Q.Find the particular solution of the differential equation: $y(1+\log x)\dfrac{dx}{dy} - x\log x = 0$, when $y = e^2$ and $x = e$.

Separate the variables, integrate both sides (using $\int\frac{1}{x\log x}dx=\log|\log x|$), then apply the initial condition.

Given: $y(1+\log x)\dfrac{dx}{dy} - x\log x = 0$, i.e.

$y(1+\log x)\frac{dx}{dy} = x\log x$

Separate variables:

$\frac{1+\log x}{x\log x}\,dx = \frac{dy}{y}$

Split the left side:

$\frac{1+\log x}{x\log x} = \frac{1}{x\log x}+\frac{1}{x}$

Integrate both sides:

$\int\left[\frac{1}{x\log x}+\frac1x\right]dx = \int\frac{dy}{y}$

For $\displaystyle\int\frac{1}{x\log x}\,dx$, let $u=\log x$, $du=\dfrac{dx}{x}$: this gives $\displaystyle\int\frac{du}{u}=\log|\log x|$.

$\log|\log x| + \log|x| = \log|y| + C$

$\log|x\log x| = \log|y|+C \implies x\log x = k\,y \quad (\text{where } k=e^{C})$

Apply the initial condition $y=e^2$ when $x=e$: since $\log e=1$, …