Question:

Step 1: Rearrange the given differential equation to the standard form $\dfrac{dy}{dx}$. The given differential equation is $(x^2-3y^2)\,dx+3xy\,dy=0$. We can rewrite it to express $\dfrac{dy}{dx}$. $$ (x^2-3y^2)\,dx = -3xy\,dy $$ $$ \dfrac{dy}{dx} = \dfrac{-(x^2-3y^2)}{3xy} = \dfrac{3y^2-x^2}{3xy} $$ This is a homogeneous differential equation since all terms in the numerator and denominator have the same degree (degree 2). Step 2: Solve the homogeneous differential equation using the substitution $y=vx$. Let $y=vx$, which implies $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$. Substitute this into the differential equation: $$ v + x\dfrac{dv}{dx} = \dfrac{3(vx)^2-x^2}{3x(vx)} $$ $$ v + x\dfrac{dv}{dx} = \dfrac{3v^2x^2-x^2}{3vx^2} $$ $$ v + x\dfrac{dv}{dx} = \dfrac{x^2(3v^2-1)}{x^2(3v)} $$ $$ v + x\dfrac{dv}{dx} = \dfrac{3v^2-1}{3v} $$ Now, separate the variables $v$ and $x$: $$ x\dfrac{dv}{dx} = \dfrac{3v^2-1}{3v} - v $$ $$ x\dfrac{dv}{dx} = \dfrac{3v^2-1-3v^2}{3v} $$ $$ x\dfrac{dv}{dx} = \dfrac{-1}{3v} $$ Separate variables: $$ 3v\,dv = -\dfrac{1}{x}\,dx $$ Step 3: Integrate both sides of the separated equation. Integrate both sides of the equation $3v\,dv = -\dfrac{1}{x}\,dx$: $$ \int 3v\,dv = \int -\dfrac{1}{x}\,dx $$ $$ \dfrac{3v^2}{2} = -\ln|x| + C $$ Now, substitute back $v=\dfrac{y}{x}$: $$ \dfrac{3}{2}\left(\dfrac{y}{x}\right)^2 = -\ln|x| + C $$ $$ \dfrac{3y^2}{2x^2} = -\ln|x| + C $$ Step 4: Apply the initial condition $y(1)=1$ to find the constant $C$. Given that $y(1)=1$, substitute $x=1$ and $y=1$ into the general solution: $$ \dfrac{3(1)^2}{2(1)^2} = -\ln|1| + C $$ $$ \dfrac{3}{2} = -0 + C $$ Thus, the constant $C = \dfrac{3}{2}$. Step 5: Write the particular solution using the value of $C$. Substitute $C=\dfrac{3}{2}$ back into the general solution: $$ \dfrac{3y^2}{2x^2} = -\ln|x| + \dfrac{3}{2} $$ To simplify, multiply the entire equation by $2x^2$: $$ 3y^2 = -2x^2\ln|x| + 3x^2 $$ Rearranging the terms, the particular solution is: $$ 3y^2 = 3x^2 - 2x^2\ln|x| $$ Step 6: Evaluate $6y^2(e)$ by substituting $x=e$ into the particular solution. We need to find the value of $6y^2(e)$. First, find $3y^2(e)$ by setting $x=e$: $$ 3y^2(e) = 3(e)^2 - 2(e)^2\ln|e| $$ Since $\ln|e| = \ln e = 1$: $$ 3y^2(e) = 3e^2 - 2e^2(1) $$ $$ 3y^2(e) = 3e^2 - 2e^2 $$ $$ 3y^2(e) = e^2 $$ Now, multiply by 2 to find $6y^2(e)$: $$ 6y^2(e) = 2 \times (3y^2(e)) = 2 \times e^2 $$ $$ 6y^2(e) = 2e^2 $$ Step 7: Match the result with the given options. The calculated value of $6y^2(e)$ is $2e^2$. Comparing this with the given options: Option 1: $3e^2$ Option 2: $e^2$ Option 3: $2e^2$ Option 4: $\dfrac{3e^2}{2}$ The result matches Option 3. The final answer is $\boxed{2e^2}$.