I have the differential equation $R\frac{dQ}{dt}+\frac{Q(t)}{C}=V(t)$ where $R,C\in\mathbb R$ and $Q,V$ are functions of $t$. If I take the laplace transform of the differential equations I get:

$$\mathscr Q = \frac{\mathscr V+RQ_0}{Rs+\frac{1}{C}} = \frac{\mathscr V}{Rs+\frac{1}{C}}+\frac{Q_0}{s+\frac{1}{RC}},$$ where $\mathscr {Q,V}$ are the Laplace transforms of $Q,V$ respectively. The term $\frac{Q_0}{s+\frac{1}{RC}}$ is of course very easy to solve ($Q_0=Q(0)$). But I'm having trouble with the inverse Laplace transform of $\frac{\mathscr V}{Rs+\frac{1}{C}}$. Can anyone help me please? Thanks!


1 Answer 1


Write it as $$ \frac{\mathscr V}{Rs+\frac{1}{C}} = \frac{1}{R} \frac{\mathscr{V}}{s+1/(RC)}, $$ or, better, just take the original term and write it as $$ \frac{\mathscr V+RQ_0}{Rs+\frac{1}{C}} = \frac{\mathscr V/R+Q_0}{s+\frac{1}{CR}}, $$ then invert, by recalling that $$ \mathcal{L}(be^{-at})(s) = \int_0^{\infty} e^{-st} be^{-at} \, dt = \frac{b}{s+a}, $$ and you have the convolution theorem, so the answer is $$ Q(0)e^{-t/(CR)}+\int_0^t e^{-(t-\tau)/(CR)} V(\tau) \, d\tau $$

Probably the better way to do it is with the integrating factor $e^{t/(CR)}/R$, which gives you $$ (e^{t/(CR)}Q(t))' = e^{t/(CR)}V(t)/R, $$ which you can the integrate directly.

  • $\begingroup$ I'm sorry, I'm not seeing it :( $\endgroup$ Apr 2, 2015 at 0:39
  • 1
    $\begingroup$ Oh, $\mathscr V$'s a function... very well, I'll add some more detail. $\endgroup$
    – Chappers
    Apr 2, 2015 at 1:22
  • $\begingroup$ Yes, I was checking that convolution theorem. Though it is a bit unusual to give an answer in terms of the convolution of two functions. And it isn't very clear to work with. Thanks for your help. $\endgroup$ Apr 2, 2015 at 1:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .