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 ______.

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Differential Equations

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)

(\frac{y}{x})^{3}

. If this curve also passes through the point

(α, 610)

in the first quadrant, then

α

is equal to ______.

Answer: 6

Key concept: Differential equations from area condition.

From the area condition:

\int_{3}^{x} y d t = (\frac{y}{x})^{3}

Differentiating both sides with respect to

x

y = 3 (\frac{y}{x})^{2} \cdot \frac{x y ^{'} - y}{x ^{2}}

Simplifying:

3 x y \frac{d y}{d x} = 3 y^{2} + x^{4}

Substituting

t = y^{2}

, we get the linear DE:

\frac{d t}{d x} - \frac{2 t}{x} = \frac{2 x ^{3}}{3}

Solving with integrating factor

x^{- 2}

\frac{y ^{2}}{x ^{2}} = \frac{x ^{2}}{3} + C

Using

(3, 3)

C = - 2

So:

3 y^{2} = x^{4} - 6 x^{2}

Substituting

y = 610

x^{4} - 6 x^{2} - 1080 = 0

Solving:

x = 6

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Integer answer