The voltage of a capacitor can be described with the differential equation $ \frac {du} {dt} + \frac {1} {RC} u = 0$ where the voltage is u(t) at the time t.
Solve the differential equation:
Don't really know how to solve this one. Would appreciate tips/hints on how to tackle differential equations like this in general.
$\endgroup$ 43 Answers
$\begingroup$$$\frac {du} {dt} + \frac {1} {RC} u = 0$$ $$\frac {du} {dt} =- \frac {u} {RC} $$ $$\frac {du} {u} =- \frac {dt} {RC} $$ Integrate $$\int \frac {du} {u} =-\int \frac {dt} {RC} $$ $$\ln(u)=- \frac {t} {RC} +K $$ $$u(t)=Ke^{- \frac {t} {RC}} $$
$\endgroup$ $\begingroup$Hint:This is equivalent to $\frac{(\frac{du}{dt})}{u}=-1/RC$. Now LHS is just $\frac{d \log u}{dt}$.
$\endgroup$ $\begingroup$Assuming $R$ and $C$ are constant this has the solution
$$u(t) = e^{-t\frac{1}{RC}}$$
This is a first-order homogeneous ODE, for which general solutions are easily availble. Khan Academy has a good tutorial, for example.
$\endgroup$
