Solve the following differential equation: `(1+y^2)tan^(-1)dx+2y(1+x^2)dy=0`

Explanation

Given differential equation is
`" "(1+y^(2))tan^(-1)xdx+2y(1+x^(2))dy=0`
`rArr" "(1+y^(2))tan^(-1)xdx=-2y(1+x^(2))dy`
`rArr" "(tan^(-1)xdx)/(1+x^(2))=-(2y)/(1+y^(2))dy`
On integrating both sides, we get
`" "int(tan^(-1)x)/(1+x^(2))dx=-int(2y)/(1+y^(2))dy`
Put `tan^(-1)x=t` in LHS, we get
`" "(1)/(1+x^(2))dx=dt`
and put `1+y^(2)=u` in RHS, we get
`" "2ydy=du`
`rArr" "inttdt=-int(1)/(u)durArr(t^(2))/(2)=-logu+C`
`rArr" "(1)/(2)(tan^(-1)x)^(2)=-log(1+y^(2))+C`
`rArr" "(1)/(2)(tan^(-1)x)^(2)+log(1+y^(2))=C`