# Simmultanous equations

Discussion in 'Math' started by jagjit Sehra, Mar 4, 2008.

1. ### jagjit Sehra Thread Starter Member

Feb 29, 2008
25
0
Hi Folks,
Got a question regarding simmultanous equations. I am studying Electronics and have been looing at a certain type of response. The crtically damped response.

vC = Aexp(mt) + Btexp(mt)
where m is one value lets say -1. vC has intial

value of 2.

The problem is to obtain the values for A and B.

The derivative of vC with respect time.

dvC/dt = mAexp(mt) + Bexp(mt) + mBtexp(mt)

(differentiated the second term using the product rule)

assuming t = 0
Putting values in.

2 = Aemt + Btemt
0 = mAemt + Bemt + mBtemt

2 = A + 0 A must equal 2
0 = -A + B + 0

0 = -2 + B B must equal 2

Therefore A and B both equal 2. Is that correct the way I have done it! It doesnt look right!

2. ### Mark44 Well-Known Member

Nov 26, 2007
626
1
It's not clear to me what you're doing. At the beginning you say assume that m = -1, but you don't use that value in your later work.

After taking the derivative, it looks like you are setting d(VC)/dt to 0, but you don't mention that this is given information.

Using all the information that you have, does A = B = 2 make sense? That is, do these values satisfy both the original equation and its derivative when t=0, and presumably for m=-1?

Mark

3. ### Mark44 Well-Known Member

Nov 26, 2007
626
1
If you have these equations:

V(t) = A*exp(-t) + B*t*exp(-t)
V'(t) = -A*exp(-t) -B*t*exp(-t) + B*exp(-t)

with these initial conditions:
V(0) = 2, V'(0) = 0

Then, yes, your work is correct; namely, that A = B = 2.

I used your assumed value of m = 1.
Mark

4. ### jagjit Sehra Thread Starter Member

Feb 29, 2008
25
0
Thanks Mark for confirming that. I must also apologise for bieng so vague with the description of the problem. I must say I have found this foroumn very usefull and hopefully one day day I can help somone aswell!