Find $\vec{r}$ satisfying the equation $\frac{d^2\vec{r}}{dt^2}+\omega^2\vec{r}=0$

Question

Problem : Find $\vec{r}$ satisfying the equation $\dfrac{d^2\vec{r}}{dt^2}+\omega^2\vec{r}=0$, where $\omega$ is a constant different from zero. Given that $\vec{r}=(1,1,1)$ and $\dfrac{d\vec{r}}{dt}=(0,0,1)$ when $t=0$.
I stuck to find $\vec r$.
I have tried in the following way:
$\dfrac{d^2\vec{r}}{dt^2}+\omega^2\vec{r}=0\implies 2\dfrac{d\vec{r}}{dt}\dfrac{d^2\vec{r}}{dt^2}+2\omega^2\vec{r}\dfrac{d\vec{r}}{dt}=0\implies \left( \dfrac{d\vec{r}}{dt}\right)^2+ \omega^2 \vec{r}^2=C$.

Answer 1

$$\frac{d^2 \vec r}{dt^2}+w^2 \vec r=0 ~~~~(1)$$ $\vec r=x\vec i+ y \vec j+z \vec k$, so there will be three un coupled PDEs like $$\frac{d^2 f}{dt^2}+w^2 f=0, f=x,y,z~~~(2)$$ Here, check the solution of (2) is $f=A \sin wt+B \cos wt$ Finally $$\vec r(t)= (A_1 \sin wt +B_1 \cos wt)\vec i +(A_2 \sin wt + B_2 \cos wt) \vec j+ (A_3 \sin wt +B_3 \cos wt) \vec k$$ If $\vec r(0)=\vec i + \vec j+ \vec k$ and $\frac{d\vec r}{dt}=\vec k$, then $A_1=A_2=0,A_3=1/w$, $B_1=B_2=B_3=1.$ Finally, $$\vec r(t)= \cos wt \vec i+ \cos wt \vec j+ (\frac{1}{w} \sin wt + \cos wt) \vec k$$