Question

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

Answer 1

On rearranging, we have

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

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

Integrating on both sides, we get

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

$\therefore \log y=\int \frac{1+\log x}{x\log x}dx$

Put $z=x\log x⟹dz=(1+\log x)dx$

$\therefore \log y=\int \frac{1}{z}dz$

$\therefore \log y=\log z+k$

$\therefore y=z{e}^k$

$\therefore y=cx\log x$

Now, substituting $x=e$ and $y=e^2$, we have

$e^2=ce\log e$

$\therefore c=e$

Hence, required particular solution is $y=e\left(x\log x\right)$