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Mathematics Question on Differential equations

The solution of the differential equation dydx=1x+y2\frac{dy}{dx}=\frac{1}{x+{{y}^{2}}} is

A

y=−x2−2x−2+cexy=-{{x}^{2}}-2x-2+c{{e}^{x}}

B

y=x2+2x+2−cexy={{x}^{2}}+2x+2-c{{e}^{x}}

C

x=−y2−2y+2−ceyx=-{{y}^{2}}-2y+2-c{{e}^{y}}

D

x=−y2−2y−2+ceyx=-{{y}^{2}}-2y-2+c{{e}^{y}}

Answer

x=−y2−2y−2+ceyx=-{{y}^{2}}-2y-2+c{{e}^{y}}

Explanation

Solution

Given differential equation is dydx=1x+y2\frac{dy}{dx}=\frac{1}{x+{{y}^{2}}} ⇒\Rightarrow dxdy−x=y2\frac{dx}{dy}-x={{y}^{2}} Here, P=−1,Q=y2P=-1,Q={{y}^{2}} If =e∫−1 dy=e−y={{e}^{\int{-1}\,dy}}={{e}^{-y}} ∴\therefore Solution is xe−y=∫e−yy2dyx{{e}^{-y}}=\int{{{e}^{-y}}{{y}^{2}}dy} =−e−yy2+∫2e−yydy=-{{e}^{-y}}{{y}^{2}}+\int{2{{e}^{-y}}ydy} =−e−yy2+2[−e−yy+∫e−ydy]+c=-{{e}^{-y}}{{y}^{2}}+2[-{{e}^{-y}}y+\int{{{e}^{-y}}dy]}+c =−e−yy2+2[−e−yy−e−y]+c=-{{e}^{-y}}{{y}^{2}}+2[-{{e}^{-y}}y-{{e}^{-y}}]+c ⇒\Rightarrow xe−y=e−y(−y2−2y−2)+cx{{e}^{-y}}={{e}^{-y}}(-{{y}^{2}}-2y-2)+c ⇒\Rightarrow x=−y2−2y−2+ceyx=-{{y}^{2}}-2y-2+c{{e}^{y}}