Uniqueness of solutions of $y''+y=0$ Recently I was asked about the uniqueness of the solutions of the equation $y'=y$. It could be obtained by multiply $e^{-x}$. 
This time I was wondering about the same question regarding a different equation. We know that the solutions of $y''+y=0$ are of the form $a\sin(x)+b\cos(x) $. 
How can we explain that these are the ONLY solutions?
 A: We shall show that every solution of the equation $y''+y=0$ is of the form $y(x)=a\cos x+b\sin x$, for some $a,b\in\mathbb R$.
In particular, we shall show that $$y(x)=y(0)\cos x+y'(0)\sin x$$ or equivalently
$\,z(x)\equiv 0,\,$
where $z(x)=y(x)-\big(y(0)\cos x+y'(0)\sin x\big)$.
First of all, it can be readily shown that
$$
z''(x)+z(x)=0,
$$
and
$$
z(0)=z'(0)=0. \tag{1}
$$
Hence
$$
0=z'(z''+z)=z'z''+zz'=\frac{1}{2}\big((z')^2+z^2\big)'
$$
and thus $w(x)=\big(z(x)\big)^2+\big(z'(x)\big)^2$ is a constant function, and in particular $w(x)=w(0)$
$$
\big(z(x)\big)^2+\big(z'(x)\big)^2=w(x)=w(0)=\big(z(0)\big)^2+\big(z'(0)\big)^2=0.
$$
Thus $z\equiv 0$.
A: Consider the change of variable $$y' = x$$
Then you get the following ODE system:
$$\begin{cases}x' =-y\\y'=x\end{cases}$$
which is of the form
$$\begin{pmatrix}x'\\y' \end{pmatrix} = \begin{pmatrix}0&-1\\1&0 \end{pmatrix}\begin{pmatrix}x\\y \end{pmatrix} = A\begin{pmatrix}x'\\y' \end{pmatrix}$$
where $A$ is the fundamental matrix. Therefore, you have a linear homogeneous differential equation with constant coefficients which is explicitly solvable. You can find a proof here with some examples.
A: You can also solve it easily
$$y''=-y$$
for $y \neq 0$
$$ \frac {y''}{y}=-1$$
$$ \left (\frac {y'}{y} \right)'+\left( \frac {y'}{y}\right )^2=-1$$
Substitute $z=\frac {y'}y$
$$z'+z^2=-1$$
That's separable
$$\int \frac {dz}{z^2+1}=-x+K \implies \arctan(z)=K-x $$
$$y'=y\tan (K-x)$$
$$y'\cos(K-x)-y\sin(K-x)=0$$
$$y=C_1\cos(x+C_2)$$
A: Using $z(t)=y(t)\cos (t+a)-y'(t)\sin (t+a)$ you get
$$
z'(t)=-(y(t)+y''(t))\sin(t+a)=0
$$
so that $z$ is constant without deviation, with the constant uniquely determined by the initial condition, so that for all $a$
$$
y_0\cos(a)-y'_0\sin(a)=y(t)\cos (t+a)-y'(t)\sin (t+a).
$$
Now substitute $a=-t$ (or $t=-a$) to get, also uniquely, 
$$y(t)=y_0\cos(t)+y'_0\sin(t).$$
