Question

Let a curve $y=y(x)$ pass through the point $(3,3)$ and the area of the region under this curve, above the $x$-axis and between the abscissae 3 and $x(>3)$ be ${\left(\frac{y}{x}\right)}^3$. If this curve also passes through the point $(\alpha ,6\sqrt{10})$ in the first quadrant, then $\alpha$ is equal to ______

Answer 1

Answer: 6

$x^4=3yx\cdot y^{\prime }-3y^2$
$\Rightarrow 3xy\frac{dy}{dx}=3y^2+x^4$
Put $y^2=t,y\frac{dy}{dx}=\frac{1}{2}\frac{dt}{dx}$
$\frac{dt}{dx}-\frac{2}{x}t=\frac{2}{3}{x}^3$
$\therefore \frac{t}{x^2}=\frac{x^2}{3}+C$
$\Rightarrow \frac{y^2}{x^2}=\frac{x^2}{3}-2$
Put $(3,3),C=-2$
$\therefore \frac{y^2}{x^2}=\frac{x^2}{3}-2$
$3y^2=x^4-6x^2$
$x^4-6x^2=1080$
$\therefore x=6$