How do I write the equations for three inductors connected together in state space form?
You need to use Kirchhoff's current law on the node with voltage \$V_T\$:
$$
i_1 = i_2 + i_3
$$
Taking the time derivative at both sides results in
$$
\frac{\mathrm{d}i_1}{\mathrm{d}t} =
\frac{\mathrm{d}i_2}{\mathrm{d}t} + \frac{\mathrm{d}i_3}{\mathrm{d}t}
$$
You now have a set of 4 equations in 4 unknowns (\$\frac{\mathrm{d}i_1}{\mathrm{d}t}\$, \$\frac{\mathrm{d}i_2}{\mathrm{d}t}\$, \$\frac{\mathrm{d}i_3}{\mathrm{d}t}\$, and \$V_T\$):
$$
\begin{cases}
L_1 \frac{\mathrm{d}i_1}{\mathrm{d}t} &amp; = V_T - V_1 \\
L_2 \frac{\mathrm{d}i_2}{\mathrm{d}t} &amp; = V_2 - V_T \\
L_3 \frac{\mathrm{d}i_3}{\mathrm{d}t} &amp; = V_3 - V_T \\
\frac{\mathrm{d}i_1}{\mathrm{d}t} &amp; =
\frac{\mathrm{d}i_2}{\mathrm{d}t} + \frac{\mathrm{d}i_3}{\mathrm{d}t}
\end{cases}
$$

After eliminating \$\frac{\mathrm{d}i_1}{\mathrm{d}t}\$ and \$V_T\$, you get
$$
\begin{bmatrix}
\frac{\mathrm{d}i_2}{\mathrm{d}t} \\ \frac{\mathrm{d}i_3}{\mathrm{d}t}
\end{bmatrix} =
\frac{1}{L_1L_2+L_1L_3+L_2L_3}
\begin{bmatrix}
-L_3 &amp; L_1+L_3 &amp; -L_1 \\
-L_2 &amp; -L_1 &amp; L_1+L_2
\end{bmatrix}
\begin{bmatrix}
V_1 \\ V_2 \\ V_3
\end{bmatrix}
$$
which is a state equation of the form \$\frac{\mathrm{d}x}{\mathrm{d}t}=Ax+Bu\$. The state variables are \$x_1=i_2\$ and \$x_2=i_3\$. The inputs are \$u_1=V_1\$, \$u_2=V_2\$, and \$u_3=V_3\$. The matrix \$A\$ is a zero matrix and the matrix \$B\$ is
$$
B =
\frac{1}{L_1L_2+L_1L_3+L_2L_3}
\begin{bmatrix}
-L_3 &amp; L_1+L_3 &amp; -L_1 \\
-L_2 &amp; -L_1 &amp; L_1+L_2
\end{bmatrix}
$$