The solution of differential equation `(dy)/(dx)=(1+y^(2))/(1+x^(2))"is"`

Options

A. `y=tan^(-1)x`
B. `y=x=k(1+xy)`
C. `x=tan^(-1)y`
D. `tan(xy)=k`

Explanation

Given that, `" "(dy)/(dx)=(1+y^(2))/(1+x^(2))`
`rArr" "(dy)/(1+y^(2))=(dx)/(1+x^(2))`
On integrating both sides, we get
`" "tan^(-1)y=tan^(-1)x+C`
`rArr" "tan^(-1)y-tan^(-1)x=C`
`rArr" "tan^(-1)((y-x)/(1+xy))=C`
`rArr" "(y-x)/(1+xy)=tanC`
`rArr" "y-x=tanc(1+xy)`
`rArr" "y-x=K(1+xy)`
where, `" "k=tanC`