Solve the following differential equation: `(x^3+y^3)dy-x^2y dx=0`

Explanation

The given differential equation may be written as
`(dy)/(dx) =(x^(2)y)/(x^(3)+y^(3))`. ………………….(i)
On dividing the Nr and Dr of RHS of (i) by `x^(3)`, we get
`(dy)/(dx)=(y/x)/(1+(y/x)^(3))=f(y/x)`
So, the given differential equation is homogeneous.
Putting `v=vx` and `(dy)/(dx)=v+x(dv)/(dx)` in (i), we get
`v+x(dv)/(dx)=(vx^(3))/(x^(3)+v^(3)x^(3))`
`rArr v+x(dv)/(dx)=v/(1+v^(3))`
`rArr x(dv)/(dx)=(v/(1+v^(3))-v)=-v^(4)/(1+v^(3))`
`rArr ((1+v^(3))/(v^(4)))dv=-1/xdx`
`int(1/v^(4)+1/v)dv+int1/xdx=C`
`rArr -1/(3v^(3))+log|vx|=C`
`rArr -x^(3)/(3y^(3))+log|y|=C`, which is the required solution.