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

Explanation

Given,`(1+x)(1+y^2)dx+(1+y)(1+x^2)dy=0`

`(1+x)(1+y^2)dx=-(1+y)(1+x^2)dy`

=`(1+x)/(1+x^2)dx=-(1+y)/(1+y^2)dy`

Integrating both sides we get

`int(1+x)/(1+x^2)dx=-int(1+y)/(1+y^2)dy`

=`int(1)/(1+x^2)dx+int x/(1+x^2)dx=-int(1)/(1+y^2)dy-inty/(1+y^2)dy`

Making `1+x^2=t` and `1+y^2=u`

Then,`2xdx=dt` and `2ydy=du`

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