The solution of dydx=3x2y−y33y2x−x3 is

Put y=vx. Then dydx=v+xdvdxGiven equation is dydx=3x2y−y33y2x−x3⇒v+xdvdx=3x3v−v3x33x3v2−x3⇒xdvdx=3v−v33v2−1−v=3v−v3−3v3+v3v2−1=4v−4v33v2−1=−4v3−v3v2−1⇒3v2−1v3−vdv=−4dxx⇒∫3v2−1v3−vdv=−4∫dxx⇒log⁡v3−v=−4log⁡x+log⁡c⇒x4v3−v=c⇒xy3−x3=c

The solution of dydx=3x2y−y33y2x−x3 is

Put y=vx. Then dydx=v+xdvdxGiven equation is dydx=3x2y−y33y2x−x3⇒v+xdvdx=3x3v−v3x33x3v2−x3⇒xdvdx=3v−v33v2−1−v=3v−v3−3v3+v3v2−1=4v−4v33v2−1=−4v3−v3v2−1⇒3v2−1v3−vdv=−4dxx⇒∫3v2−1v3−vdv=−4∫dxx⇒log⁡v3−v=−4log⁡x+log⁡c⇒x4v3−v=c⇒xy3−x3=c

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The solution of $\frac{d y}{d x} = \frac{3 x^{2} y - y^{3}}{3 y^{2} x - x^{3}}$ is

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$y (y^{3} - x^{3}) = c$

$y (y^{3} + x^{3}) = c$

$x (y^{3} - x^{3}) = c$

$x (y^{3} + x^{3}) = c$

answer is A.

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Detailed Solution

Put $y = v x . Then \frac{d y}{d x} = v + x \frac{d v}{d x}$

Given equation is $\frac{d y}{d x} = \frac{3 x^{2} y - y^{3}}{3 y^{2} x - x^{3}} \Rightarrow v + x \frac{d v}{d x} = \frac{3 x^{3} v - v^{3} x^{3}}{3 x^{3} v^{2} - x^{3}}$

$\begin{array}{l} \Rightarrow x \frac{d v}{d x} = \frac{3 v - v^{3}}{3 v^{2} - 1} - v = \frac{3 v - v^{3} - 3 v^{3} + v}{3 v^{2} - 1} = \frac{4 v - 4 v^{3}}{3 v^{2} - 1} = \frac{- 4 (v^{3} - v)}{3 v^{2} - 1} \Rightarrow \frac{3 v^{2} - 1}{v^{3} - v} d v = - 4 \frac{d x}{x} \\ \Rightarrow \int \frac{3 v^{2} - 1}{v^{3} - v} d v = - 4 \int \frac{d x}{x} \Rightarrow \log (v^{3} - v) = - 4 \log x + \log c \Rightarrow x^{4} (v^{3} - v) = c \Rightarrow x (y^{3} - x^{3}) = c \end{array}$