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(B) Given the differential equation: $2x^{2} \frac{dy}{dx} - 2xy + 3y^{2} = 0$.Divide by $2x^{2}y^{2}$ to transform it into a Bernoulli equation or su

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$\frac{2}{3}$

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(B) Given the differential equation: $2x^{2} \frac{dy}{dx} - 2xy + 3y^{2} = 0$.Divide by $2x^{2}y^{2}$ to transform it into a Bernoulli equation or substitute $y = vx$:$\frac{dy}{dx} - \frac{y}{x} = -\frac{3}{2} \left(\frac{y}{x}\right)^{2}$.Let $v = \frac{y}{x}$,then $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $v + x \frac{dv}{dx} - v = -\frac{3}{2} v^{2}$.$x \frac{dv}{dx} = -\frac{3}{2} v^{2}$.Separating variables: $\frac{dv}{v^{2}} = -\frac{3}{2} \frac{dx}{x}$.Integrating both sides: $-\frac{1}{v} = -\frac{3}{2} \ln|x| + C$.Substituting $v = \frac{y}{x}$ back: $-\frac{x}{y} = -\frac{3}{2} \ln|x| + C$.Using the condition $y(e) = \frac{e}{3}$: $-\frac{e}{e/3} = -\frac{3}{2} \ln(e) + C \implies -3 = -\frac{3}{2} + C \implies C = -\frac{3}{2}$.So,$-\frac{x}{y} = -\frac{3}{2} \ln|x| - \frac{3}{2}$.For $x = 1$: $-\frac{1}{y} = -\frac{3}{2} \ln(1) - \frac{3}{2} \implies -\frac{1}{y} = -\frac{3}{2} \implies y = \frac{2}{3}$.

If $y = y(x)$ is the solution of the differential equation $2x^{2} \frac{dy}{dx} - 2xy + 3y^{2} = 0$ such that $y(e) = \frac{e}{3}$,then $y(1)$ is equal to

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