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) \[x={{y}^{2}}+2y+2-cey\]
Correct Answer: D
Solution :
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 \[x{{e}^{-y}}=\int{{{e}^{-y}}{{y}^{2}}dy}\] \[=-{{e}^{-y}}{{y}^{2}}+\int{2{{e}^{-y}}ydy}\] \[=-{{e}^{-y}}{{y}^{2}}+2[-{{e}^{-y}}y+\int{{{e}^{-y}}dy]}+c\] \[=-{{e}^{-y}}{{y}^{2}}+2[-{{e}^{-y}}y-{{e}^{-y}}]+c\] \[\Rightarrow \] \[x{{e}^{-y}}={{e}^{-y}}(-{{y}^{2}}-2y-2)+c\] \[\Rightarrow \] \[x=-{{y}^{2}}-2y-2+c{{e}^{y}}\]
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