Solve the differential equation: `(1+y^2)(1+logx)dx+x dy=0` given that when `x=1,\ y=1.`

Answer

`- tan ^{-1} y= frac{(1+ log x)^{2}}{2}- ( frac{1}{2}+(pi) / 4 )`

Explanation

We have,

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


`(1+y^2)(1+ log x) d x+x d y =0`


`(1+y^2)(1+ log x) d x =-x d y`


`(1+log x)dx= frac{-x}{1+y^2} dy`


`frac{(1+ log x)}{x} d x= frac{-d y}{1+y^{2}}`


Integrating both the sides, we have


` int frac{(1+ log x)}{x} d x=- int frac{d y}{1+y^{2}}`


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