Showing if $f + f'' = 0$ and if $f(0)=a$ and $f'(0)=b$, then $f(x) = b \sin x + a \cos x$ I'm reading Spivak's Calculus, 4th edition Chapter 15. On page 313, there is a proof I'm not quite understanding. I will quote:

If $f$ has a second derivative everywhere and $f + f'' = 0$, $f(0)=a$, $f'(0)=b$, then $f = b \cdot \sin + a \cdot \cos$

First, I'd like ask how I'm supposed to interpret the equals sign here. Clearly it means the function equals $b \cdot \sin + a \cdot \cos$, which I understand and am convinced (by the proof that follows), but does it also mean that this is the unique and only form it can take on?
If the answer is yes, then I'm confused about why the proof adequately shows that this is the unique form it can take on. I will quote the proof in the book:

Let $g(x) = f(x) - b \sin x - a \cos x$.
Then $g'(x) = f'(x) - b \cos x + a \sin x$
$g''(x) = f''(x) + b \sin x + a \cos x$
Consequently,
  $g'' + g = 0$
$g(0)=0$
$g'(0)=0$
which shows that $0=g(x)=f(x) - b\sin x - a\cos x$ for all $x$

Side note: He uses a Lemma, that if $f'''+f = 0$ and $f(0)=0$ and $f'(0)=0$, then $f=0$ to derive the last step in the proof above.
Question: How do we know there's not a more general solution to this differential equation? In other words, how do we know that $f(x)$ can't equal $b \sin x - a \cos x + \text{<Insert a function that is neither $\sin$ or $\cos$ such that the conditions required hold>}$?
After all, he started the proof by letting $g(x)$ equal to $f(x) - b\sin x - a\cos x$.
 A: What you describe is essentially what he has done.
Suppose $f(x)=b \sin x + a \cos x + g(x)$, where $g(x)$ is your function to be inserted.
Then $f'(x)=b \cos x - a \sin x + g'(x)$
and $f''(x)=-b \sin x - a \cos x + g''(x)$
So $f''(x)+f(x)=-b \sin x - a \cos x + g''(x)+b \sin x + a \cos x + g(x)$
or $f''(x)+f(x)=g''(x)+ g(x)$
We know that $f(x)$ obeys $f''(x)+f(x)=0$, so $0=g''(x)+ g(x)$
So far, no problem. But then we consider $f(0)=b \sin 0 + a \cos 0 + g(0)$
Since $f(0)=a$, we have $a=0+a+g(0) \Rightarrow g(0)=0$
Consider also $f'(0)=b \cos 0 - a \sin 0 + g'(0)$
Since $f'(0)=b$, we have $b=b-0+g'(0) \Rightarrow g'(0)=0$
He then uses his Lemma to prove that $g(x)=0$ for all values of $x$.
Thus your suggested alternative function is shown to be $b \sin x + a \cos x + 0$: no different to the suggested form at all!
A: If 
$$
g''+g=0, \quad\text{and}\quad g(0)=g'(0)=0,
$$
then
$$
0=g'g''+gg'=\frac{1}{2}\big(g^2+(g')^2\big)',
$$
and thus
$$
0=\int_0^x(g'g''+gg')\,dx=\frac{1}{2}\big(g^2(x)+(g')^2(x)\big)-
\frac{1}{2}\big(g^2(0)+(g')^2(0)\big)
\\=\frac{1}{2}\big(g^2(x)+(g')^2(x)\big)
$$
and hence $g(x)=g'(x)=0$, for all $x$.
A: How do we know there's not a more general solution to this differential equation?  Because that is precisely what you showed! You had no restriction on $f$ in the beginning other than the IVP. Then you proved that $f$ is none other than that trig function. So, if $h$ is another hypothetical solution then again you show that it too equals the sine and cosine combination above.
Also, it is a universal fact that solutions to regular enough then solutions are unique.
A: $$f''+f=0$$
Multipy by $ \mu(x)=\sin x$:
$$f''\sin x+f \sin x=0$$
$$f''\sin x \color{red}{+f' \cos x -f' \cos x}+f \sin x=0$$
$$(f' \sin x)' -(f\cos x)'=0$$
After integration:
$$f' \sin x -f\cos x=c_1$$
$$\left (\dfrac f {\sin x}\right )'= \dfrac {c_1}{\sin^2 x}$$
$$\left (\dfrac f {\sin x}\right )=-{c_1}{\cot x}+c_2$$
Therefore:
$$f(x)={c_1}{\cos x}+c_2 \sin x$$
Apply initial conditions.
$$f(0)=0 \implies c_1=a, f'(0)=b \implies c_2=b$$
$$f(x)=a{\cos x}+b \sin x$$
