The solution of x2dydx=x2+xy+y2

dydx= 1+(yx)+(yx)2Put y= vx⇒ v+xdvdx⇒ v+xdvdx=1+v+v2⇒ ∫11+v2dv= ∫1xdx⇒ Tan−1(v)=logx+c⇒ Tan−1(yx)=logx+c

The solution of x2dydx=x2+xy+y2

dydx= 1+(yx)+(yx)2Put y= vx⇒ v+xdvdx⇒ v+xdvdx=1+v+v2⇒ ∫11+v2dv= ∫1xdx⇒ Tan−1(v)=logx+c⇒ Tan−1(yx)=logx+c

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The solution of $x^{2} \frac{dy}{dx} = x^{2} + xy + y^{2}$

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$\log x + {Tan}^{- 1} (\frac{x}{y}) = c$

${Tan}^{- 1} (\frac{y}{x}) = \log x + c$

$\log x + {Tan}^{- 1} (\frac{y}{x}) = c$

${Tan}^{- 1} (\frac{x}{y}) = \log x + c$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{d y}{d x} = 1 + (\frac{y}{x}) + {(\frac{y}{x})}^{2} \\ P u t y = v x \Rightarrow v + x \frac{d v}{d x} \\ \Rightarrow v + x \frac{d v}{d x} = 1 + v + v^{2} \\ \Rightarrow \int \frac{1}{1 + v^{2}} d v = \int \frac{1}{x} d x \\ \Rightarrow T a n^{- 1} (v) = \log x + c \\ \Rightarrow T a n^{- 1} (\frac{y}{x}) = \log x + c \end{array}$