# Finding a general solution to a first order linear differential equation, using the integration factor method

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?

• When integrating $\int e^{{t} / {\tau}} \frac{E_0}{\tau} \ dt$ be sure to put in an arbitrary constant. (This is one of the reasons the integration constant $+C$ is drummed into us early in our calculus education!) – Simon S Jun 8 '15 at 18:35
• Wow that was such a school boy error. Thank you @SimonS . – HBeel Jun 8 '15 at 18:38
• I assume that adding a constant when working out $v(t)$ won't add anything to the general solution and just make the calculation longer, so I can let the constant of integration $0$ there ? – HBeel Jun 8 '15 at 18:39
• I'm not following you. See below. – Simon S Jun 8 '15 at 18:41

## 1 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 $C = E_0C'$)