# How do you solve y=lny' given y(2)=0?

Dec 6, 2016

$y = \ln \left(\frac{1}{3 - x}\right)$

#### Explanation:

A am assuming that $y$ is a function of $x$, ie $y = f \left(x\right)$

$y = \ln y '$
$\therefore \ln \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = y$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{y}$

We now have a first order separable differential equation, so we can separate the variables to get:

$\setminus \setminus \setminus \setminus \setminus \setminus \int \frac{1}{e} ^ y \mathrm{dy} = \int \mathrm{dx}$
$\therefore \int {e}^{-} y \mathrm{dy} = \int \mathrm{dx}$

We can now integrate to get:

$- {e}^{-} y = x + c$
$\therefore {e}^{-} y = - x - c$

Using $y \left(2\right) = 0 \implies$

${e}^{0} = - 2 + c$
$\therefore x = 1 + 2 = 3$

$\therefore {e}^{-} y = 3 - x$
$\therefore - y = \ln \left(3 - x\right)$
$\therefore y = - \ln \left(3 - x\right)$
$\therefore y = \ln \left(\frac{1}{3 - x}\right)$

CHECK

$y = \ln \left(\frac{1}{3 - x}\right)$
$\therefore y \left(2\right) = \ln \left(\frac{1}{1}\right) = 0$, so the initial condition is met

And

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\left(\frac{1}{3 - x}\right)} \cdot \left(- {\left(3 - x\right)}^{-} 2\right) \left(- 1\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \left(3 - x\right) \cdot \frac{1}{3 - x} ^ 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3 - x}$
$\therefore \ln \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = y$, so the DE condition is met

Confirming our solution is correct.