#x[x^2+y^2]dy/dx=y[y^2-x^2]#, #rArr# #dy/dx=[y^3-yx^2]/[x^3+y^2]#.......#[1]#
Dividing numerator and denominator on the right hand side of .....#[1]# by #x^3# and substituting #v=y/x# will give a homogeneous equation in #v# of the form.......
#dy/dx=[v^3-v]/[1+v^2]# . From #v=y/x#, #rArr# #y=vx#.......#[2]#
Differentiating.....#[2]# w.r.t #x# [implicitly], #dy/dx = x[dv]/[dx]+v# and sustituting for #dy/dx# we obtain,
#x[dv]/[dx] # = #[ v^3-v]/[1+v^2]-v#, i.e, #x[dv]/[dx]=[-2v]/[1+v^2]# which will yield
#dx/x=-[1+v^2]/ [2v]dv# , therefore, #-dx/x=[1/[2v]+v^2/[2v]]dv#.
Integrating both sides, #-lnx=1/2ln v+v^2/4 + C# and multiplying both sides by #2# and let #C# =#lnA# will give,
#-lnx^2=lnv+v^2/2+ln A#, tidying up, #ln[Ax^2v]=-v^2/2#,
#Ax^2v# = #e^-[v^2/2]#[ where #1/A# is equal to another constant #B # given in the answer ] and so,
Substituting back for #v=y/x#, we will obtain the answer given above. I hope this was both helpful and correct, and will ask for it to be checked.
