From Mathematica:DSolve[y′[x]=(y[x]2−x2)2xy[x],y[x],x]{{y[x]→−−x2+xC[1]−−−−−−−−−−√},{y[x]→−x2+xC[1]−−−−−−−−−−√}}Take the second solution and square each side:y[x]2=−x2+xC[1]Move the -x^2 from the RHS to the LHS.y[x]2+x2=xC[1]If the same procedure is applied to my answer then the following is the result:y[x]2+x2= 2cx+2C[1]where C[1] is zero. I am not sure that their expression is totally correct.
