Solving the system of ODE's $f'(t)=kg(t)e^{i\delta t}, g(t)=-kf(t)e^{-i\delta t}$ So I want to solve a system of differential equations with the two equations
$$f'(t)=kg(t)e^{i\delta t}$$
$$g'(t)=-kf(t)e^{-i\delta t}$$
With $k,\delta$ being real valued constants.
I'm unfortunately not very experienced with coupled differential equations like this. My initial thought was to perhaps use the laplace transform, but since I have products of two functions of $t$ on the RHS, I'm not quite sure how I'd go about transforming these. Any help here would be deeply appreciated.
 A: Building on @Hans Engler a bit, you need to rearrange the first equation slightly to get a usable form.
Observe that
$$f'(t)=kg(t)e^{i\delta t} \implies \frac{f'(t)e^{-i\delta t}}{k} = g(t) $$
Differentiating both sides:
$$ g'(t) = \frac{f''(t)e^{-i\delta t}-i\delta f'(t)e^{-i\delta t}}{k} $$
Therefore, subbing into the second equation:
$$\frac{f''(t)e^{-i\delta t}-i\delta f'(t)e^{-i\delta t}}{k}=-kf(t)e^{-i\delta t} \implies$$
$$f''(t)-i\delta f'(t)=-k^2f(t) \implies f''(t)-i\delta f'(t)+k^2f(t) = 0$$
This is a homogenous, linear differential equation. Representing it as a set of differential operators (this is like the characteristic equation, but with the linear algebra justification):
$$(D^2-i\delta D+k^2)f(t) = 0$$
Through the quadratic formula, you can deduce that this is equivalent to:
$$(D-\frac{i}{2}(\delta+\sqrt{\delta^2+4k^2}))(D-\frac{i}{2}(\delta-\sqrt{\delta^2+4k^2}))f= 0$$
The kernel of these operators together, and therefore the set of possible solutions for $f$ is the set of functions
$$Ae^{\frac{i}{2}(\delta+\sqrt{\delta^2+4k^2})t}+Be^{\frac{i}{2}(\delta-\sqrt{\delta^2+4k^2})t} = e^{\frac{i\delta}{2}}(Ae^{\frac{i}{2}(\sqrt{\delta^2+4k^2})t}+Be^{\frac{i}{2}(-\sqrt{\delta^2+4k^2})t})$$
Where $A$ and $B$ are arbitrary real or complex numbers.
$e^{\frac{i\delta}{2}}$ can be expressed as $\cos(\frac{\delta}{2})+i\sin(\frac{\delta}{2}) $, while $Ae^{\frac{i}{2}(\sqrt{\delta^2+4k^2})t}+Be^{\frac{i}{2}(-\sqrt{\delta^2+4k^2})t}$ is equal to $C_1\cos(\frac{\sqrt{\delta^2+4k^2}}{2}t)+C_2\sin(\frac{\sqrt{\delta^2+4k^2}}{2}t)$ for $C_1 = A+B$ and $C_2 = i(A-B)$. Therefore:
$$f(t) = (\cos(\frac{\delta}{2})+i\sin(\frac{\delta}{2}))(C_1\cos(\frac{\sqrt{\delta^2+4k^2}}{2}t)+C_2\sin(\frac{\sqrt{\delta^2+4k^2}}{2}t))$$
$$f'(t) = \frac{\sqrt{\delta^2+4k^2}}{2}(\cos(\frac{\delta}{2})+i\sin(\frac{\delta}{2}))(-C_1\sin(\frac{\sqrt{\delta^2+4k^2}}{2}t)+C_2\cos(\frac{\sqrt{\delta^2+4k^2}}{2}t))$$
Substituting back into the first equation:
$$\frac{e^{-i\delta t}}{k}\frac{\sqrt{\delta^2+4k^2}}{2}(e^{i\frac{\delta}{2}})(-C_1\sin(\frac{\sqrt{\delta^2+4k^2}}{2}t)+C_2\cos(\frac{\sqrt{\delta^2+4k^2}}{2}t)) = g(t) \implies$$
$$\frac{e^{i\frac{\delta}{2}}}{k}\frac{\sqrt{\delta^2+4k^2}}{2}(-C_1\sin(\frac{\sqrt{\delta^2+4k^2}}{2}t)+C_2\cos(\frac{\sqrt{\delta^2+4k^2}}{2}t))(\cos(\delta t)-i \sin(\delta t)) = g(t)$$
A: Alternatively, define new functions $\tilde{f}$ and $\tilde{g}$ via
$$
{f}(t) = \tilde{f}(t)e^{i\delta t/2},
~~~~~~~{g}(t) = \tilde{g}(t)e^{-i\delta t/2}.
$$
Plugging these into the differential equation yields a coupled set of homogeneous first-order differential equations with constant coefficients which can be written as a matrix equation:
$$
\begin{bmatrix}
\tilde{f}'(t) \\
\tilde{g}'(t)
\end{bmatrix}
=
\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}
\begin{bmatrix}
\tilde{f}(t) \\
\tilde{g}(t)
\end{bmatrix},
$$
where $a$, $b$, $c$, and $d$ depend on $\delta$ and $k$. Diagonalizing the matrix will yield the general solution to the differential equation in the usual way. You can then back-substitute the transformations to get the expressions for $f$ and $g$.
A: Applying the Laplace Transform we have
$$
\cases{
sF(s) = k G(s-i\delta)+f(0)\\
sG(s) = -k F(s+i\delta)+g(0)
}
$$
but $(s-i\delta)G(s-i\delta) = -k F(s)+g(0)\Rightarrow G(s-i\delta) = \frac{g(0)-k F(s)}{s-i\delta}$
$$
F(s) = \frac{k}{s+k^2}g(0)+\frac{s-i\delta}{s+k^2}f(0)
$$
and inverting we have
$$
f(t) = f(0)D(t)+e^{-k^2 t} (k (g(0)-f(0) k)-i \delta  f(0))
$$
here $D(t)$ is the Dirac delta function. The procedure to obtain $g(t)$ follows the same steps.
