The solution of dydx=y2−2xyx2−2xy is

dydx=y2−2xyx2−2xyLet y=vx⇒dydx=v+xdvdx⇒x⋅dvdx=v2−2v1−2v−v=3v2−3v1−2v∫2v−1v2−vdv=∫−3xdx⇒log⁡v2−v=−3log⁡x+log⁡c⇒v2−v=cx3⇒y2x2−yx=cx3⇒y2−xy=cx⇒xy(y−x)=c

The solution of dydx=y2−2xyx2−2xy is

dydx=y2−2xyx2−2xyLet y=vx⇒dydx=v+xdvdx⇒x⋅dvdx=v2−2v1−2v−v=3v2−3v1−2v∫2v−1v2−vdv=∫−3xdx⇒log⁡v2−v=−3log⁡x+log⁡c⇒v2−v=cx3⇒y2x2−yx=cx3⇒y2−xy=cx⇒xy(y−x)=c

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The solution of $\frac{d y}{d x} = \frac{y^{2} - 2 x y}{x^{2} - 2 x y}$ is

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$x y (x - y) = c$

$x y (y + x) = c$

$x y (y - x) = c$

$y (y - x) = c$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{d y}{d x} = \frac{y^{2} - 2 x y}{x^{2} - 2 x y} \\ L e t y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x} \\ \Rightarrow x \cdot \frac{d v}{d x} = \frac{v^{2} - 2 v}{1 - 2 v} - v = \frac{3 v^{2} - 3 v}{1 - 2 v} \\ \int \frac{2 v - 1}{v^{2} - v} d v = \int - \frac{3}{x} d x \\ \Rightarrow \log (v^{2} - v) = - 3 \log x + \log c \\ \Rightarrow v^{2} - v = \frac{c}{x^{3}} \\ \Rightarrow \frac{y^{2}}{x^{2}} - \frac{y}{x} = \frac{c}{x^{3}} \\ \Rightarrow y^{2} - x y = \frac{c}{x} \\ \Rightarrow x y (y - x) = c \end{array}$