Solve the differential equation `x^2dy+y(x+y)dx=0,` given that `y=1\ w h e n\ x=1.`

Answer

`1/2log(y/(y+2x))=-logx+logc`

Explanation

We are given


`x^2dy+y(x+y)dx=0`


Simplifying,


`x^2dy+(xy+y^2)dx=0`


`rArr x^2dy=-(xy+y^2)dx`


`rArr x^2(dy)/(dx)=-xy-
y^2`


`rArr (dy)/(dx)=-x/x^2y-
...