The solution of dydx=y2xy-x2 is

dxdy=xy−x2y2Put x= vydxdy=v+ydvdy⇒ v+ydvdy=vy2−y2v2y2⇒ v+y dvdy=vy2−y2v2y2⇒ ∫−1v2 dv= ∫1y dy⇒ 1v= log (y)+log k⇒ yx= log (ky)⇒ ky= eyx

The solution of dydx=y2xy-x2 is

dxdy=xy−x2y2Put x= vydxdy=v+ydvdy⇒ v+ydvdy=vy2−y2v2y2⇒ v+y dvdy=vy2−y2v2y2⇒ ∫−1v2 dv= ∫1y dy⇒ 1v= log (y)+log k⇒ yx= log (ky)⇒ ky= eyx

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

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$e^{y / x} = ky$

$e^{- y / x} = kx$

$e^{y / x} = kx$

$e^{- y / x} = ky$

answer is B.

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Detailed Solution

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