Answer 1

(1+y²)dx−(tan⁻¹y−x)dy=0

`=>(1+y^2)dx/dy=tan^−1 y−x`

 

`⇒(1+y^2)dx/dy+x=tan^−1 y`

`⇒dx/dy+1/(1+y^2)x=tan−1y/(1+y^2)`

∴ Integrating factor (IF)= `e^(int 1/(1+y^2)dy)`

`=e^(tav^-1y)`