Question

I have a circuit and I'm trying to find the general solution for a homogeneous equation.

My differential equation is:

q"+8q'+25q=75

Now, I know that you need to equate it to you so the equation would be: q"+8q'+25q=0

Letting q=e^kt and so on and so forth I've got a gen sol : C1e^(-4t)cos3t+C2e^(-4t)sin3t

I've been told that initially the charge on the capacitor and the current in the circuit are zero. So, q(0)=0 and q'(0)=0

How do I find the constants using these initial conditions? I'm looking for guidelines and not the actual answer. Thanks for any help in advance.

Answer 1

Well, you aren't quite done, after you find the homogenous solution you need to find the particular solution. Which is in a form C. So:
\[y_P(x)=C;y'_P(x)=0;y''_P(x)=0\]
Plugging this in your see.
\[25C=75 \implies C=3\]
So your solution would be:
\[y(t)=y_H(t)+y_P(t)=c_1e^{-4t}\cos(3t)+c_2e^{-4t}\sin(3t)\]
To solve for the constants you simply plug in 0 wherever you see t. Then set your y(t)=q(0)=0.
Then differentiate it and do the same thing:
y'(t)=q'(0)=0
You'll get a system of equations. Solve for the constants using linear algebra.