Given $f''(x) = -4f(x)$ for all $x \in$ $\mathbb{R}$ and $f(0) = f'(0) = 1$, how do I find $f(x)$ Let $f$ be a function so that $f''(x) = -4f(x)$ for all $x \in \mathbb{R}$ and $f(0) = f'(0) = 1$.
i) Prove that $2f(x)\sin(2x)+f'(x)\cos(2x)=1$ for all $x \in \mathbb{R}$
ii) Prove that $2f(x)\cos(2x)-f'(x)\sin(2x)=2$ for all $x \in \mathbb{R}$
iii) Using (i) and (ii), prove that $f(x)=\frac12\sin(2x)+\cos(2x)$ for all $x \in \mathbb{R}$

If we put $t=2x$ and if prime denotes derivative with respect to $t$ then it is easy to see that we have $f''+f=0$ and this has a unique solution $f= f(0)\cos t + f'(0)\sin t$ and note that given conditions imply that $f(0)=1,f'(0)=1/2$ so that $f(x) = \cos 2x +(1/2)\sin 2x$.
