I did my calculations with infinite numerical precision. This can be done by keeping all numerical quantities as rational numbers (fractions), not floating point numbers. In that case there is only one solution which satisfies the relevant laws (KCL, KVL, Ohm's law).

Yes what happened in my case was i was looking at the wrong schematic again. This caused the sense terminals for the dependent voltage source to be swapped, which of course changes everything.

Just one question now:
Did you get whole number solutions?

 

I did it here.
https://forum.allaboutcircuits.com/threads/how-to-solve-the-next-circuit.184688/page-2#post-1705777

I believe it's correct as my calculation and simulation matches. Just wonder why people got many different results which is almost 42 when Vx is swapped.

Ok well either that cant be right or there are at least two solutions to this circuit.

Just one question:
Did you get whole number solutions for all the nodes?
Oh i checked your link and you seem to have the same solutions set that i have. What changed?
Are you now saying that 42 was with the wrong circuit? That's ok, just wondering now.

 

Here is the solution set in symbolic terms however i wont say which nodes are which just yet but i have a feeling the TS left. Also, note i left out part of the circuit you know what part that was.
E3 is the 40 volt constant voltage source.

v1=(R2*(I1*R1*R4*R5+E1*R4*R5+I1*R1*R3*R5+E1*R3*R5+I1*R1*R3*R4+E1*R3*R4-E3*R1*R3))/(R2*R4*R5+R1*R4*R5+R2*R3*R5+R1*R3*R5-3*R1*R2*R5+R2*R3*R4+R1*R3*R4+R1*R2*R3)

v2=-(3*R2*(I1*R1*R4*R5+E1*R4*R5+I1*R1*R3*R5+E1*R3*R5+I1*R1*R3*R4+E1*R3*R4-E3*R1*R3))/(R2*R4*R5+R1*R4*R5+R2*R3*R5+R1*R3*R5-3*R1*R2*R5+R2*R3*R4+R1*R3*R4+R1*R2*R3)

v3=-(R4*(3*I1*R1*R2*R5+3*E1*R2*R5-I1*R1*R2*R3-E3*R2*R3-E1*R2*R3-E3*R1*R3))/(R2*R4*R5+R1*R4*R5+R2*R3*R5+R1*R3*R5-3*R1*R2*R5+R2*R3*R4+R1*R3*R4+R1*R2*R3)

v4=E1

v5=(I1*R1*R2*R4*R5+E3*R2*R4*R5+E1*R2*R4*R5+E3*R1*R4*R5+I1*R1*R2*R3*R5+E3*R2*R3*R5+E1*R2*R3*R5+E3*R1*R3*R5-3*E3*R1*R2*R5+I1*R1*R2*R3*R4+E3*R2*R3*R4+E1*R2*R3*R4+E3*R1*R3*R4)/(R2*R4*R5+R1*R4*R5+R2*R3*R5+R1*R3*R5-3*R1*R2*R5+R2*R3*R4+R1*R3*R4+R1*R2*R3)

 

Yes what happened in my case was i was looking at the wrong schematic again. This caused the sense terminals for the dependent voltage source to be swapped, which of course changes everything.

Just one question now:
Did you get whole number solutions?

For the circuit shown in post #1, with the signs for Vx and the controlled source shown, I get the whole number solution shown in post #21.

If the sign of Vx is reversed, or the polarity of the controlled source is reversed, I get the solution shown in post #30.

 

Ok well either that cant be right or there are at least two solutions to this circuit.

Just one question:
Did you get whole number solutions for all the nodes?
Oh i checked your link and you seem to have the same solutions set that i have. What changed?
Are you now saying that 42 was with the wrong circuit? That's ok, just wondering now.

This is what I got. The node notation is as in my simulation.
For orginal circuit:
VA = 470, VB = -300
If Vx is swapped:
VA = 470/33, VB = 460/11

Are you now saying that 42 was with the wrong circuit? That's ok, just wondering now.

If you swap Vx polarity then the asnwer very close to 42 but your result is not exactly right. I think you made mistake somewhere, not too important though.

 

This is what I got. The node notation is as in my simulation.
For orginal circuit:
VA = 470, VB = -300
If Vx is swapped:
VA = 470/33, VB = 460/11

If you swap Vx polarity then the asnwer very close to 42 but your result is not exactly right. I think you made mistake somewhere, not too important though.

Yes at first i read the schematic wrong seeing the values as not being what they should have been.
After correcting, i get the same values as you and Electrician.
There is one catch however, and that is that in the original circuit there may be positive feedback and in the real world this would have to be looked into. With the gain of the dependent source equal to 4.00 we get a stable solution set, but with a little change in that gain we get very large voltages at the nodes.
Solving for the problem gain, it comes out to 4.25 which is rather close to 4.00 so a sensitivity analysis would have to be performed to make sure small component value changes dont cause massive instability.
The sensitivity factor for the gain itself is very high, a change of -1 percent causes a node voltage change of some 20 percent.
Leaving the gain of the dependent source a variable, the simplified node voltages are:
[470/(4*G1+17),(470*G1+470)/(4*G1+17),(210*G1+540)/(4*G1+17),-15,(160*G1+1150)/(4*G1+17)]
and in that set to get the original circuit solutions set G1=-4, and to get the swapped terminals solution set G1=4. To see it go nuts set G1=-4.24 or something like that.

 

For the circuit shown in post #1, with the signs for Vx and the controlled source shown, I get the whole number solution shown in post #21.

If the sign of Vx is reversed, or the polarity of the controlled source is reversed, I get the solution shown in post #30.

Yes thanks i think we all have the same results. The only catch is the gain of the dependent source. If that goes just a little bit higher, we get a massive instability.

Leaving the gain of the dependent source a variable, the simplified node voltages are:
[470/(4*G1+17),(470*G1+470)/(4*G1+17),(210*G1+540)/(4*G1+17),-15,(160*G1+1150)/(4*G1+17)]
and in that set to get the original circuit solutions set G1=-4, and to get the swapped terminals solution set G1=4. To see it go nuts set G1=-4.24 or something like that. The theoretical gain that causes infinite response is -4.25 bur circuit simulators may just show a very high number like 1e20 or something.
In fact, a change in gain of -1 percent leads to node voltage change of around 20 percent.
Actually it is easy to see the first node voltage above go to infinity, just solve for zero in the denominator.
4*G1+17=0
and keep in mind a gain of -4 is the same as a gain of +4 in the original circuit.
So a gain of -4.25 is the same as a gain of +4.25 in the original circuit.
So the gain sign has to change in those expressions to match the original circuit, and not changing the sign means we are working with the original circuit with the two sense inputs swapped.

If you dont like being bothered by the sign swapping, we can just set G2=-G1 and get:
[470/(17-4*G2),(470-470*G2)/(17-4*G2),(540-210*G2)/(17-4*G2),-15,(1150-160*G2)/(17-4*G2)]

and in that set the sign of G2 matches the sign of the gain of the original circuit.

 

There is one catch however, and that is that in the original circuit there may be positive feedback and in the real world this would have to be looked into. With the gain of the dependent source equal to 4.00 we get a stable solution set, but with a little change in that gain we get very large voltages at the nodes.

This reminds me of this thread where The Electrician used the same method to conclude that circuit is DC stable when A is smaller than 7.051.
Still hope The Electrician could comment on this and give a final conclusion.
https://forum.allaboutcircuits.com/threads/cascaded-op-amp-circuit-2.180576/#post-1649491

 

This reminds me of this thread where The Electrician used the same method to conclude that circuit is DC stable when A is smaller than 7.051.
Still hope The Electrician could comment on this and give a final conclusion.
https://forum.allaboutcircuits.com/threads/cascaded-op-amp-circuit-2.180576/#post-1649491

That would also be the way a root locus is done. Vary some parameter and see if the roots reache or cross the jw axis. If it reaches the jw axis it's an oscillator, but if it crosses the jw axis it is an oscillator that also ramps up or down to infinity which of course means the circuit does not work.
With an actual real life oscillator, part of the problem is to try to maintain the root or roots on the jw axis. In practices it will vary on and off the jw axis slightly.