Let a curve y = y(x) pass through the point ( | vedclass

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(C) Given that the area under the curve from $3$ to $x$ is $\int_{3}^{x} y(t) dt = \left(\frac{y}{x}\right)^{3}$.Differentiating both sides with respe

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$6$

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(C) Given that the area under the curve from $3$ to $x$ is $\int_{3}^{x} y(t) dt = \left(\frac{y}{x}ight)^{3}$.Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus:$y = \frac{d}{dx} \left( \frac{y^3}{x^3} ight) = \frac{3y^2 y' x^3 - 3x^2 y^3}{x^6} = \frac{3y^2 y' x - 3y^3}{x^4}$.Multiplying by $x^4$:$x^4 y = 3xy^2 y' - 3y^3$.Divide by $y$ (assuming $y 
eq 0$):$x^4 = 3xy y' - 3y^2$.Let $t = y^2$,then $dt/dx = 2y y'$. Substituting this:$x^4 = \frac{3}{2} x \frac{dt}{dx} - 3t$.Rearranging into a linear differential equation:$\frac{dt}{dx} - \frac{2}{x} t = \frac{2}{3} x^3$.The integrating factor is $IF = e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = \frac{1}{x^2}$.Multiplying by $IF$:$\frac{d}{dx} \left( \frac{t}{x^2} ight) = \frac{2}{3} x$.Integrating both sides:$\frac{t}{x^2} = \frac{x^2}{3} + C$.Since the curve passes through $(3,3)$,$t = 3^2 = 9$ at $x=3$:$\frac{9}{9} = \frac{9}{3} + C \implies 1 = 3 + C \implies C = -2$.So,$\frac{y^2}{x^2} = \frac{x^2}{3} - 2$.For the point $(\alpha, 6\sqrt{10})$:$\frac{(6\sqrt{10})^2}{\alpha^2} = \frac{\alpha^2}{3} - 2 \implies \frac{360}{\alpha^2} = \frac{\alpha^2 - 6}{3}$.$1080 = \alpha^4 - 6\alpha^2 \implies \alpha^4 - 6\alpha^2 - 1080 = 0$.Let $u = \alpha^2$: $u^2 - 6u - 1080 = 0 \implies (u - 36)(u + 30) = 0$.Since $\alpha^2 > 0$,$u = 36$,so $\alpha = 6$.

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