Could you help me with a question? I get stuck at ii),
Define the function
$$I(x):=\frac{1}{\pi} \int^\pi_0 \cos(x\sin\theta) d\theta$$
i) Via application of Leibniz rule (or otherwise) calculate $I'$ and $I''$.
ii) Thus, determine non-zero value(s) of $k$ for which $I$ will be a solution to the differential equation
$$k^2x^2I''+xI'+x^2I=0$$
iii) Write down the values of $I(0)$ and $I'(0)$.
So far what I have is $$I'(x)=\frac{1}{\pi} \int^\pi_0 -\sin\theta \sin(x\sin\theta) d\theta$$ $$I''(x)=\frac{1}{\pi}\int^\pi_0 -\sin^2\theta \cos(x\sin\theta) d\theta$$
No idea where to go from there :S
You only have to integrate $I^{\prime}(x)$ by parts to get there. $$\begin{align}I^{\prime}(x)&=\frac1{\pi}\int_0^{\pi}\left(\frac d{d\theta}\cos\theta\right)\sin(x\sin\theta)d\theta\\ &=\frac1{\pi}\left[(\cos\theta)\sin(x\sin\theta)\right]_0^{\pi}-\frac1{\pi}\int_0^{\pi}\cos\theta\cos(x\sin\theta)\cdot x\cos\theta\,d\theta\\ &=-\frac x{\pi}\int_0^{\pi}\cos^2\theta\cos(x\sin\theta)d\theta\end{align}$$ And now you're all set: $$k^2x^2I^{\prime\prime}+xI^{\prime}+x^2I=\frac{x^2}{\pi}\int_0^{\pi}\left[-k^2\sin^2\theta-\cos^2\theta+1\right]\cos(x\sin\theta)d\theta=0$$ So we want $$-k^2\sin^2\theta-\cos^2\theta+1=(1-k^2)\sin^2\theta=0$$ Thus $k=\pm1$. The last part is straightforward $$I(0)=\frac1{\pi}\int_0^{\pi}1\cdot d\theta=1$$ $$I^{\prime}(0)=\frac1{\pi}\int_0^{\pi}0\cdot d\theta=0$$