Let y = y (x) be the solution curve of the differential equation...

Given differential equation can be rewritten as,

$\dfrac{dy}{dx} - \dfrac{y}{x} = {y}^{3}\left(1 + \log_{e}x\right)$

$\dfrac{1}{{y}^{3}}\dfrac{dy}{dx} - \dfrac{1}{{xy}^{2}} = 1 + \log_{e}x$

Let \left(- \dfrac{1}{{y}^{2}}\right) = t ⇒ \dfrac{2}{{y}^{3}}\dfrac{dy}{dx} = \dfrac{dt}{dx}$</p><p>$∴ \dfrac{dt}{2dx} + \dfrac{t}{x} = \left(1 + \log_{e}x\right)$</p><p>$⇒ \dfrac{dt}{dx} + \dfrac{2t}{x} = 2\left(1 + \log_{e}x\right) . . . . . . . . . . ( 1 )$</p><p>We know the solution of the differential equation$\dfrac{dy}{dx} + Py = Q$is given by,</p><p>${ye}^{∫ Pdx} = ∫ {Qe}^{∫ Pdx} + C ( WhereI . F . = {e}^{∫ Pdx} )$</p><p>Therefore, on solving equation(1) we get,</p><p>I.F.$= {e}^{∫ \dfrac{2}{x}dx} = {x}^{2}$</p><p>$\dfrac{- {x}^{2}}{{y}^{2}} = \dfrac{2}{3}\left(\left(1 + \log_{e}x\right){x}^{3} - \dfrac{{x}^{3}}{3}\right) + C . . . . . . ( 2 )$</p><p>Given,$y ( 1 ) = 3$</p><p>$⇒ \dfrac{1}{9} = - 2\left(\dfrac{1}{3} - \dfrac{1}{9}\right) + C$</p><p>$∴ C = \dfrac{5}{9}$</p><p>Now putting value of C in equation(2) we get,</p><p>$\dfrac{{y}^{2}}{9} = \dfrac{{x}^{2}}{5 - 2{x}^{3}\left(2 + \log_{e}{x}^{3}\right)}$