Circuit Analysis – Finding Current and Voltage Through the Circuit

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.

My solution: Consider \$v_L\$ as the voltage on inductor (in accordance with the direction of \$i\$) and \$v_A\$ as the voltage on upper node (called here node \$A\$). Applying the KCL on this node:

$$ \frac{v_A-v_L}{4}+ \frac{v_A-2v_x}{1} + \frac{v_A}{4}=0 $$

If $$ \left\{\begin{matrix}v_L=2 \frac{\mathrm{d}i }{\mathrm{d} t} \\v_x=-4i \end{matrix}\right. $$

Then

$$ \frac{\mathrm{d}i }{\mathrm{d} t}-16i-3v_a=0 $$

Replacing \$ v_a=2\frac{\mathrm{d}i }{\mathrm{d} t}+4i \$ leads to:

$$ 5\frac{\mathrm{d}i }{\mathrm{d} t}+28i=0 $$

So, the current \$i(t)\$ is given by (in Ampere): $$i(t) = 5e^{-5.6t} A$$

Remebering that \$ v_x=-4i \$, this voltage is given by (in Volt):

$$v_x(t) =-20e^{-5.6t} $$