i tried to solve differential equation without independent variable
$yy"+y'^2-1=0$
using substitution :
$y'=p$ and
$y"=\frac{dp}{dx}=p \frac{dp}{dy}$
after substitution:
$yp\frac{dp}{dy}+p^2-1=0$
then doing separation variable
this is part that i got confused:
$\frac{p dp}{p^2-1}=- \frac{dy}{y}$
and
$\frac{p dp}{1-p^2}= \frac{dy}{y}$
different place to put minus
i solve both but the gave different result, the first one i got right result, but from second one i got result that i cannot integrate it.
first one:
$yp\frac{dp}{dy}+p^2-1=0$
$\frac{2p}{p^2-1}dp+2 \frac{dy}{y}=0$
$\log (p^2-1)+2\log y= C_1$
$y \sqrt{p^2-1}=C$
$p^2=1+\frac{c}{y^2}$
here if i substitute back $y'=p $, i can integrate y and got the result
second one:
$yp \frac{dp}{dy}=1-p^2$
$\frac{p}{1-p^2}dp=\frac{dy}{y}$
then substitute $1-p^2=u $
$\sqrt{1-p^2}=y.C$
$p=\sqrt{1-y^2C}$
clearly if i substitute back $y'=p $ , i cannot integrate y to get y
so my question : are both of my approach wrong/right? and which one is right? and how to know which approach is right or wrong? thanks!!
In this answer I'll give a different appoach (count it as the third) just in case you interest.
Note that since $(yy')' = y'^2 + yy''$ the differential equation becomes $$(yy')' = 1$$ This yield the solution $$yy' = x + C$$ which can be rerwiten a $y dy = x dx + C dx$ and hence $$\frac{1}{2} y^2 = \frac{1}{2} x^2 + Cx + C_1 $$ for arbitrary constants $C$ and $C_1$.