<s>[INST] Easy way to solve this non-linear second degree DY $\sqrt{y}\;y''=1$? I prooceeded by integrating both sides $$y'=\int y^{-\frac{1}{2}} dx=\cdots$$
so I got $(y')^{2}+C y' - \frac{1}{2} \sqrt{y} = 0$ but I am thinking that I am proceeding the wrong or the hard way. Some easy to solve this kind of 2nd-degree DYs?
Page 633 on the book I have been reading earlier.
 [/INST] $$y'=\int y^{-\frac{1}{2}} dx=...$$
$$y''= y^{-\frac{1}{2}} $$
$$y'y''= y'y^{-\frac{1}{2}} $$
$$\int y'y'' dx=\int y^{-\frac{1}{2}} y'dx$$
$$\frac {y'^{2}}{2} = 2y^{\frac{1}{2}} +k $$
$$y'^{2} = 4y^{\frac{1}{2}} +2k=4y^{\frac{1}{2}} +c $$
$$y' = \sqrt{4y^{\frac{1}{2}} +c} $$
$$\frac{y'}{\sqrt{4y^{\frac{1}{2}} +c}} = 1 $$
$$\int \frac{y'}{\sqrt{4y^{\frac{1}{2}} +c}} dx=\int dx$$
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
</s>