Let a curve y=y(x) pass through (3,3), and the area under the curve from 3 to x (x3) be y…

Question

Let a curve $y=y(x)$ pass through $(3,3)$, and the area under the curve from $3$ to $x$ ($x>3$) be $\dfrac{y^3(x)}{x^3}\cdot\ (	ext{schematic})$. If it also passes through $(\alpha,6\sqrt{10})$, then $\alpha$ is \_\_\_\_\_\_.

Answer: 6

Accepted Answer

Differentiating the area relation twice leads to $3xy\dfrac{dy}{dx}=x^4+3y^2$, a Bernoulli-type equation. With $t=y^2$, solving gives $y^2 = \dfrac{x^4}{3}+Cx^2$; using $(3,3)$: $C=-2$.

$$y^2 = \frac{x^4}{3}-2x^2.$$

Through $(\alpha,6\sqrt{10})$: $\dfrac{\alpha^4-6\alpha^2}{3}=360 \implies \alpha^4-6\alpha^2-1080=0 \implies \alpha=6$.

Solutions · 1