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

Answer

`y=(log sin x)/(1+x^2)+c/(1+x^2)`

Explanation

We have,


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


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


Comparing this equation


`(dy)/(dx)+Py=Q`


Then, `P=(2x)/(1+x^2), quad Q=cotx/(1+x^2)`


`I.F.=e^(int P dx) `


`=e^(int (2x)/(1+x^2) dx)=e^(log (1+x^2))`


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