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$$y'=y^2$$
This differential equation can be solved by dividing both sides with $y^2$ and integrating both sides to obtain $$y=\frac{1}{c_1-x}$$

But if I differentiate both sides instead, I get $$y''=2yy'=2y^3=2y'^\frac{3}{2}$$
This is a Bernoulli DE that can be solved with the substitution $u=y'^\frac{-1}{2}$ and results in
$$y'=\frac{1}{(-x+c_1)^2}$$
$$y=\frac{1}{c_1-x}+c_2$$

Is the method of differentiating both sides of the DE correct?
It resulted in an extra constant and I'm not sure what went wrong.

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    $\begingroup$ Compare this with how $x=5 \implies x^2=25$, but $x^2=25$ does not imply $x=5$. $\endgroup$
    – Joe
    Dec 11, 2020 at 21:41
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    $\begingroup$ Nothing went wrong, it's just that differentiating both sides is not an invertible operation. You can't distinguish $y' = y^2$ from $y' = y^2 + c$. $\endgroup$
    – Qiaochu Yuan
    Dec 11, 2020 at 21:43

1 Answer 1

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What surprises you? Differentiating the differential equation $y'= 0$ introduces another constant: you have linear solutions in addition to the constants.

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  • $\begingroup$ Does this mean that differentiating both sides of the DE is not a good idea? $\endgroup$
    – helpme
    Dec 12, 2020 at 13:40
  • $\begingroup$ It's a good idea if it helps you find a solution. Just be sure to check that your solution satisfies the original equation. (I suspect that in most cases this strategy will not make the problem easier than solving the original equation.) $\endgroup$
    – Ethan Bolker
    Dec 12, 2020 at 13:53
  • $\begingroup$ Ah, when I substitute y with the extra constant, I can see that the original DE is satisfied when c2=0. Thank you! $\endgroup$
    – helpme
    Dec 12, 2020 at 20:31

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