Show that this is a solution
$$i(t)\text{:=}\frac{U_m \sin (t \omega -\varphi )}{Z}$$
to
$$i'+\frac{i R}{L}=\frac{U_m \sin (t \varphi )}{L}$$
given:
$$\varphi =\tan ^{-1}\left(\frac{\text{L$\omega $}}{R}\right)$$
$$Z=\sqrt{L^2 \omega ^2+R^2}$$
This is what I have done so far.
$$\frac{d\frac{U_m \sin (t \omega -\varphi )}{Z}}{dt}$$
$$\sin (\varphi )=\frac{L \omega }{\sqrt{L^2 \omega ^2+R^2}}$$
$$\cos (\varphi )=\frac{R}{\sqrt{L^2 \omega ^2+R^2}}$$
$$\frac{U_m \sin (t \omega -\varphi )}{Z}=\frac{U_m (\cos (\varphi ) \sin (t \omega )-\sin (\varphi ) \cos (t \omega ))}{Z}=\frac{U_m \left(\frac{R \sin (t \omega )}{\sqrt{L^2 \omega ^2+R^2}}-\frac{(L \omega ) \cos (t \omega )}{\sqrt{L^2 \omega ^2+R^2}}\right)}{Z}=\frac{U_m (R \sin (t \omega )-L \omega \cos (t \omega ))}{\sqrt{L^2 \omega ^2+R^2} \sqrt{L^2 \omega ^2+R^2}}=\frac{U_m (R \sin (t \omega )-L \omega \cos (t \omega ))}{L^2 \omega ^2+R^2}$$
$$i'=\frac{\left(\omega U_m\right) \cos (\varphi -t \omega )}{Z}=\frac{\left(\omega U_m\right) (\sin (\varphi ) \sin (t \omega )+\cos (\varphi ) \cos (t \omega ))}{Z}=\frac{\left(\omega U_m\right) \left(\frac{(L \omega ) \sin (t \omega )}{\sqrt{L^2 \omega ^2+R^2}}+\frac{R \cos (t \omega )}{\sqrt{L^2 \omega ^2+R^2}}\right)}{\sqrt{L^2 \omega ^2+R^2}}=\frac{\omega U_m (L \omega \sin (t \omega )+R \cos (t \omega ))}{L^2 \omega ^2+R^2}$$
Now Im stuck, how do I continue?