Let y = y x be the solution curve of the differential equation d y d x = y x 1 + x 2 1 + log e x , x > 0…

Given differential equation can be rewritten as,

$\frac{dy}{dx}-\frac{y}{x}=y^3\left(1+{\log }_{e}x\right)$

$\frac{1}{{\mathrm{y}}^3}\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{1}{{\mathrm{xy}}^2}=1+{\log }_e\mathrm{x}$

Let $\left(-\frac{1}{{\mathrm{y}}^2}\right)=\mathrm{t}\Rightarrow \frac{2}{{\mathrm{y}}^3}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{dx}}$

$\therefore \frac{\mathrm{dt}}{2\mathrm{dx}}+\frac{\mathrm{t}}{\mathrm{x}}=\left(1+{\log }_e\mathrm{x}\right)$

$\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}+\frac{2\mathrm{t}}{\mathrm{x}}=2\left(1+{\log }_e\mathrm{x}\right) ..........\left(1\right)$

We know the solution of the differential equation $\frac{dy}{dx}+Py=Q$ is given by,

${\mathrm{ye}}^{\int \mathrm{Pdx}}=\int {\mathrm{Qe}}^{\int \mathrm{Pdx}}+\mathrm{C} (\mathrm{Where} \mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{Pdx}})$

Therefore, on solving equation(1) we get,

I.F.$={\mathrm{e}}^{\int \frac{2}{x}dx}=x^2$

$\frac{-x^2}{y^2}=\frac{2}{3}\left(\left(1+{\log }_{e}x\right)x^3-\frac{x^3}{3}\right)+C ......\left(2\right)$

Given,$y\left(1\right)=3$

$\Rightarrow \frac{1}{9}=-2\left(\frac{1}{3}-\frac{1}{9}\right)+C$

$\therefore C=\frac{5}{9}$

Now putting value of C in equation(2) we get,

$\frac{y^2}{9}=\frac{x^2}{5-2x^3\left(2+{\log }_e{x}^3\right)}$