Find the particular solution of the differential equation `(1+e^(2x))dy+(1+y^2)e^x dx=0,` given that `y=1` when `x=0.`

Answer

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

Explanation

We are given

`(1+e^{2x} ) d y+ (1+y^2 ) e^x d x=0`


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


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


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


Integrating both sides,

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


Let `t=e^x`

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