Solve each of the following initial value problem: `(1+y^2)dx=(tan^(-1)y-x)dy , y(0)=0`

Explanation

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

`frac{d x}{d y}=frac{tan ^{-1} y}{1+y^{2}}-frac{x}{1+y^{2}}`

`frac{d x}{d y}+frac{x}{1+y^{2}}=frac{tan ^{-1} y}{1+y^{2}}`


Hence, IF `=e^{int frac{1}{1+y^{2}} cdot d y}`


`=>e^{tan -1 y}`


Hence the above differential equation changes to
`e^{tan ^{-1} y} cdot frac{d x}{d y}+frac{x e^{tan ^{-1} y}}{1+y^{2}}=frac{e^{tan ^{-1} y} tan ^{-1} y}{1+y^{2}}`


`e^{tan ^{-1} y} cdot d x+frac{x e^{tan ^{-1} y}}{1+y^{2}} d y=frac{e^{tan ^{-1} y} tan ^{-1} y}{1+y^{2}} d y`


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