is it possible to integrate this function to get x(t) and y(t)?

Question

say you have a function as below;

$d^2V(t)/dt = -B^2V(t)$

B is a constant

Initial conditions $V_x(0) = V$, $V_y(0) = 0$

I can't see how to integrate to get x(t) and y(t);

I ended up with $x(t) = dVx(t)/dt = -B^2V(t)x$, when I integrated RHS with respect to dt and LHS with respect to dx. Any help here is appreciated......

EDIT: The question has been asked, here it is. How I came to the beginning equation above is in the comments section, though I used G for $B_o$ and $B$ is not same as $B_o$ also because I just used it to represent a chunk;

Answer 1

This is just the standard ODE for sin and cos. Note that $\sin'(bx)=b\cos(bx)\\ \cos'(bx) = -b\sin(bx)\\ \sin''(bx)=b\cos'(bx)=-b^2\sin(bx)\\ \cos''(bx) = -b\sin'(bx)=-b^2\cos(bx) $.

To get $b^2$ instead of $-b^2$, use $\sinh$ and $\cosh$.

Answer 2

It's not totally clear to me whether your problem is to solve the equation $$m\frac{\text d\mathbf v}{\text dt} = q\mathbf v\times\mathbf B$$ or something else. So I will assume this is the problem you are trying to solve, and I will further assume that $\mathbf B$ is a constant vector in both space and time.

Under the above assumptions, the RHS of the equation is linear in $\mathbf v$, and therefore there exists a $3\times3$ matrix, say $\star\mathbf B$ (the $\star$ here can be interpreted as the Hodge star operator) such that $\mathbf v\times\mathbf B = (\star\mathbf B)\mathbf v$. The equation then becomes $$m\frac{\text d\mathbf v}{\text dt} = q(\star\mathbf B)\mathbf v,$$ which integrates to $$\mathbf v(t) = e^{\frac qm\star\mathbf Bt}\mathbf v_0,$$ where $\mathbf v_0$ is the initial condition on the velocity. To get the trajectory one has then to simply integrate $\mathbf v(t)$ with the initial condition on the position to get $\mathbf r(t)$.