Differential equations for a transformer
You have two unknowns, \$i_1, i_2\$ and two equations, so your equations are solvable.

Solving these equations are generally done using the Laplace transform.

$$\left\{ \begin{align}
i_1R_1 + L_1\frac{di_1}{dt} - M\frac{di_2}{dt} &amp;= v_1(t)\\
-M\frac{di_1}{dt} + i_2R_2 + L_2\frac{di_2}{dt} &amp;= 0
\end{align}\right.$$

Leads to

$$\left\{ \begin{align}
(R_1+L_1s)&amp;\cdot I_1 &amp;- Ms\cdot I_2 &amp; = V_1(s)\\
-Ms&amp;\cdot I_1 &amp;+ (R_2+L_2s)\cdot I_2 &amp; = 0
\end{align}\right.$$

$$I_2 = \frac{\left|\begin{matrix}
R_1+L_1s &amp; V_1(s) \\
-Ms &amp; 0
\end{matrix}\right|
}{\left|\begin{matrix}
R_1+L_1s &amp; -Ms \\
-Ms &amp; R_2+L_2s
\end{matrix}\right|
}=\frac{Ms\cdot V_1(s)}{(R_1+L_1s)(R_2+L_2s)-M^2s^2}$$

If you prefer differential equations you can always go back using:

$$\begin{align}
\left[(R_1+L_1s)(R_2+L_2s)-M^2s^2\right]\cdot I_2(s) &amp;= Ms\cdot V_1(s)\\
&amp;\Downarrow\\
\left[R_1R_2 + (R_1L_2 + R_2L_1)s + (L_1L_2-M^2)s^2\right]\cdot I_2(s) &amp;= Ms\cdot V_1(s)\\
&amp;\Downarrow \mathcal{L}^{-1}\\
R_1R_2\cdot i_2(t)+(R_1L_2+R_2L_1)\frac{di_2}{dt}+(L_1L_2-M^2)\frac{d^2i_2}{dt^2} &amp;= M\frac{dv_1}{dt}
\end{align}$$