Electronic – Trying to solve a circuit containing 2 current sources and 1 voltage source and resistors: what am I doing wrong

The superposition theorem just says to replace remaining voltage sources with shorts and current sources with opens and evaluate. Then just sum everything up. I don't think any of your results are correct, yours or the ones you think may be right. But perhaps I just can't read your problem with understanding. I can read the circuit, though. Just to recall, here it is:

^{simulate this circuit – Schematic created using CircuitLab}

Let's focus on \$V_1\$, shorting out \$V_2\$ to produce this for step 1:

^{simulate this circuit}

We can now lay out these currents:

\$I_{R_1} = \frac{V_1}{R_1}\$

\$I_{R_2} = \frac{V_1}{R_2}\$

\$I_{R_3} = 0\$

Note that \$R_3\$ is shorted out by the replaced \$V_2\$ and so its current must be zero.

Now let's focus on \$V_2\$, shorting out \$V_1\$ to produce this for step 2:

^{simulate this circuit}

We can now lay out these currents:

\$I_{R_1} = 0\$

\$I_{R_2} = -\frac{V_2}{R_2}\$

\$I_{R_3} = \frac{V_2}{R_3}\$

Note that \$R_1\$ is shorted out by the replaced \$V_1\$ and so its current must be zero. Also notice that the direction of the current in \$R_2\$ is the opposite to the earlier direction. So here we use an opposing sign. Just be consistent about this.

Now we can simply take the two above cases and sum them up together as though they are happening simultaneously:

\$I_{R_1} = \frac{V_1}{R_1} + 0=\frac{V_1}{R_1}\$

\$I_{R_2} = \frac{V_1}{R_2}-\frac{V_2}{R_2}= \frac{V_1-V_2}{R_2}\$

\$I_{R_3} = 0 + \frac{V_2}{R_3}=\frac{V_2}{R_3}\$

The current in \$V_1\$ will be the sum of the two returning currents from \$R_1\$ and \$R_2\$ or else it will be the sum of the currents through \$R_1\$, \$V_2\$, and \$R_3\$, depending on which way you'd rather look. Either way, it has to be the same.

Just by way of making sense of things, it should be clear that since \$V_1\$ is directly across \$R_1\$ that the final summed current through \$R_1\$ must simply be \$\frac{V_1}{R_1}\$. And it is. Good. Similarly, it should be clear that since \$V_2\$ is directly across \$R_3\$ that the final summed current through \$R_3\$ must simply be \$\frac{V_2}{R_3}\$. And it is. Also good.

Back to business. Easiest way to get \$I_{V_1}\$ is to sum the returning currents in \$R_1\$ and \$R_2\$:

\$I_{V_1}= I_{R_1}+I_{R_2}= \frac{V_1}{R_1}+\frac{V_1-V_2}{R_2}\$

I don't recall seeing that answer in the stuff you provided. Maybe I just didn't see it.

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.