Proof that if $ f''(x) + Cf(x) = 0$ then $f(x) = Acos(\sqrt{C}x) + Bsin(\sqrt{C}x)$ [solved] The actual task is to proof:
$f''(x) + 25f(x) = 0$ then $f(x) = A\cos(5x) + B\sin(5x)$ for $A, B \in \mathbb{R}$
I tried to proof that for $g(x) = A\cos(5x)$ and $h(x) = B\sin(5x)$, but I can't get anywhere with that. 
Regards,
Marco
 A: Consider the initial-value problem
$$
f'' + Cf = 0,\qquad
f(0) = A,\quad f'(0) = B\sqrt{C}.
\tag{1}
$$
Here's a sketch (parts of which you already know) of a proof that the unique solution of (1) is
$$
f(x) = A\cos(\sqrt{C}x) + B\sin(\sqrt{C}x).
$$
Throughout, "the ODE" refers to the differential equation $f'' + Cf = 0$ in (1).


*

*If $f$ and $g$ are solutions of the ODE, so is every linear combination of $f$ and $g$.

*The functions $c(x) = \cos(\sqrt{C}x)$ and $s(x) = \sin(\sqrt{C}x)$ are solutions of the ODE.

*If $A = B = 0$, the unique soluton of (1) is the zero function.
Proof: Suppose $g$ is a solution of (1) with $A = B = 0$. Since $0 = g'' + g$, multiplying by $2g'$ gives
$$
  0 = 2g'\, g'' + 2g\, g' = \bigl[(g')^{2} + g^{2}\bigr]'.
  $$
By the identity theorem, the function in square brackets is constant, and by the initial conditions, this constant is $0$. In particular, $g = 0$ as a function.

*The unique solution of (1) is $f(x) = A\cos(\sqrt{C}x) + B\sin(\sqrt{C}x)$.
Proof: Let $f$ be an arbitrary solution of (1). The function
$$
  g(x) = f(x) - \bigl[A\cos(\sqrt{C}x) + B\sin(\sqrt{C}x)\bigr]
  $$
satisfies the ODE with initial conditions $g(0) = g'(0) = 0$.
A: Hint: We know that by the chain rule of differentiation: $$\frac {d}{dx}(\cos ax) = a (-\sin ax) $$ and $$ \frac {d}{dx}(\sin ax)=a (\cos ax) $$ Can you take it from here?
A: $$f''(x)+Cf(x)=0$$
The procedure here is look for solutions of the form $ce^{\lambda x}$ i.e. we get by plugging into our equation and dividing by $ce^{\lambda x}$
$$\lambda^2+C=0$$
i.e. $\lambda=\pm i \sqrt{C}$. Now these two roots give you the solutions $f_1(x)=c_1 \exp(-i\sqrt{C}x)$ and $f_2(x)=c_2 \exp(i\sqrt{C}x)$. Now add these two solutions and use Euler's formula. You get 
$$f(x)=(c_1+c_2) \cos(\sqrt{C}x)+i(-c_1+c_2) \sin(\sqrt{C}x)$$
And now just define $A=c_1+c_2$ and $B=i(c_2-c_1)$ to get the solution in your form.
A: First, note that both $g$ and $h$ are indeed solutions. (This is easy to see since second derivitive of sin or cos will give you itself back with a min sign.) Then, by linearity, sum of solutions is a solution as well. Thus $g+h$ is a solution.
Are you looking to prove that the span of $g,h$ are the only solutions?
A: One can use reduction of order. Let $f(x) = y(x)\cos{cx}$. Then
$$ f''+ c^2 f = y''\cos{cx} + 2cy'\sin{cx}-c^2 y\cos{cx} + c^2 y\cos{cx} = y''\cos{cx} + 2cy'\sin{cx}. $$
This is a first-order differential equation for $y'$; we now look for the integrating factor, $ \exp{( \int 2c\tan{cx} \, dx )} = (\sec{cx})^2 $, and the equation becomes.
$$ ( \cos^2{cx} \, y')' = 0 $$
The unique type of solution to $F'=0$ is a constant, so the unique way to integrate this equation is
$$ y' = A\sec^2{cx}. $$
Again, this equation has a unique solution up to a constant, by direct integration:
$$ y = A\tan{cx}+B, $$
and then $f(x) = A\sin{cx}+B\cos{cx}$. This works on any interval where $\cos{cx} \neq 0$, but we can refine it a bit to include the whole line by invoking continuity (or doing the same thing with $f(x)\sin{cx}$ and stitching together on intervals where they agree).
