# 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.