Why solve cannot solve this system of diff eq with 3 eq?

[Cornel]

Hi,

I am making something wrong? Are only 3 differential equations...

But using laplace command this system of differential eq can be solved:

See MCP9 file attached.

[Werner_E · Accepted solution]

It may be because of the dependency of the three function in equation #2

This works (derivatives of vc1 equals derivative of vc3:

v.C3(0) cannot be chosen freely!

[Cornel]

This is the circuit:
@ttokoro 

[LucMeekes]

The situation of your circuit is impossible.

To the left of the switch, at t=-0 you have a capacitor C3 charged to 5.7 V, in parallel to the series connection of a diode, that you model with a (constant) voltage source of 0.6 V, and a capacitor C1.

This means that the voltage across C1 must be 5.7-0.6=5.1 V, not the 0 V that you set it.

Success!
Luc

[ttokoro]

[Cornel]

As in your suggested case:

This means that in this case voltage at t=0 for C3 to be 5.7V but also we need to force that voltage at t=0 for C1 to be 5V.

But in this case:

We could let voltage at t=0 for C3 to be 5.7V, but also we could let that voltage at t=0 for C1 to be 0V.

What do you think, what is the good approach?

[Cornel]

For my and my checking it seems that the variant with laplace is ok, but then this means that solve command cannot solve such a system of diff eq while laplace can

[Werner_E]

Your second equation clearly states that v3(t)=v1(t)+VD.

Given that VD is constant and not dependent on t this means that you can't demand an independent initial condition v3(0)=... because v3(0) must equal v1(0)+vD as in my approach above - I forgot to use "simplify so the expression looked far too complicated 😉

Your second equation also means that v3'(t)=v1'(t) and v3' can be replaced by v1' as I did.

So if your equation is correct, you just have a system with two functions to solve for. The third function v3 must be derived as shown above.