Absolute value in a differential equation

[Mr Davis 97]

Homework Statement

$\displaystyle (x+3)\frac{dy}{dx} = y - 2$, where x is not 3 and y is not 2.

Homework Equations

The Attempt at a Solution

$\displaystyle (x+3)\frac{dy}{dx} = y - 2$
$\displaystyle \frac{dy}{y-2} = \frac{dx}{x+3}$
$\displaystyle \int \frac{dy}{y-2} = \int \frac{dx}{x+3}$
$\ln|y-2| = \ln|x+3| + c$
$\displaystyle |y-2| = e^{\ln |x+3|+c}$
$\displaystyle |y-2| = ke^{\ln |x+3|}$, where $k=e^c$
$\displaystyle |y-2| = k|x+3|$

This is as far as I can get. I am not sure how I can get rid of the absolute value signs in order to obtain a single function.

 

[andrewkirk]

You can't, because there are two functions, not one, so you have to choose one of them - sometimes called 'choosing a branch'.
To find out what the two functions are, write the right-hand side, which is a well-defined function, as f(x), and then explicitly write out the two possibilities created by the absolute value function on the left hand side.

Rearranging, you will get two different expressions for y in terms of f(x), corresponding to the two different branches of the absolute value.

 

[Mr Davis 97]

You can't, because there are two functions, not one, so you have to choose one of them - sometimes called 'choosing a branch'.
To find out what the two functions are, write the right-hand side, which is a well-defined function, as f(x), and then explicitly write out the two possibilities created by the absolute value function on the left hand side.

Rearranging, you will get two different expressions for y in terms of f(x), corresponding to the two different branches of the absolute value.

So are the two possible functions

$y = 2 + k|x+3|$
and
$y = 2-k|x+3|$
?

So I have to choose one of these as a solution?

 

[Mark44]

So are the two possible functions

$y = 2 + k|x+3|$
and
$y = 2-k|x+3|$
?

So I have to choose one of these as a solution?

Without an initial condition you can't know which is the correct solution. In that case, you can leave your solution as $y = 2 \pm |x + 3|$, with k being a positive constant.

 

[Mr Davis 97]

Without an initial condition you can't know which is the correct solution. In that case, you can leave your solution as $y = 2 \pm |x + 3|$, with k being a positive constant.

Given that the initial condition is (x = -1, y = 1), how do you go from $y = 2 \pm k|x+3|$ to a specific solution?

 

[Samy_A]

Given that the initial condition is (x = -1, y = 1), how do you go from $y = 2 \pm k|x+3|$ to a specific solution?

Plug in the initial condition in each possible solution and see what you get.