Second-order linear voltage equation

[ongxom]

Homework Statement

v_s = 300 (t>0)
v_s = 150 (t<0)
v(0⁺)≠0
i(0⁺)≠0
R=50ohm
L=10H
C=1/100F

Determine v(t) when t>0

Homework Equations

The Attempt at a Solution

I wrote node voltage equation
(1/R)*v+(1/L)∫(v-v_s)dt+C.(dv/dt)=0

First I do is get derivation to two sides of the equation to escape the integral, after that, replace all known datas, I got the equation
v''+2v+10v=10v_s
So we have A(s)=s²+2s+10=0 ⇔ s=-1±3i
v(t)=v_n+v_p
which v_n = 150V (cause we have t>0)
→ v(t)=150 + e$^{-t}$.(Acos3t+Bsin3t)

I am stucking on the way to find A and B constants, all I can do is below
v(0⁺)=150+A
v(0⁺)'=-A+3B

 

[rude man]

EDIT: determine the initial conditions v (0+) and dv/dt (0+) and then solve your differential equation with those two initial conditions.

You should be able to determine v(0+) and v'(0+) by inspecting the diagram. v(0+) is fairly obvious (voltage cannot change instantaneously across a capacitor) but v'(0+) requires some thought.

Hint: i = C dv/dt. What is initial current thru the capacitor?

Second hint: your equation to solve is
v'' + 2v' + 10 = 10Vs = 3000. The 150V for t < 0 does not come into the equation directly as you have shown.

Your equation for v is actually almost correct except for the "150". But you need to work on the initial conditions. When you have those you can determine A and B.

 

[ongxom]

So my correct equation for v(t) should be v(t)= 300 + e$^{-t}$.(Asin3t+Bcos3t), right ?
First hint : i(0⁺)=C.d(v0⁺)/dt ?
Second hint : I think the equation should be v'' + 2v' + 10v = 3000, but what is the reason to solve it ?

 

[rude man]

So my correct equation for v(t) should be v(t)= 300 + e$^{-t}$.(Asin3t+Bcos3t), right ?

right

First hint : i(0⁺)=C.d(v0⁺)/dt ?

can you use that to determine v'(0+)? Think about what the currents are doing going thru L, R and C just after the switch is thrown at t = 0+.

Second hint : I think the equation should be v'' + 2v' + 10v = 3000, but what is the reason to solve it ?

Well, by now you already have solved part of it. I just meant that's the full equation you need to solve. And you haven't until you figured out A and B.

Now you need to finish up by solving for A and B given the initial conditions on v and v'.

 

[ongxom]

right

can you use that to determine v'(0+)? Think about what the currents are doing going thru L, R and C just after the switch is thrown at t = 0+.

I am stucking at this point. Have no idea for it .

 

[rude man]

I am stucking at this point. Have no idea for it .

1. Have you come up with v(0+) yet?

2. What is the current thru the inductor just before switching the source from 150V to 300V, i.e. t = 0-?
What is the current thru R at that time?
What is the current into C? (Obviously zero since the source had been = 150V for a long time).

OK, so now we switch the source voltage from 150V to 300V and it's t = t(0+). What now is the current thru L and R? And C?
Then, i_C = C dv/dt. Doesn't that tell you what v'(0+) = dv/dt at t = t(0+) must be?