Well, this question may also require some physics knowledge of electric circuit to solve:
The First equation:
$R \times i_k + L \times \frac{di_k}{dt} = U_m \times \sin(\omega t + \varphi_{0u})$
where
$i_k$ is instantaneous value of short-circuit current.
$\varphi_{0u}$ is phase of power source voltage when short-circuit happened.
The Second equation:
$i_k = \frac{U_m}{Z}\sin(\omega t + \varphi_{0u}-\varphi)-\frac{U_m}{Z}\sin(\varphi_{0u}-\varphi)\times e^{-\frac{R}{L}\times t}$
where
$Z = R+jX$
$\varphi$ is the phase that current lags behind voltage.
Here is the equivalent circuit:
According to my book, the first equation can be transformed to the second one, I tried to proof that but I got stuck after a few steps.
Since it is simple ODE, you can solve it and get $$i_k(t) = c_1 e^{-(R t)/L} + \dfrac{U}{(L^2 \omega^2 + R^2)}\left(R \sin(t \omega + ϕ_{0u}) - L \omega \cos(t \omega + ϕ_{0u})\right)$$
Since $Z = R + jX = R + j \omega L $, so $|Z|^2 = R^2 + \omega^2L^2$
So you get, $$i_k(t) = c_1 e^{-(R t)/L} + \dfrac{U}{|Z|^2}\left(R \sin(t \omega + ϕ_{0u}) - L \omega \cos(t \omega + ϕ_{0u})\right) $$
Let $\cos \phi = \dfrac{R}{|Z|}$ and $\sin \phi = \dfrac{X}{|Z|}$
$$i_k(t) = c_1 e^{-(R t)/L} + \dfrac{U}{|Z|}\left( \sin(t \omega + ϕ_{0u} - \phi) \right) $$
At $t = 0$.
$$i_k(0) = c_1 + \dfrac{U}{|Z|}\sin( ϕ_{0u} - \phi)$$
I think at $t = 0$ current should be zero, so
$$c_1 = - \dfrac{U}{|Z|}\sin( ϕ_{0u} - \phi)$$.
$$i_k(t) = \dfrac{U}{|Z|}\left( \sin(t \omega + ϕ_{0u} - \phi) \right) - \dfrac{U}{|Z|}\sin( ϕ_{0u} - \phi) e^{-(R t)/L} $$