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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?

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    $\begingroup$ You divided by $y$, which doesn’t work if $y=0$ $\endgroup$ Sep 11, 2020 at 16:51
  • $\begingroup$ @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? $\endgroup$
    – actinidia
    Sep 11, 2020 at 16:52

2 Answers 2

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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.

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Is there a general way to solve these that won't require me to check restricted solutions afterwards

Yes integrating factor method is easier here: $$y'+5y=0$$ $$(ye^{5x})'=0$$ $$ye^{5x}=C$$ $$y=Ce^{-5x}$$

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