How to find the general solution for #xy (dy/dx - 1)= x^2 + y^2# ?

Question

Homogeneous differential equations

1038 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-20T12:16:01

#y/x=ln|y+x|+C#

Given

#xy(y'-1)=x^2+y^2#

Rearrange:

#y'=1+x/y+y/x#

Apply the substitution #y(x)=xv(x)#:

#v+xv'=1+1/v+v#

Simplify and collect like terms:

#v/(v+1)v'=1/x#

Form the integral:

#int(1-1/(v+1))dv=int1/xdx#

Integrate term by term:

#v-ln|v+1|=ln|x|+C#

Simplify:

#v=ln|xv+x|+C#

Reverse the substitution:

#y/x=ln|y+x|+C#