Circuit Analysis – Find the Thévenin Equivalent Between Terminals A and B

To calculate the Thevenin voltage the port 2 has to be opened, thus the current is zero: \$ i_2=0 \$.
You get: \$ 0.5⋅u_1+2⋅10^{-4}⋅u_2=0 \to u_2= -2500⋅u_1 \to u_2=10mV*2500=25V\$.
To calculate the Thevenin resistance, do the same but with \$ u_2=0 \to i_2=0.5u_1 \to i_2=5mA\$.
Then, since the voltage drop is all on the Thevenin resistance $$ R_{th}=\frac{u_2}{i_2} \to \frac{25}{5m}=5k\Omega$$
Notice it is the \$(y_{22})^{-1}\$.

I really think nidhin's answer is the best way to go. Use the Laplace transform, calculate \$I_2(s)\$ and then convert it back into a differential equation if you're not interested in the solution.

I just did it for you to show you what you're up against if you try to avoid Laplace transforms:

The Laplace transform gives the solution (I used a CAS for this):

\$I_2\cdot (a\cdot s^3 + b\cdot s^2 + c\cdot s + d) = V_1\cdot (e\cdot s^3 + f\cdot s^2)\$

Yielding the differential equation of the form:

\$a\frac{d^3i_2}{dt^3} + b\frac{d^2i_2}{dt^2} + c\frac{di_2}{dt} + d\cdot i_2 = e\frac{d^3v_1}{dt^3} + f\frac{d^2v_1}{dt^2}\$

Where the coefficients are (I hope I copied them correctly):

\$a = C_1C_2L(R_3R_4 + R_2R_4 + R_1R_4 + R_1R_2)\$

\$b = C_1C_2R_3(R_2R_4 + R_1R_4 + R_1R_2) + (C_1+C_2)L(R_2 + R_3) + L(C_2R_4 + C_1R_1)\$

\$c = C_2R_3(R_4 + R_2) + C_1R_3(R_2 + R_1)\$

\$d = R_3\$

\$e = C_1C_2L(R_2+R_3+R_4)\$

\$f = C_1(C_2R_2R_3+L)\$

I would not advise doing this without transforming to Laplace. While it is possible to solve it without, it can be very hard to see how the equations need to be substituted.