Differential equation: capacitor
Solution 1
$$\frac {du} {dt} + \frac {1} {RC} u = 0$$ $$\frac {du} {dt} = \frac {u} {RC} $$ $$\frac {du} {u} = \frac {dt} {RC} $$ Integrate $$\int \frac {du} {u} =\int \frac {dt} {RC} $$ $$\ln(u)= \frac {t} {RC} +K $$ $$u(t)=Ke^{ \frac {t} {RC}} $$
Solution 2
Hint:This is equivalent to $\frac{(\frac{du}{dt})}{u}=1/RC$. Now LHS is just $\frac{d \log u}{dt}$.
Solution 3
Assuming $R$ and $C$ are constant this has the solution
$$u(t) = e^{t\frac{1}{RC}}$$
This is a firstorder homogeneous ODE, for which general solutions are easily availble. Khan Academy has a good tutorial, for example.
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gbgult
Updated on July 23, 2022Comments

gbgult over 1 year
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.

Paul over 5 yearsThis is the most basic first order homogeneous constant coefficient differential equation. Look in any engineering maths textbook.

snulty over 5 yearsA usual idea for a homogenous equation with constant coefficients is to try an exponential solution $u=e^{\lambda t}$. On the other hand since it's first order (one derivative) you can try separate the variables, bring $u$'s to one side $t$'s to the other

gbgult over 5 yearsYeah, I misread the question and didn't realise that R and C were constants. Thus my confusion. I'll leave the question be though because someone might find it useful.

hardmath over 5 yearsIf you want to leave it, that's fine by me. It is worth noting that this example exhibits exponential decay because $\lambda = 1/RC$ is negative. Whatever (voltage) charge is initial placed on the capacitor, it will be discharged more quickly with lower resistance $R$.
