Question

Let $Y = Y(X)$ be a curve lying in the first quadrant such that the area enclosed by the line $Y - y = Y'(x) (X - x)$ and the coordinate axes, where $(x, y)$ is any point on the curve, is always$\frac{-y^2}{2Y'(x)} + 1, \quad Y'(x) \neq 0.$If $Y(1) = 1$, then $12Y(2)$ equals ______.

Answer 1

The line $Y - y = Y'(x)(X - x)$ represents the tangent to $Y = Y(X)$ at $(x, y)$.

The area $A = -\frac{Y(x)^2}{2Y'(x)} + 1$ relates $y$ and $Y'(x)$ on the curve.

Differentiate with respect to $x$ and solve for $Y(x)$ using the initial condition $Y(1) = 1$.

$1 = \frac{2}{3} + c$

$c = \frac{1}{3}$

$Y = \frac{2}{3} \times \frac{1}{X} + \frac{1}{3}X^2$

$12Y(2) = \frac{5}{3} \times 12 = 20$