Differential equation: capacitor [closed]

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.

closed as off-topic by Namaste, Tom-Tom, John B, Ethan Bolker, Chris GodsilMar 7 '18 at 0:00

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• This is the most basic first order homogeneous constant coefficient differential equation. Look in any engineering maths textbook. – Paul Mar 6 '18 at 10:01
• A 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 – snulty Mar 6 '18 at 10:01
• Yeah, 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. – gbgult Mar 6 '18 at 10:05
• If 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 $R$. – hardmath Mar 6 '18 at 22:59

$$\frac {du} {dt} + \frac {1} {RC} u = 0$$ $$\frac {du} {dt} =- \frac {u} {RC}$$ $$\frac {du} {u} =- \frac {dt} {RC}$$ Integrate $$\int \frac {du} {u} =-\int \frac {dt} {RC}$$ $$\ln(u)=- \frac {t} {RC} +K$$ $$u(t)=Ke^{- \frac {t} {RC}}$$
Hint:This is equivalent to $\frac{(\frac{du}{dt})}{u}=-1/RC$. Now LHS is just $\frac{d \log u}{dt}$.
Assuming $R$ and $C$ are constant this has the solution
$$u(t) = e^{-t\frac{1}{RC}}$$