I am trying to solve a very simple ODE using a integrating factor, I have a solution but its trivial and I'm unsure how to find the general solution.
$$\tau \frac{dV}{dt} = E_0 – V$$
Where $E_0$ and $\tau$ are constants. I do the following
\begin{align}
\tau \frac{dV}{dt} &= E_0 – V &\implies \\
\frac{dV}{dt} + \frac{V}{\tau} &= \frac{E_0}{\tau}
\end{align}
Then the integrating factor is $e^{v(t)}$ where $v(t) = \int\frac{dt}{\tau} = \frac t\tau$ so the integrating factor is $e^{{t} / {\tau}}$. Then the solution is given by $$V(t) =e^{{-t} / {\tau}} \int e^{{t} / {\tau}} \frac{E_0}{\tau} =e^{{-t} / {\tau}} e^{{t} / {\tau}} E_0 = E_0 $$
This is a solution since $$\tau \frac{dE_0}{dt} = 0 = E_0 – E_0$$
But I was expecting a more general solution such as $$V = C + Ee^{-t / \tau}$$ without having to "guess" it, can somebody please give me some direction?
Best Answer
Following from above, the general solution is
$$V(t) =e^{{-t} / {\tau}} \int e^{{t} / {\tau}} \frac{E_0}{\tau} \ dt = E_0 e^{-t/\tau} \left( e^{t/\tau} + C' \right) = E_0 + Ce^{-t/\tau} $$
(where $C = E_0C'$)