# Proving the Form of a Homogenous LC Circuit?

1. May 21, 2013

1. The problem statement, all variables and given/known data

I have a circuit with input source x(t), which contains also an inductor and a capacitor in series which I have found to be related to the output voltage y(t) (across the capacitor) like so: LC*d2y/dt2 + y(t) = x(t). I have also found its roots through the quadratic equation to be complex, and equal to +/- the imaginary unit j.

The question asks me to show its homogenous solution has the form K1ej*a*t + K2ej*b*t, where a and b are the roots I found above.

2. Relevant equations

Unsure. See above for my progress so far.

3. The attempt at a solution

My solution has led me here, having gotten the above equation and its roots. I just do not know how to show that I can express the solution to the differential equation in the requested form. Any help would be greatly appreciated!

2. May 21, 2013

### tiny-tim

If you have a homogeneous quadratic equation in y(t),

write D = d/dt, and convert the equation into an ordinary quadratic equation in D …

(D - a)(D - b)(y) = 0​

then put (D - b)(y) = z, so (D - a)(z) = 0

3. May 21, 2013

I follow, but that does not appear to express it in the form K1*e^(j*a*t) + K2*e^(j*b*t). I don't think it makes it any simpler either, at least not to my eyes.

4. May 21, 2013

### tiny-tim

yes it does, try it

5. May 21, 2013

Expanding, I get d^2y/dt^2 - (dy/dt)*b - (dy/dt)*a + ab.

Your z substitution is a little confusing, but then I get (dy/dt-a)*(dy/dt-b) = 0 again?

6. May 21, 2013

### tiny-tim

solve (D - a)z(t) = 0 first …

z = … ?

(and then solve the non-homogeneous equation (D - b)y(t) = z(t))