Let f be a differentiable function such that x ^ 2 f (x) - x = 4 ∫...

Given,

${x}^{2}f\left(x\right) - x = 4{∫}_{0}^{x}tf\left(t\right)dt$

Now Differentiating both side w.r.t. $x$ we get,

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

$⇒ {x}^{2}{f}^{'}\left(x\right) - 1 = 2xf\left(x\right)$

$⇒ \dfrac{dy}{dx} - \dfrac{2}{x}y = \dfrac{1}{{x}^{2}}$ \left(Lety = f\left(x\right)\dfrac{dy}{dx} = {f}^{'}\left(x\right)\right)$</p><p>Which is a linear differential equation,</p><p>So,$I . F = {e}^{∫ - \dfrac{2}{x}dx} = {e}^{- 2\ln x} = \dfrac{1}{{x}^{2}}$</p><p>Now solution of the differential equation is given by,</p><p>$\dfrac{y}{{x}^{2}} = ∫ \dfrac{1}{{x}^{4}}dx + C$</p><p>$⇒ \dfrac{y}{{x}^{2}} = - \dfrac{1}{3{x}^{3}} + C$</p><p>Now given,$f\left(1\right) = \dfrac{2}{3}$</p><p>So,$\dfrac{2}{3} = - \dfrac{1}{3} + C$</p><p>$⇒ C = 1$</p><p>Hence, the function will be,</p><p>$f\left(x\right) = - \dfrac{1}{3x} + {x}^{2}$</p><p>So, the required value is given by,</p><p>$18f\left(3\right) = 18\left[- \dfrac{1}{9} + 9\right] = 160