This is an after-class exercise in Tongji textbook, and the answer can be found in any advanced mathematics tutorial book.

Method 1:

Let y'=p(y), then y''=p'*dy/dx=p'p, so the original formula is: yp' p+p 2 = 0, that is, yp' =-p.

After separating the variables: dp/p=-dy/y, the two-side integration results in: ln|p|=-ln|y|+ln|C 1|, that is, p = c1/y.

Then dy/dx=C 1/y, and then the variables are separated: ydy=C 1dx.

The integrals on both sides are1/2y 2 = c1x+C2, that is, y 2 = c3x+C4.

The initial conditions are your own.

Method 2: y'' y+(y') 2 = 0 Deduce (yy')'=0, then yy'=C 1, then 1/2 (y 2)' = c 1.

Then (y 2)' = C2, then y 2 = C2X+C3.