Question

i

Answer 1

I cant see an easy way to do this !!!

we have\[yy''+y'^2=\frac{1}{2y}\]let \(f=yy'\) and we will get\[f'=\frac{df}{dx}=\color\red{\frac{df}{dy}\frac{dy}{dx}}=yy''+y'^2=\color\green{\frac{1}{2y}}\]it gives\[\frac{df}{dy}\frac{dy}{dx}=\frac{1}{2y}\]\[2\frac{df}{dy}yy'=1\]and we know that \(f=yy'\) so\[2fdf=dy\]integrating gives\[f^2=y+c\]\[f=\pm \sqrt{y+c}\]\[yy'=\pm \sqrt{y+c}\]and finally\[\frac{ydy}{\sqrt{y+c}}=\pm dx\]Let's see what will be others opinions on this problem...maybe there are easier ways to do this :)