Find the general solution of the differential equations:`(1+x^2)dy+2x y dx=cotx dx(x!=0)`

Explanation

Given,

`=>(1+x^2)dy+2xydx=cotxdx`

`=>(dy)/(dx)​+(2x​)/(1+x^2)y=cotx/(1+x^2)`
​
Integrating Factor is
`=>e∫^((2xdx)/(1+x^2))`

​`=>1+x^2`

So,the equation is

`⟹y(1+x^2)=∫cotxdx`

The solution of the equation is
`y(1+x^2)=logsinx+C`