# What is a solution to the differential equation xydy/dx-lnx=0 with y(1)=0?

Jan 17, 2017

${y}^{2} = {\ln}^{2} x$

#### Explanation:

$x y \frac{\mathrm{dy}}{\mathrm{dx}} - \ln x = 0$

Is a First Order Separable Differential Equation, so we can just separate the variables;

$\setminus \setminus \setminus x y \frac{\mathrm{dy}}{\mathrm{dx}} = \ln x$
$\therefore y \frac{\mathrm{dy}}{\mathrm{dx}} = \ln \frac{x}{x}$

$\int \setminus y \setminus \mathrm{dy} = \int \setminus \ln \frac{x}{x} \setminus \mathrm{dx}$

The LHS is immediately integrable, and for the RHS we use the substitution;

$u = \ln x \implies \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x}$

Which gives:

$\int \setminus y \setminus \mathrm{dy} = \int \setminus u \setminus \mathrm{du}$
$\therefore \frac{1}{2} {y}^{2} = \frac{1}{2} {u}^{2} + K$
$\therefore {y}^{2} = {u}^{2} + 2 K$
$\therefore {y}^{2} = {\ln}^{2} x + 2 K$

We are also given $y \left(1\right) = 0 \implies$

$0 = \ln 1 + 2 K$
$\therefore K = 0$

Hence the particular solution is;

${y}^{2} = {\ln}^{2} x$

Check
We can quickly validate the solution;

${y}^{2} = {\ln}^{2} x$
$2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(\ln x\right) \cdot \frac{1}{x}$
$y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\ln x}{x}$
$x y \frac{\mathrm{dy}}{\mathrm{dx}} = \ln x$
$x y \frac{\mathrm{dy}}{\mathrm{dx}} - \ln x = 0 \setminus \setminus$ QED