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

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Answer 1 · 2016-12-06T18:38:43

$y = ln(1/(3-x))$

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

$y=lny'$
$:. ln(dy/dx)=y$
$:. dy/dx=e^y$

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

$\ \ \ \ \ \ int 1/e^ydy = int dx$
$:. int e^-ydy = int dx$

We can now integrate to get:

$-e^-y = x+c$
$:. e^-y = -x-c$

Using $y(2)=0 =>$

$e^0=-2+c$
$:. x=1+2=3$

$:. e^-y = 3-x$
$:. -y = ln(3-x)$
$:. y = -ln(3-x)$
$:. y = ln(1/(3-x))$

CHECK

$y = ln(1/(3-x))$
$:. y(2) = ln(1/1)=0$ , so the initial condition is met

And

$dy/dx = 1/((1/(3-x))) * (-(3-x)^-2) (-1)$
$:. dy/dx = (3-x) * 1/(3-x)^2$
$:. dy/dx = 1/(3-x)$
$:. ln(dy/dx) = y$ , so the DE condition is met

Confirming our solution is correct.

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