Question

The solution of the differential equation $ \frac{dy}{dx}=\frac{1}{x+{{y}^{2}}} $ is

• A. $ y=-{{x}^{2}}-2x-2+c{{e}^{x}} $
• B. $ y={{x}^{2}}+2x+2-c{{e}^{x}} $
• C. $ x=-{{y}^{2}}-2y+2-c{{e}^{y}} $
• D. $ x=-{{y}^{2}}-2y-2+c{{e}^{y}} $
• E. $ x={{y}^{2}}+2y+2-cey $

Answer 1

Answer: (D)

Given differential equation is $ \frac{dy}{dx}=\frac{1}{x+{{y}^{2}}} $
$ \Rightarrow $ $ \frac{dx}{dy}-x={{y}^{2}} $ Here, $ P=-1,Q={{y}^{2}} $ If
$={{e}^{\int{-1}\,dy}}={{e}^{-y}} $
$ \therefore $ Solution is $ x{{e}^{-y}}=\int{{{e}^{-y}}{{y}^{2}}dy} $
$=-{{e}^{-y}}{{y}^{2}}+\int{2{{e}^{-y}}ydy} $
$=-{{e}^{-y}}{{y}^{2}}+2[-{{e}^{-y}}y+\int{{{e}^{-y}}dy]}+c $
$=-{{e}^{-y}}{{y}^{2}}+2[-{{e}^{-y}}y-{{e}^{-y}}]+c $
$ \Rightarrow $ $ x{{e}^{-y}}={{e}^{-y}}(-{{y}^{2}}-2y-2)+c $
$ \Rightarrow $ $ x=-{{y}^{2}}-2y-2+c{{e}^{y}} $