Strictly speaking, the circuit has 5 nodes, at points labelled a, b, c, d and f. If you use the modified nodal analysis to solve the circuit, you'll apply KCL at all nodes but one (usualy the reference one) to end up with a system of lineal independent equations. So, you'll apply KCL at nodes a, b, c and d to determine the (initially) unknown voltages at these nodes.
However, this is modified by the presence of voltage sources connected to ground at nodes a and c. Due to the presence of these voltage sources, the voltages at nodes a and c are not unknowns, so you have only 2 "real" unknowns (voltages at b and d), and you have to write only 2 KCL. If you have to choose, you would like to write KCL at b and d, because writing the KCL at a and c involves the currents flowing through the voltage sources, and you don't know how to express these currents in terms of the node voltages, so you avoid writing KCL equations at nodes having voltage sources connected to ground.
So finally you have to write the KCL at b and d to solve the circuit, and you can also get rid of the KCL at d if you add the impedances of the resistor and the capacitor together to have 10-10j Ohm.
So your first equation (KCL) is ok, and all you have to do from that is to express the currents in terms of node voltages and impedances:
$$I_1+I_2−I_3−I_4=0$$
$$I_1 = \frac{E_1-V_b}{-10j}$$
$$I_2 = \frac{E_2-V_b}{+20j}$$
$$I_3 = \frac{V_b}{10-10j}$$
$$I_4 = \frac{V_b}{10}$$
And if you substitute these currents in the first KCL and solve for Vb you obtain:
$$V_b = \frac{1-j}{1-3j} * (2E_1-E_2) = \frac{1}{2-j} * (2E_1-E_2)$$
Knowing Vb (and E1 and E2) allows you to easily determine all other circuit variable.
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.
Best Answer
Problem
You are getting the wrong answer because you have not accounted for the fact that
I_2is perceived in the negative direction from the current loop in the central cell.I_3andI_4have the same sign for a current which flows in the clockwise direction, but the sign forI_2flows in the counter-clockwise direction for the middle cell.Fixing this sign yields the correct equation.
Alternative analysis
I find it very helpful to analyze KVL as a full loop, as below. I have defined three loop currents: $$I_\alpha, I_\beta, I_\gamma$$
I wrote the loops as a function of those currents, and then I provided the transformation for the currents $$ I_1, I_2, I_3, I_4, I_5$$:
In my opinion, this is a much more reliable way to write the eqns, as it does not require you to remember to analyze each current leg for the appropriate sign.