RC Circuit Differential Equation

[sritter27]

Homework Statement

Find the general solution of $$L \frac{d^2I}{dt^2} + R \frac{dI}{dt} + \frac{I}{C} = \frac{dV}{dt}$$ given $$L = 0$$ and $$V = V_0 cos(\omega t)$$.

Homework Equations

The Attempt at a Solution

So the equation basically turns into a first-order RC circuit equation $$R \frac{dI}{dt} + \frac{I}{C} = \frac{dV}{dt}$$, but I'm not sure how to approach it to find a general solution.

The answer the book gives is $$I = Ae^{-\frac{t}{RC}} - \frac{V_0 \omega C (sin(\omega t) - \omega R C cos(\omega t))}{1 + \omega^2 R^2 C^2}$$ and I'm not sure how they came to that conclusion, so any help or nudge in the right direction would be greatly appreciated.

 

[tiny-tim]

Welcome to PF!

$$R \frac{dI}{dt} + \frac{I}{C} = \frac{dV}{dt}$$, but I'm not sure how to approach it to find a general solution.

The answer the book gives is $$I = Ae^{-\frac{t}{RC}} - \frac{V_0 \omega C (sin(\omega t) - \omega R C cos(\omega t))}{1 + \omega^2 R^2 C^2}$$ and I'm not sure how they came to that conclusion, so any help or nudge in the right direction would be greatly appreciated.

Hi sritter27! Welcome to PF!

To find the general solution of R dI/dt + I/C = -ωV₀sinωt,

assume it's of the form Acosωt + Bsinωt, and you get the given result,

except that you've copied it wrong … it's $$V_0 \omega C\frac{ (sin(\omega t) - \omega R C cos(\omega t))}{1 + \omega^2 R^2 C^2}$$

 

[sritter27]

Oh wow I should have been able to see that. My many thanks for the help!