Show that the differential equation `(x^(2)+xy)dy=(x^(2)+y^(2))dx` is homogenous and solve it.

Explanation

The given differential equation is
`(dy)/(dx) = (x^(2)+y^(2))/(x^(2)+xy)`……………..(1)
Let `F(x,y)=(x^(2)+y^(2))/(x^(2)+xy)`
Now, `F(lambdax,lambday) = ((lambdax)^(2)+(lambday)^(2))/((lambdax)^(2)+(lambdax)(lambday)) = (x^(2)+y^(2))/(x^(2)+xy) = lambda^(0).F(x,y)`.
This shows that equation (1) is a homogenous equation.
To solve it, we make the substitution as
`y=vx`
Differentiating both sides with respect to x, we get
`(dy)/(dx) =v+x(dv)/(dx)`
Substituting the value of v and `(dy)/(dx)` in equation (1), we get
`v+x(dv)/(dx)=(1+v^(2))/(1+v)`
or `x(dv)/(dx) = (1+v^(2))/(1+v)-v=(1-v)/(1+v)`
or `(1+v)/(1-v)dv=(dx)/(x)`
or `(2-1+v)/(1-v)dv =(dx)/(x)`
or `(2/(1-v)-1)dv=(dx)/(x)`
Integrating both sides, we get
`int(2/(1-v)-1)dv=int(dx)/(x)`
or `-2log(1-v)-v=logx+logC`
or `v=-2log(1-v)-logx+logC`
or `(Cx)/(x-y)^(2)=e^(y/x)`
or `(x-y)^(2) = Cx^(-y/x)`