Deriving currents of parallel mutual inductors
There are a few ways to solve this, but I guess to me the most natural one for me would be:

$$
\begin{align}
j\omega L_1I_1 + j\omega MI_2 &amp;= j\omega MI_1 + j\omega L_2I_2 \\
&amp;\Downarrow \\
L_1I_1 + MI_2 &amp;= MI_1 + L_2I_2 \\
&amp;\Downarrow \\
(L_1 - M)I_1 &amp;= (L_2 - M)I_2
\end{align}
$$

So in order to find the relationship between \$V\$ and \$I_1\$, you use the substitution

$$\begin{align}
V &amp;= j\omega L_1I_1 + j\omega MI_2\\
&amp;\Downarrow\\
V &amp;= j\omega L_1I_1 + j\omega M\cdot \frac{L_1 - M}{L_2 - M}I_1 \\
&amp;= j\omega I_1 \left(L_1 + M\frac{L_1 - M}{L_2 - M} \right) \\
&amp;= j\omega I_1 \left( \frac{L_1(L_2 - M) + M(L_1 - M)}{L_2 - M} \right) \\
&amp;= j\omega I_1 \left(\frac{L_1L_2 - M^2}{L_2 - M}\right)
\end{align}$$

You can do similar steps to find the relationship \$V\$ and \$I_2\$.