# Homework Help: Absolute value in a differential equation

1. Dec 27, 2015

### Mr Davis 97

1. The problem statement, all variables and given/known data
$\displaystyle (x+3)\frac{dy}{dx} = y - 2$, where x is not 3 and y is not 2.

2. Relevant equations

3. 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.

2. Dec 27, 2015

### 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.

3. Dec 27, 2015

### Mr Davis 97

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?

4. Dec 27, 2015

### Staff: Mentor

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.

5. Dec 28, 2015

### Mr Davis 97

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?

6. Dec 28, 2015

### Samy_A

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