# Missing solutions for the ODE $y' + 5y = 0$

Let's solve the ODE $$y' + 5y = 0$$ through separation of variables.

We get

\begin{align*} \frac 1y y' &= -5 \\ \log|y| &= -5x+c \\ |y| &= \exp(-5x+c) \\ y &= \pm \exp(-5x+c) \\ &= \pm \exp(c)\exp(-5x) \\ &= \pm c_1 \exp(-5x), c_1 > 0 \end{align*}

How come the solution where $$c_1 = 0$$ is missing?

• You divided by $y$, which doesn’t work if $y=0$ Sep 11, 2020 at 16:51
• @J.W.Tanner Oh of course! Thanks. Is there a general way to solve these that won't require me to check restricted solutions afterwards? Sep 11, 2020 at 16:52

When you solve an equation by separation of variables $$y'=f(x)g(y)$$ you could have some singular solutions (solutions that don't show up in the general integral) by taking a constant solution $$y=y_0.$$ If you substitute, you get $$0=f(x)g(y_0)$$ from which you can conclude that every zero of $$g$$ is a constant solution.
Yes integrating factor method is easier here: $$y'+5y=0$$ $$(ye^{5x})'=0$$ $$ye^{5x}=C$$ $$y=Ce^{-5x}$$