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Help me please to solve this problem: $u_{xx}+(\cos x+\cos^{2} x)u=e^{\cos x - 1}$

Thanks a lot!

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    Yes, thanks, it's ODE2012-06-22

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The starting point could be changing the dependant variable: $\cos x=t$ $\frac{du}{dx}=\frac{d(\cos x)}{dx}\frac{du}{d(\cos x)}=-\sin x\frac{du}{d(\cos x)}$ $\begin{aligned}\frac{d}{dx}\left(\frac{du}{dx}\right) & =-\cos x\frac{du}{d(\cos x)}-\sin x\frac{d(\cos x)}{dx}\frac{d^{2}u}{d(\cos x)^{2}}=-\cos x\frac{du}{d(\cos x)}+\sin^{2}x\frac{d^{2}u}{d(\cos x)^{2}}\\ & =-\cos x\frac{du}{d(\cos x)}+(1-\cos^{2}x)\frac{d^{2}u}{d(\cos x)^{2}}=-t\frac{du}{dt}+(1-t^{2})\frac{d^{2}u}{dt^{2}} \end{aligned}$ $(1-t^{2})\frac{d^{2}u}{dt^{2}}-t\frac{du}{dt}+(t+t^{2})u=e^{t-1}$ This is a second-order linear inhomogeneous equation with regular singular points at $t=\pm 1$. It does not look immediately familiar, so perhaps Frobenius method could lead to a solution for the homogeneous part and Green's function (as long as you state the boundary conditions) could be invoked for a particular solution.

EDIT: The original equation without RHS is actually a Hill equation

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    And thus $u = e^{\cos(x)-1}$ is a solution of the original differential equation.2012-06-11