How does substituting $ u = y' $ in $ y''(t) - y'(t) = y(t) y'(t) $ reduce it to $ uu' − u = yu $?

Question

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 $$ u' - u = yu?$$

Answer 1

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).

Answer 2

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.