Oscillatory system with general argument

Question

So by inspection, the solution to the following set of ODEs $$ x'(t) = f(t)y(t) $$

$$ y'(t) = -f(t)x(t) $$

Are $$ x(t) = A\cos\left(\int^t f(t')dt' \right)-B\sin\left(\int^t f(t')dt' \right) $$

$$ y(t) = B\cos\left(\int^t f(t')dt' \right)+A\sin\left(\int^t f(t')dt' \right) $$

Does anyone know how to prove this? In general this system has form $$ \begin{pmatrix} x'\\y' \end{pmatrix} = f(t) \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\begin{pmatrix}x \\y\end{pmatrix} $$

And given the eigenvalues of the matrix are $\lambda = \pm i$, you can see the oscillatory structure appear. Any comments or useful reading is appreciated, thanks!

Answer 1

Actually having thought about it I was a step behind. All you need to do is diagonalise the matrix to rewrite the system as

$$ \begin{pmatrix} i/2 & 1/2 \\ -i/2 & 1/2 \\ \end{pmatrix}\begin{pmatrix} y' \\ z' \\ \end{pmatrix} = f\begin{pmatrix} i&0 \\ 0 & -i \\ \end{pmatrix} \begin{pmatrix} i/2 & 1/2 \\ -i/2 & 1/2 \\ \end{pmatrix} \begin{pmatrix} y \\ z \\ \end{pmatrix} $$

Which can be solved exactly (say define

$$\begin{pmatrix} i/2 & 1/2 \\ -i/2 & 1/2 \\ \end{pmatrix} \begin{pmatrix}y \\ z\end{pmatrix} = \begin{pmatrix}v \\ w\end{pmatrix}$$ )

Answer 2

Since the eigenvalues are $\pm i$, you can substitute $z=x+iy$ to obtain

$$ z' = x' + iy' = f(t)(y-ix) = -if(t)(x+iy)=-ifz $$

Then obviously

$$ z = C\exp\left(-i\int f(t)\ dt\right) $$

where $C = A+iB$ is some complex number. Taking the real and imaginary parts of $z$ gives the answer.