0
$\begingroup$

$$y'=y^2$$
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.

$\endgroup$
2
  • 2
    $\begingroup$ Compare this with how $x=5 \implies x^2=25$, but $x^2=25$ does not imply $x=5$. $\endgroup$ – Joe Dec 11 '20 at 21:41
  • 3
    $\begingroup$ 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 '20 at 21:43
1
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ Does this mean that differentiating both sides of the DE is not a good idea? $\endgroup$ – helpme Dec 12 '20 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 '20 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 '20 at 20:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.