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 co-ordinate axes, where (x, y) is any point on the curve, is always , Y'(X) 0. If Y(1) = 1, then 12Y(2) equals __________. [

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Solution

Q.
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 co-ordinate axes, where (x, y) is any point on the curve, is always $\frac{- y^{2}}{2 Y' (x)} + 1$ , Y'(X) $\neq$ 0. If Y(1) = 1, then 12Y(2) equals __________. [2024]

Ans.
(20)

We have, curve : Y = y(x)

Line : Y – y = Y'(x)(X – x)

Area = $\frac{1}{2} (x - \frac{y}{Y' (x)}) (y - x Y' (x))$

$\Rightarrow 1 - \frac{y^{2}}{2 Y' (x)} = \frac{(x Y' (x) - y) (y - x Y' (x))}{2 Y' (x)}$

$\Rightarrow \frac{2 Y' (x) - y^{2}}{2 Y' (x)} = \frac{(x Y' (x) - y) (y - x Y' (x))}{2 Y' (x)}$

$\Rightarrow 2 Y' (x) - y^{2} = x y Y' (x) - x^{2} {[Y' (x)]}^{2} - y^{2} + x y Y' (x)$

$\Rightarrow x^{2} {[Y' (x)]}^{2} + Y' (x) [2 - x y - x y] = 0$

$\Rightarrow Y' (x) [x^{2} Y' (x) + 2 - 2 x y] = 0$

$\Rightarrow x^{2} Y' (x) + 2 - 2 x y = 0 or Y' (x) = \frac{2 x y - 2}{x^{2}}$

$\Rightarrow \frac{d y}{d x} = (\frac{2}{x}) y - \frac{2}{x^{2}} \Rightarrow \frac{d y}{d x} - (\frac{2}{x}) y = \frac{- 2}{x^{2}}$

$I . F . = e^{- \int \frac{2}{x} d x} = e^{- 2 \log x} = \frac{1}{x^{2}}$

Solution is

$y \cdot \frac{1}{x^{2}} = \int \frac{- 2}{x^{2}} \times \frac{1}{x^{2}} d x = - 2 \int \frac{1}{x^{4}} d x or \frac{y}{x^{2}} = \frac{2}{3 x^{3}} + c$

$y = \frac{2}{3 x} + c x^{2}$ ... (i)

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

At x = 2,

$∴ y = \frac{2}{3 \times 2} + \frac{1}{3} \times 4 = \frac{1}{3} + \frac{4}{3} = \frac{5}{3} \Rightarrow 12 y = 20$ .

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