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
$x(>3)\ be\ (\frac{y}{x})^3$
. If this curve also passes through the point (α,6√10) in the first quadrant, then α is equal to _______.

Answer 1

$\int_{3}^{x} \, f(x) \,dx = \left(\frac{f(x)}{x}\right)^3$
$x^3 \cdot \int_{3}^{x} f(x) \,dx = f^3(x)$
Differentiate w.r.t. x
$x^3 f(x) +3x^2. \frac{f^3(x)}{x^3} = 3f^2(x)f'(x)$
⇒$$3y^2 \frac{dy}{dx} = x^3y + \frac{3y^3}{x}
$3xy \frac{dy}{dx}=x^4+3y^2$
Let y 2 = t
$\frac{3}{2} \frac{dt}{dx} = x^3 + \frac{3t}{x}$
$\frac{dt}{dx} - \frac{2t}{x} = \frac{2x^3}{3}$
$IF=e^{\int−\frac{2}{x} dx}=\frac{1}{x^2}$
Solution of differential equation
$t⋅\frac{1}{x^2}=\int\frac{2}{3}x\ dx$
$\frac{y^2}{x^2}=\frac{x^2}{3}+C$
$y^2=\frac{x^4}{3}+Cx^2$
Curve passes through(3,3)
$⇒C=–2$
$y^2=\frac{x^4}{3}−2x^2$
Which passes through$(α, 6\sqrt10)$
$\frac{α^4−6α^2}{3}=360$
$α^4−6α^2−1080=0$
α=6
So, the correct answer is 6.