# Second Order Cosine Differential Equation

Can one think of a solution to:

$$\lambda f''(x)=f(x)\cos x$$

s.t. $f(0)=0$ and $f(\frac \pi 2)=0$, $\lambda>0$?

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The solutions are special functions known as the Mathieu functions. The more general Mathieu equation is: $$f''(y)+(a-2q\cos(2y))f(y)=0$$ So in your case $a=0,\ q=1/2\lambda, \ x = y/2$, and the solution is: $$C_1 \text{MathieuCos}(0,\ 2/\lambda,\ x/2) + C_2 \text{MathieuSin}(0,\ 2/\lambda,\ x/2)$$ If $y$ is very very small you can approximate $\cos(y)\sim 1$, and get the usual harmonic solutions. This leads to their computation as a Fourier series, an example of which can be found here.

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Thanks!! I understand it cannot be expressed in a closed form, but can it be expressed as an integral or sum? –  Troy McClure Jan 9 '13 at 19:39
@TroyMcClure - not really, the're notoriously difficult to express. A Fourier Series exists, but I don't know about a closed form expression for a power series.You can search here for more, but there's not very much: dlmf.nist.gov/28 –  nbubis Jan 9 '13 at 19:49