How to find the general solution of $u''(a)+u(a)=0$?

Question

When I get to know the solution of $u''(a)+u(a)=0$, I always come with the solution with exponential but to get to the solution $u(a)=A \cos(a-b)$ how should I proceed? And the where this $b$ has come from?

Answer 1

Note that exponential and sines and closely related. One has the following equations: $$\cos x = \dfrac{e^{ix} + e^{-ix}}{2}; \quad \sin x = \dfrac{e^{ix} - e^{-ix}}{2i}.$$

Note that $A\cos(x - b)$ can be decomposed as:

$$A\cos(x - b) = A\cos x\cos b + A\sin x\sin b.$$

Let $C_1 = A\cos b$ and $C_2 = A\sin b$. This then gives the RHS to be

$$C_1\cos x + C_2\sin x.$$

Now, you can convert it to exponential using the above equations.
Moreover, you can start with an exponential solution of the form $C_1e^{ix} + C_2e^{-ix}$ and recover $A$ and $b$.
(Well, you can only recover $b$ up to a multiple of $2\pi$ but that is all that's necessary.)