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How do I go about solving this question?:

Find the general solution of the non-homogeneous ODE

$y''+\frac 12y'+\frac{1}{16}y=cos(\frac x4)$.

Solving the homogeneous equation, I get the real root $\lambda$ = $\frac{-1}{4}$.

This gives the general solution $y_h=e^{\frac{-x}{4}}(C_1 + C_2x)$.

What do I do next?

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  • $\begingroup$ look at the solution to $y'' = 0.$ see what happens. $\endgroup$ – abel May 4 '15 at 19:26
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yes and the homogeneous equation has the solution $C_1xe^{-1/4x}+C_2e^{-1/4x}$

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  • $\begingroup$ Could you briefly explain why there is an additional x for the constant? I know it does, I just don't know why. $\endgroup$ – Leia Hassan May 4 '15 at 18:11
  • $\begingroup$ the solutions must be linear independent $\endgroup$ – Dr. Sonnhard Graubner May 4 '15 at 18:20
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When the auxiliary equation has a double root $\lambda$ as you have here, the complementary function must be of the form $$(Ax+B)e^{\lambda x}$$ Your next step is to find the particular integral which must be of the form$$y=P\cos(\frac x4)+Q\sin(\frac x4)$$ Substitute this expression into the entire differential equation to find the constants $P$ and $Q$ and then you're done.

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