$\sin(t)$ solution of $\ddot{x}=-x$ Let $\sin(t)$ be the solution of the differential equation $\ddot{x}=-x$.
Why is the general solution of $\ddot{x}=-x$, $x(t)=\lambda \sin(t) + \mu \cos(t)$, with $\lambda, \mu \in \mathbb{R}$?
 A: Let $x\colon\mathbb{R}\longrightarrow\mathbb{R}$ be a solution of $x''=-x$. Then\begin{align}x''=-x&\iff x''+x=0\\&\implies x'x''+xx'=0\\&\iff2(x'x''+xx')=0\\&\iff\bigl((x')^2+x^2\bigr)'=0\\&\iff(x')^2+x^2\text{ is constant.}\end{align}So, if we assume that $x(0)=x'(0)=0$, the conclusion is that $x$ must be the null function.
Otherwise, let $y(t)=x(t)-x(0)\cos(t)-x'(0)\sin(t)$. Then


*

*$y''=-y$;

*$y(0)=0$;

*$y'(0)=0$.


Therefore, $y\equiv0$, and this means that$$x(t)=x(0)\cos(t)+x'(0)\sin(t).$$
A: Because also $\cos t$ is a solution of the same equation and, since $\sin t$ and $\cos t$ are linearly independent solutions of a second order differential equation, the general solution is a linear combination:
$\lambda \sin t + \mu \cos t$ 
A: Who says $\sin(t)$ is the solution of $\ddot x = -x$? 
It is a solution, not the solution. 
You can see that the general solution confirms that $\sin(t)$ is a solution, because if you set $\lambda=1, \mu=0$ in the general solution,
the resulting function is $x(t) = \sin(t).$
