Find the equation of a curve passing through `(0,1)` and having gradient `(1(y+y^3))/(1+x+x y)a t(x , y)`

Answer

`xy+tan^(-1)y=pi/4`

Explanation

We have `(dy)/(dx)=(-y+y^(3))/(1+x(1+y^(2))`
or `(dx)/(dy) = -(1+x(1+y^(3)))/(1+x(1+y^(2))`
`=-1/(y(y^(2)+1))+x(1+y^(2))/(y(1+y^(2))]`
`therefore (dx)/(dy) + x/y=-1/(y(1+y^(2))`
I.F. `=e^(int(dy)/y)=e^(logy)=y`.
Hence, the solution is
`x.y=-int1/(y(1+y^(2)).ydy+C`
`=-int(dy)/(1+y^(2))+C=-tan^(-1)y+C`
or `xy+tan^(-1)y=C`.
This passes through (0,1). Therefore, `tan^(-1)1=C, i.e., C=pi/4`.
Thus, the equation of the curve is `xy+tan^(-1)y=pi/4`