y'=\int y^{-\frac{1}{2}} dx=...
y''= y^{-\frac{1}{2}} y'y''= y'y^{-\frac{1}{2}}
\int y'y'' dx=\int y^{-\frac{1}{2}} y'dx
\frac {y'^{2}}{2} = 2y^{\frac{1}{2}} +k
y'^{2} = 4y^{\frac{1}{2}} +2k=4y^{\frac{1}{2}} +c
y' = \sqrt{4y^{\frac{1}{2}} +c}
\frac{y'}{\sqrt{4y^{\frac{1}{2}} +c}} = 1 \int \frac{y'}{\sqrt{4y^{\frac{1}{2}} +c}} dx=\int dx
If you select $ u^{2}= 4y^{\frac{1}{2}} +c $
$ y= \frac{(u^{2}-c)^{2}}{16} $
y'= uu'\frac{(u^{2}-c)}{4}
\int \frac{u'u\frac{(u^{2}-c)}{4}}{u} dx=\int dx
\int \frac{u'u\frac{(u^{2}-c)}{4}}{u} dx=x+c_1
\int \frac{u'u^{2}}{4} dx -\int \frac{u'c}{4} dx=x+c_1
$\frac{u^{3}}{12} -\frac{cu}{4} =x+c_1$
After solving cubic equation you must put $ u= \sqrt{4y^{\frac{1}{2}} +c} $
then you must find y depend on X