Solution of ydydx2+2xdydx=y is

x=12p−p2y Differentiating both sides w·r·t yp=dydx1p=12p−p2+y−12p2−12dpdy12p+p2=−y1+p22p2dpdy→1+p22p=−y1+p22p2dpdy−dyy=dpp⇒py=C i.e. ydy=cdx⇒y2=2Cx+C2.

Solution of ydydx2+2xdydx=y is

x=12p−p2y Differentiating both sides w·r·t yp=dydx1p=12p−p2+y−12p2−12dpdy12p+p2=−y1+p22p2dpdy→1+p22p=−y1+p22p2dpdy−dyy=dpp⇒py=C i.e. ydy=cdx⇒y2=2Cx+C2.

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Solution of $y {(\frac{d y}{d x})}^{2} + 2 x \frac{d y}{d x} = y$ is

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$y^{2} = 2 C x + C^{2}$

$x y = C x^{2} + C^{2}$

$y = 2 C x + C^{2}$

$x^{2} = 2 C y + C^{2}$

answer is C.

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Detailed Solution

$x = (\frac{1}{2 p} - \frac{p}{2}) y$ Differentiating both sides w·r·t

$y (p = \frac{d y}{d x})$

$\begin{array}{l} \frac{1}{p} = \frac{1}{2 p} - \frac{p}{2} + y (- \frac{1}{2 p^{2}} - \frac{1}{2}) \frac{d p}{d y} \\ \frac{1}{2 p} + \frac{p}{2} = - y (\frac{1 + p^{2}}{2 p^{2}}) \frac{d p}{d y} \\ \to \frac{1 + p^{2}}{2 p} = - y (\frac{1 + p^{2}}{2 p^{2}}) \frac{d p}{d y} \\ - \frac{d y}{y} = \frac{d p}{p} \Rightarrow p y = C \\ i.e. y d y = c d x \Rightarrow y^{2} = 2 C x + C^{2} . \end{array}$