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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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1. This looks like an ODE, not a PDE (there's only one independent variable). 2. What are the boundary conditions that you're referring to in the title? –  Hans Lundmark Jun 11 '12 at 19:36
    
Yes, thanks, it's ODE –  Lilly Jun 22 '12 at 16:41

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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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Maple 16 comes up with the following general solution for the transformed equation: $$u \left( t \right) ={{\rm e}^{t}}{\it HeunC} \left( 4,-1/2,-1/2,-2,3/8 ,1/2+1/2\,t \right) {c_1}+{{\rm e}^{t}}{\it HeunC} \left( 4,1/2,- 1/2,-2,3/8,1/2+1/2\,t \right) \sqrt {2+2\,t}{c_2}+{{\rm e}^{t-1}} $$ Note in particular that $u = e^{t-1}$ is a solution. –  Robert Israel Jun 11 '12 at 22:43
    
And thus $u = e^{\cos(x)-1}$ is a solution of the original differential equation. –  Robert Israel Jun 11 '12 at 22:45

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