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Higher order of derivatives: variation of parameters?

$$y'''+y' = \tan t, \quad 0 < t< \pi.$$

Use variation of parameters to determine the general solution of the given differential equation.

I know the roots of the characteristic equation are $0$, $-i$, and $i$. But I'm really confused about how to approach the question. I tried doing the Wronskian but because it has a double root, it gave me a weird answer. Can someone please help me?

By the way, the answer of the question is $$y= c_1 + c_2 \cos t + c_3 \sin t - \ln \cos t - \sin t \ln(\sec t + \tan t).$$

share|cite|improve this question look at page 9, it provides the answer and how to go about it. – diimension Nov 19 '12 at 9:21

First, let $f = y'$ to transform the equation into a second order equation $$ f'' + f = \tan(t). $$ The characteristic equation is $\lambda^2 + 1 = 0$ with roots $\lambda = \pm i$. The Wronskian will be $$ \left| \begin{array}{cc} \sin(t) & \cos(t) \\ \cos(t) & -\sin(t) \end{array} \right| = -1. $$ Solve this equation using variation of parameters to find the general form of $f$ and then integrate to find $y$.

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