How Does Charge in a Capacitor Satisfy the Given Differential Equation?

[TFM]

Well,

$$p(t) = \frac{1}{RC}$$

$$q(t) = \frac{V_0}{R} sin\omega t$$

So:

$$F = \int \frac{1}{RC}dt$$

$$F = \frac{t}{RC}$$

so:

$$e^{\frac{t}{RC}}x = \int e^{\frac{t}{RC}}\frac{V_0}{R} sin\omega t dt + C$$

TFM

 

[alphysicist]

In the last line you want a Q instead of an x.

 

[TFM]

Oops...

$$e^{\frac{t}{RC}}Q = \int e^{\frac{t}{RC}}\frac{V_0}{R} sin\omega t dt + C$$

So do I integrate the right side now, with what was given in the question?

TFM

 

[alphysicist]

Oops...

$$e^{\frac{t}{RC}}Q = \int e^{\frac{t}{RC}}\frac{V_0}{R} sin\omega t dt + C$$

So do I integrate the right side now, with what was given in the question?

TFM

Yes.

 

[TFM]

So:

$$e^{\frac{t}{RC}}Q = \int e^{\frac{t}{RC}}\frac{V_0}{R} sin\omega t dt + C$$

$$e^{\frac{t}{RC}}Q = \frac{V_0}{R}\int e^{\frac{t}{RC}} sin\omega t dt + C$$

Using the replacement formula:

$$\int e^{t/\alpha} sin\omega t dt = \alpha e^{t/\alpha}\frac{(sin \omega t - \alpha \omega cos \omega t)}{1 + \alpha ^2\omega^2} + Constant$$

With $$\alpha = RC$$

Gives:

$$e^{t/RC}q = \frac{V_0}{R} \frac{RCe^{t/RC}sin\omega t-RC\omega cos\omega t}{1+(RC)^2\omega^2} + Constant$$

TFM

 

[alphysicist]

So:

$$e^{\frac{t}{RC}}Q = \int e^{\frac{t}{RC}}\frac{V_0}{R} sin\omega t dt + C$$

$$e^{\frac{t}{RC}}Q = \frac{V_0}{R}\int e^{\frac{t}{RC}} sin\omega t dt + C$$

Using the replacement formula:

$$\int e^{t/\alpha} sin\omega t dt = \alpha e^{t/\alpha}\frac{(sin \omega t - \alpha \omega cos \omega t)}{1 + \alpha ^2\omega^2} + Constant$$

With $$\alpha = RC$$

Gives:

$$e^{t/RC}q = \frac{V_0}{R} \frac{RCe^{t/RC}sin\omega t-RC\omega cos\omega t}{1+(RC)^2\omega^2} + Constant$$

TFM

Looks good so far (except keep the parenthesis that are in the formula around the $(\sin\omega t-\alpha\omega \cos\omega t)$).

 

[TFM]

So:

$$e^{t/RC}q = \frac{V_0}{R} \frac{RCe^{t/RC}(sin\omega t-RC\omega cos\omega t)}{1+(RC)^2\omega^2} + Constant$$

So should I rearrange to get q:

$$q = \frac{\frac{V_0}{R} \frac{RCe^{t/RC}(sin\omega t-RC\omega cos\omega t)}{1+(RC)^2\omega^2} + Constant}{e^{t/RC}}$$

??

TFM

 

[alphysicist]

So:

$$e^{t/RC}q = \frac{V_0}{R} \frac{RCe^{t/RC}(sin\omega t-RC\omega cos\omega t)}{1+(RC)^2\omega^2} + Constant$$

So should I rearrange to get q:

$$q = \frac{\frac{V_0}{R} \frac{RCe^{t/RC}(sin\omega t-RC\omega cos\omega t)}{1+(RC)^2\omega^2} + Constant}{e^{t/RC}}$$

??

TFM

Okay; now I would suggest you write the right hand side as three terms (one with a sine, one with a cosine, and one with the constant). Once you have that, you can determine the constant.