Differential equation $x'' = -kx$

Question

Initial condition $x(0)=1$, $k>0$

\begin{eqnarray} \lambda^2=-k \end{eqnarray} \begin{eqnarray} \lambda=\pm\sqrt{-k} \end{eqnarray} I have no idea how to erase this $k$

what should be the next step?

Answer 1

$$\begin{eqnarray} λ^{2}=-k \end{eqnarray}\\ \lambda = \pm i\sqrt k\\ x=c_1 \cos \sqrt k x +c_2 \sin \sqrt k x\\ x(0)=1\implies c_1=1\\x= \cos \sqrt k x + c\sin \sqrt k x $$

Answer 2

It's a linear differential equation : $$x'' = -kx$$ $$x'' +kx=0$$ $$R^2-i^2k=0 \implies R= \pm i\sqrt k$$ $$x(t)=c_1 \cos( \sqrt k t)+c_2 \sin ( \sqrt k t)$$ Initial condition $$x(0)=1 \implies c_1=1$$ $$\boxed {x(t)=\cos( \sqrt k t)+c_2 \sin ( \sqrt k t)}$$