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(C) The given differential equation is $(x^2-3y^2)dx+3xydy=0$.This can be rewritten as $\frac{dy}{dx} = \frac{3y^2-x^2}{3xy} = \frac{y}{x} - \frac{1}{

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$2e^2$

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(C) The given differential equation is $(x^2-3y^2)dx+3xydy=0$.This can be rewritten as $\frac{dy}{dx} = \frac{3y^2-x^2}{3xy} = \frac{y}{x} - \frac{1}{3}\frac{x}{y}$.Let $y=vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting this into the equation: $v + x\frac{dv}{dx} = v - \frac{1}{3v}$.This simplifies to $x\frac{dv}{dx} = -\frac{1}{3v}$,or $3vdv = -\frac{dx}{x}$.Integrating both sides: $\int 3vdv = -\int \frac{dx}{x} \Rightarrow \frac{3v^2}{2} = -\ln|x| + C$.Substituting $v = \frac{y}{x}$: $\frac{3y^2}{2x^2} = -\ln|x| + C$.Given $y(1)=1$,we have $\frac{3(1)^2}{2(1)^2} = -\ln(1) + C \Rightarrow C = \frac{3}{2}$.Thus,$\frac{3y^2}{2x^2} = -\ln|x| + \frac{3}{2} \Rightarrow 3y^2 = 2x^2(\frac{3}{2} - \ln|x|) = 3x^2 - 2x^2\ln|x|$.At $x=e$,$3y^2(e) = 3e^2 - 2e^2\ln(e) = 3e^2 - 2e^2 = e^2$.Therefore,$6y^2(e) = 2(3y^2(e)) = 2e^2$.

Let $y=y(x)$ be the solution of the differential equation $(x^2-3y^2)dx+3xydy=0$ with $y(1)=1$. Then $6y^2(e)$ is equal to $......$

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