The solution of dydx+y2+y+1x2+x+1=0 is

Given equation is dydx+y2+y+1x2+x+1=0⇒1x2+x+1dx+1y2+y+1dy=0⇒∫1(x+1/2)2+(3/2)2dx+∫1(y+1/2)2+(3/2)2dy=0⇒23Tan−1⁡x+1/23/2+23Tan−1⁡y+1/23/2=23c⇒Tan−1⁡2x+13+Tan−1⁡2y+13=c.

The solution of dydx+y2+y+1x2+x+1=0 is

Given equation is dydx+y2+y+1x2+x+1=0⇒1x2+x+1dx+1y2+y+1dy=0⇒∫1(x+1/2)2+(3/2)2dx+∫1(y+1/2)2+(3/2)2dy=0⇒23Tan−1⁡x+1/23/2+23Tan−1⁡y+1/23/2=23c⇒Tan−1⁡2x+13+Tan−1⁡2y+13=c.

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The solution of $\frac{d y}{d x} + \frac{y^{2} + y + 1}{x^{2} + x + 1} = 0$ is

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${Tanh}^{- 1} \frac{2 x + 1}{\sqrt{3}} + {Tanh}^{- 1} \frac{2 y + 1}{\sqrt{3}} = c .$

${Tan}^{- 1} \frac{2 x + 1}{\sqrt{3}} + {Tan}^{- 1} \frac{2 y + 1}{\sqrt{3}} = c .$

${Tan}^{- 1} \frac{2 x - 1}{\sqrt{3}} + {Tan}^{- 1} \frac{2 y - 1}{\sqrt{3}} = c .$

${Tan}^{- 1} \frac{2 x + 1}{\sqrt{3}} - {Tan}^{- 1} \frac{2 y + 1}{\sqrt{3}} = c .$

answer is B.

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Detailed Solution

Given equation is $\frac{d y}{d x} + \frac{y^{2} + y + 1}{x^{2} + x + 1} = 0 \Rightarrow \frac{1}{x^{2} + x + 1} d x + \frac{1}{y^{2} + y + 1} d y = 0$

$\begin{array}{l} \Rightarrow \int \frac{1}{(x + 1 / 2)^{2} + (\sqrt{3} / 2)^{2}} d x + \int \frac{1}{(y + 1 / 2)^{2} + (\sqrt{3} / 2)^{2}} d y = 0 \\ \Rightarrow \frac{2}{\sqrt{3}} {Tan}^{- 1} \frac{x + 1 / 2}{\sqrt{3} / 2} + \frac{2}{\sqrt{3}} {Tan}^{- 1} \frac{y + 1 / 2}{\sqrt{3} / 2} = \frac{2}{\sqrt{3}} c \Rightarrow {Tan}^{- 1} \frac{2 x + 1}{\sqrt{3}} + {Tan}^{- 1} \frac{2 y + 1}{\sqrt{3}} = c . \end{array}$