Consider the circuit.

The voltage drop across a resistor is
$V_R$
.
The voltage drop across a capacitor is
$V_C=\frac{Q}{C}= \frac{1}{C} \int I dt$
.
The voltage drop across an inductor is
$V_L=L \frac{dI}{dt}$
.
Using Kirchoff's Law for the left hand loop gives
$E-V_L-V_{R_1}=0$
. (1)
The voltages across the resistor and inductor respectively are
$V_{R_1}=R_1(I_1-I_2), , \: V_L=L \frac{dI_1}{dt}$
. Substituting these into (1) gives
$E- L \frac{dI_1}{dt}-R(I_1-I_2)=0$
.
Hence
$50- 0.5 \frac{dI_1}{dt}-200(I_1-I_2)=0$
. (2)
Applying Kirchoff's Voltage Law to the right hand loop
$V_R+V_C+V_{R_2}$
hence
$\frac{1}{C} \int I_2 dt+I_2R_2+R_1(I_1-I_2)=0 \rightarrow \frac{1}{C} \int I_2 dt-R_1I_1+I_2(R_1+R_2)=0$
.
Let
$Q=\frac{dI_2}{dt}$
and
$I_2=\frac{dQ}{dt}$
.
$\frac{Q}{C}-R_1I_1+\frac{dQ}{dt}(R_1+R_2)=0$
.
Using the values from the diagram gives
$\frac{1}{50 \times 10^{-6}} \int I_2 dt-200I_1+500 \frac{dQ}{dt}=0 \rightarrow \frac{dQ}{dt}-0.4I_1+40Q=0 \rightarrow \frac{dQ}{dt}=-40Q+0.4I_1$
. (3)
Use
$I_2=\frac{dQ}{dt}$
in (2) to give
$50- 0.5 \frac{dI_1}{dt}-200(I_1-\frac{dQ}{dt})=0 \rightarrow 50-0.5 \frac{dI_1}{dt}-200I_1+200 \frac{dQ}{dt}=0$
.
Substitution of (3) into the last equation results in the equation
$50-0.5 \frac{dI_1}{dt}-120I_1+80Q=0 \rightarrow \frac{dI_1}{dt}=-240I_1+1600Q+100$
.
We now have the equations
$\frac{dQ}{dt}=-40Q+0.4I_1$

$\frac{dI_1}{dt}=-240I_1+1600Q+100$

We can write this in matrix form as
$\begin{pmatrix} \frac{dI_1}{dt}\\ \frac{dQ}{dt}\end{pmatrix} = \left( \begin{array}{ccc} -240 & -1600 \\ 0.4 & -40 \end{array} \right) \begin{pmatrix} I_1\\ Q \end{pmatrix} + \begin{pmatrix} 100 \\ 0\end{pmatrix}$

The eigenvalues and eigenvectors are
$\lambda_1=-80, \: \lambda_2=-200$
,
$\mathbf{v}_1 = \begin{pmatrix}-100\\1\end{pmatrix}, \: \mathbf{v}_2 = \begin{pmatrix}-400\\1\end{pmatrix}$
.
The solution to the homogeneous system is then
$\begin{pmatrix} I_1\\ Q \end{pmatrix}=Ae^{-80t}\begin{pmatrix}-100\\1\end{pmatrix}+Be^{200t}\begin{pmatrix}-400\\1\end{pmatrix}$
.
To find a particular solution, assume
$\frac{dI_1}{dt}=\frac{dQ}{dt}=0$
. Then
$0=-240I_1-1600Q+100$
.
$0=0.4I_1-40Q$

We get
$I_1=1/4, \: Q=1/400$

The general solution is then
$\begin{pmatrix} I_1\\ Q \end{pmatrix}=Ae^{-80t}\begin{pmatrix}-100\\1\end{pmatrix}+Be^{200t}\begin{pmatrix}-400\\1\end{pmatrix}+\begin{pmatrix}1/4\\1/400\end{pmatrix}$
.
This can be fitted to any initial conditions.