '12 at 23:34

You need to take into consideration the characteristic polynomial of the homogeneous differential equation. In your case, the polynomial is $$p(\lambda) = \lambda^2 - 4 = 0.$$ The roots are $\lambda = \pm 2$. Since in your right hand side, you have the expression $Ce^{2x}$, and $2$ is a root of the characteristic equation, you need to modify your guess to be $y_p = Axe^{2x}$. More details can be found here.