I am asked to find the general solution of $$R\frac{dq(t)}{dt}+\frac{q(t)}{C}-V_0=0$$

I re-arrange so it is in the correct format.


Integrating factor => $e^{\int{p(x)dx}}$

In this case $p(x)=\frac{1}{CR}$


Multiply through by the IF


Product rule states that => $u\frac{dv}{dx}+v\frac{du}{dx}=\frac{d}{dt}(uv)$ so

$$e^{\frac{t}{CR}}\frac{dq(t)}{dt}+e^{\frac{t}{CR}}\frac{1}{CR}{q(t)}=\frac{d}{dt}(e^{\frac{t}{CR}}q(t))$$ $$\frac{d}{dt}(e^{\frac{t}{CR}}q(t))=e^{\frac{t}{CR}}\frac{V_0}{R}$$

Integrating both sides

$$\int{\frac{d}{dt}(e^{\frac{t}{CR}}q(t))dt=\int{e^{\frac{t}{CR}}\frac{V_0}{R}}}dt$$ $$e^{\frac{t}{CR}}q(t)=\frac{V_0}{R}\int{e^{\frac{t}{CR}}}dt$$

That last step I wasn't sure about. Is $V_0$ a constant in this case? I thought it was dependent on t as well.. Anyway

$$e^{\frac{t}{CR}}q(t)=\frac{V_0}{R}\cdot{\frac{1}{(\frac{1}{CR})}}\cdot{e}^{\frac{t}{CR}}+K$$ $$e^{\frac{t}{CR}}q(t)=\frac{V_0CR}{R}\cdot{e}^{\frac{t}{CR}}+K$$ $$e^{\frac{t}{CR}}q(t)=V_0C\cdot{e}^{\frac{t}{CR}}+K$$ Dividing by $e^{\frac{t}{CR}}$ $$q(t)=V_0C+\frac{K}{e^{\frac{t}{CR}}}$$ (K=the unknown constant)

I have an initial condition that states when t=0, q(t)=0 and am asked to find the particular solution.

$$q(0)=0$$ $$0=V_0C+Ke^{-\frac{0}{CR}}$$ $$=V_0C+Ke^0$$ $$=V_0C+K$$ $$K=-V_0C$$

Subbing K back in

$$q(t)=V_0C+(-V_0Ce^{-\frac{t}{CR}})$$ $$q(t)=V_0C-V_0Ce^{-\frac{t}{CR}}$$

Would this seem correct?

  • $\begingroup$ You said I could write the last line in a cleaner way, could you show me what you mean? Or do you mean $q(t)=V_0C+Ke^{-\frac{t}{CR}}$ $\endgroup$ – user88720 Apr 29 '14 at 23:51
  • $\begingroup$ Okay thanks. I also have an initial condition that says when t=0, q(t)=0. I have put that in the end of my question above although I'm not sure if it is correct $\endgroup$ – user88720 Apr 30 '14 at 1:17
  • $\begingroup$ Sorry when you say recall I'm not sure what you mean. I try to upvote on comments that are helpful although I'm not sure how to accept a comment as an answer $\endgroup$ – user88720 Apr 30 '14 at 1:32

Your solution is perfectly fine.

For the IC, you should get:

$$\large q(t) = C V_0 \left(1-e^{-\frac{t}{C R}}\right)$$

Note: I would recommend solving it using Separation of Variables and comparing the two and convincing yourself why the two are the same.

| cite | improve this answer | |
  • $\begingroup$ Always nice to be able to confirm an OP's work, and expand on it! $\endgroup$ – amWhy Apr 30 '14 at 13:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.