Question

${y}^{2}+{x}^{2} \frac{ dy}{dx} = xy \frac{dy}{dx}$
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Answer 1

$\sf\huge\bold\pink{Answer:}$

${y}^{2} + {x}^{2} \frac{dy}{dx} = xy\frac{dy}{dx} \\ {y}^{2} + 2x =x\frac{dy}{dx}.\frac{dy}{dx}+\frac{dy}{dx}\\ {y}^{2} + 2x =\frac{{d}^{2}y}{d{x}^{2}}+x\frac{dy}{dx}$

Answer 2

Given,

$\small\longrightarrow y^2+x^2\,\dfrac{dy}{dx}=xy\,\dfrac{dy}{dx}$

Dividing each term by $x^2,$

$\small\longrightarrow\dfrac{y^2}{x^2}+\dfrac{dy}{dx}=\dfrac{y}{x}\cdot\dfrac{dy}{dx}$

$\small\longrightarrow\left(\dfrac{y}{x}\right)^2=\dfrac{y}{x}\cdot\dfrac{dy}{dx}-\dfrac{dy}{dx}$

$\small\longrightarrow\left(\dfrac{y}{x}\right)^2=\left(\dfrac{y}{x}-1\right)\dfrac{dy}{dx}\quad\quad\dots(1)$

Substitute,

$\small\longrightarrow u=\dfrac{y}{x}$

$\small\longrightarrow y=ux$

$\small\longrightarrow\dfrac{dy}{dx}=u+x\,\dfrac{du}{dx}$

Then (1) becomes,

$\small\longrightarrow u^2=\left(u-1\right)\left(u+x\,\dfrac{du}{dx}\right)$

$\small\longrightarrow x\,\dfrac{du}{dx}=\dfrac{u^2}{u-1}-u$

$\small\longrightarrow x\,\dfrac{du}{dx}=\dfrac{u}{u-1}$

$\small\longrightarrow\dfrac{dx}{x}=\dfrac{u-1}{u}\,du$

Integrating,

$\small\displaystyle\longrightarrow\int\dfrac{dx}{x}=\int\left(1-\dfrac{1}{u}\right)\,du$

$\small\longrightarrow\ln|x|=u-\ln|u|+\ln|C|$

Undoing substitution $u=\dfrac{y}{x},$

$\small\longrightarrow\ln|x|=\dfrac{y}{x}-\ln\left|\dfrac{y}{x}\right|+\ln|C|$

$\small\longrightarrow\ln|y|=\dfrac{y}{x}+\ln|C|$

$\small\text{ \longrightarrow\underline{\underline{y=Ce^{\frac{y}{x}}}} }$