The given differential equation is:

$x \frac{dy}{dx} - y = 3$

This is a first-order linear differential equation that we can solve directly without using integration to find the relationship between $x$ and $y$. Let's start by simplifying and analyzing the equation.

We can rearrange it as follows:

$x \frac{dy}{dx} = y + 3$

Then,

$\frac{dy}{dx} = \frac{y + 3}{x}$

To understand why the relationship includes absolute values, we notice that for any solution to the differential equation, the equation must hold for both positive and negative values of $x$. In this particular case, absolute values appear because the function $x$ can take both positive and negative values, while maintaining a consistent relationship between $x$ and $y$.

Given the initial condition $x = 1$ when $y = -2$, you can substitute these values into the final relationship. When solving, you would find that $|y + 3| = |x|$ ensures the equation is valid for all signs of $x$ and $y + 3$. The absolute value reflects the fact that the relationship is symmetric and that flipping the signs of $x$ or $y + 3$ doesn't change the validity of the equation.

Therefore, the correct relationship includes absolute values to account for the fact that the magnitude (not just the sign) of $x$ and $y + 3$ is what determines the equality.

Would you like more details, or do you have any questions?

Here are five related questions for further exploration:

1. What is the general solution of a first-order linear differential equation?
2. How does the introduction of initial conditions affect the solution of a differential equation?
3. Why do absolute values often appear in solutions involving square roots?
4. Can you have a solution to a differential equation that does not satisfy a given initial condition?
5. How can phase plane analysis be used to visualize the solutions of differential equations?

Tip: When analyzing equations with absolute values, remember that they represent two cases: one where the expression inside is positive and another where it's negative.