Let f be a differentiable function such that x 2 f ( x ) - x = 4 0 x t f ( t ) ? d t , f ( 1 ) = 2 3 . Then 18 f ( 3 ) is equal to [2023]

Question

Let $f$ be a differentiable function such that $x^{2} f (x) - x = 4 \int_{0}^{x} t f (t) d t, f (1) = \frac{2}{3} .$ Then $18 f (3)$ is equal to [2023]

Smart Achievers · Accepted Answer

(3)

We have, $x^{2} f (x) - x = 4 \int_{0}^{x} t f (t) d t, f (1) = \frac{2}{3} .$

$\Rightarrow 2 x f (x) + x^{2} f' (x) - 1 = 4 x f (x)$

$\Rightarrow x^{2} \frac{d y}{d x} - 2 x y = 1 [f (x) = y, f' (x) = \frac{d y}{d x}]$

$\Rightarrow \frac{d y}{d x} - \frac{2 y}{x} = \frac{1}{x^{2}}, which is a linear differential equation.$

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

$∴$ Solution is given by

$\frac{y}{x^{2}} = \int \frac{1}{x^{4}} d x + C \Rightarrow \frac{y}{x^{2}} = - \frac{1}{3 x^{3}} + C$ ...(i)

$At x = 1, y = \frac{2}{3}$ $[∵ f (1) = \frac{2}{3}]$

$\frac{2}{3} = \frac{- 1}{3} + C \Rightarrow C = 1$

$∴$ $y = \frac{- 1}{3 x} + x^{2} \Rightarrow f (x) = \frac{- 1}{3 x} + x^{2}$

$Now, 18 f (3) = 18 [\frac{- 1}{9} + 9] = 160$

Solution

Q.
Let $f$ be a differentiable function such that $x^{2} f (x) - x = 4 \int_{0}^{x} t f (t) d t, f (1) = \frac{2}{3} .$ Then $18 f (3)$ is equal to [2023]

1 180

2 210

3 160

4 150

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