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I am trying to understand a substitution step at http://www.ams.sunysb.edu/~jiao/teaching/ams501_fall11/notes/hiorder_nonlinear.pdf (page 2).

Here is the step I am unable to understand:

... we obtain $$ y''(t) - y'(t) = y(t) y'(t). $$ This is an automanous equation. The substitution $ u = y' $ reduces the order to first-order equation $$ uu' − u = yu. $$

Shouldn't the substitution $ u(t) = y'(t) $ reduce $$ y''(t) - y'(t) = y(t) y'(t) $$ to $$ u'(t) - u(t) = y(t) u(t), $$ that is $$ u' - u = yu?$$

Lone Learner
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    It's not $u(t)=y'(t)$. If that were the case, you would be right. But it is $u(y)=y(t)$; and your link explains clearly how to evaluate $y''(t)$ in that case. – TonyK Nov 07 '21 at 18:55
  • See also https://math.stackexchange.com/questions/4227031/under-what-circumstances-is-ddotx-dotx-fracd-dotxd, https://math.stackexchange.com/questions/3065331/showing-ddotx-frac-mathrmd-mathrmdx-frac12-dotx2 – Lutz Lehmann Nov 07 '21 at 22:26

2 Answers2

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If you read the first page of the notes: $$ y''(x) = \frac{du}{dx}=\frac{du}{dy}\frac{dy}{dx}=u'(y)u(y)\tag{1} $$

The substitution used is $u(y)=y'(x)$. In your example on page 2, replace $x$ with $t$ in (1).

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It's $u'=\dfrac {du}{dy}$. You can also integrate without substitution: $$y''(t) - y'(t) = y(t) y'(t)$$ $$y''(t) - y'(t) = \dfrac 12 (y^2(t))'$$ Integrate: $$y'-y=\dfrac { y^2}2+C_1$$ This DE is separable.

user577215664
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