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!
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
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
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!