# Differentiating a 1st order ODE: really dumb question

1. Oct 15, 2012

### cepheid

Staff Emeritus
Suppose I have a really simple first order linear ODE like:$$\dot{\omega} = -k\omega$$ where k is some constant, ω(t) is a function of time that I want to solve for, and the overdot denotes the derivative w.r.t. time. This is really easy to solve, and we all know that with the initial condition ω(0) = ω0, the solution is given by ω(t) = ω0exp(-kt). Now, I was thinking. You can find a solution that makes this equation true. If this equation is true, then if I differentiate both sides, then the result should also be true right :uhh: (realizes there may be some sketchy reasoning here). So if I differentiate both sides, I get$$\ddot{\omega} = -k\dot{\omega}$$If the first ODE was true given the solution above, this ODE should also be true given that solution. And that is the case, because for that solution, $\ddot{\omega} = k^2\omega = -k\dot{\omega}$. Now, if I use the symbol $\alpha$ to denote $\dot{\omega}$ then the equation is $$\dot{\alpha} = -k\alpha$$The solution to *this* equation, assuming an initial condition of α(0) = α0, is, of course, α(t) = α0exp(-kt). Now, if I solve for ω(t) by integrating α(t), still assuming that ω(0) = ω0, then I get:$$\omega(t) - \omega_0 = \int_0^t \alpha_0\exp(-kt^\prime)\,dt^\prime$$ $$\omega(t) = \omega_0 + \frac{\alpha_0}{k}[1 - \exp(-kt)]$$So, this solution doesn't have the same form as the original ω(t). What stupid thing have I overlooked here?

2. Oct 15, 2012

### D H

Staff Emeritus
What you've overlooked is that $\alpha_0$ is not a free variable. It is instead given by $\alpha_0 = -k\omega_0$. Substitute this into your second $\omega(t)$ and tada! you recover the first one.

3. Oct 15, 2012

### cepheid

Staff Emeritus
Thanks for the help! This occurred to me on the walk home. I kind of wish it had occurred to me before posting, but posting definitely helped me sort things out.