Question

For the differential equation find the general solution of
$y.\log ydx-xdy=0$

Answer 1

Given differential equation is

$y.\log ydx-xdy=0$

$\frac{dy}{y\log y}=\frac{dx}{x}$

Integrating both sides
We get,

$\int \frac{dy}{y\log y}=\int \frac{dx}{x}$ (i)

Let $u=\log y$
Differentiating $u$ w.r.t. $y$
we get,

$\frac{du}{dy}=\log y$

$dy=\frac{du}{y}$

Putting value of $dy$ in $\left(i\right)$ then we get,

$\int \frac{ydu}{y.u}=\int \frac{dx}{x}$

$\int \frac{du}{u}=\int \frac{dx}{x}$

$\log |u|=\log |x|+\log c$

Now, putting $u=\log y$ then we get,

$\log \left(\log y\right)=\log x+\log c$

$\log \left(\log y\right)=\log cx$

$(\log ab=\log a+\log b)$

$\log y=cx$

$y=e^{cx}$

Final Answer:
Hence, the required general solution is

$y=e^{cx}$