Let's start by solving the differential equation $x \frac{dy}{dx} - y = 3$ given the initial condition $x = 1$ when $y = -2$.

Step 1: Rearrange the differential equation

We can rewrite the differential equation as: $x \frac{dy}{dx} = y + 3$

Step 2: Separate the variables

To solve the differential equation, we'll separate the variables: $\frac{dy}{y + 3} = \frac{dx}{x}$

Step 3: Integrate both sides

Next, integrate both sides: $\int \frac{dy}{y + 3} = \int \frac{dx}{x}$

The integral of the left side is: $\ln |y + 3|$ And the integral of the right side is: $\ln |x|$

So, we have: $\ln |y + 3| = \ln |x| + C$ where $C$ is the constant of integration.

Step 4: Solve for $y$

Exponentiating both sides to remove the logarithms: $|y + 3| = |x| \cdot e^C$

Since $e^C$ is a positive constant, let's denote it as $k$: $|y + 3| = k |x|$

Step 5: Apply the initial condition

We know that when $x = 1$, $y = -2$. Substituting these into the equation: $|-2 + 3| = k |1|$ $|1| = k \times 1 \implies k = 1$

So, the relationship becomes: $|y + 3| = |x|$

Why Absolute Value?

The absolute value arises naturally from the integration process because the logarithm of a negative value is not defined within the real numbers, so we consider the absolute value to ensure we stay within real numbers. This also accounts for the fact that $y + 3$ and $x$ could be either positive or negative, depending on the values of $x$ and $y$.

Thus, the correct relationship is $|y + 3| = |x|$, which corresponds to option (a).

Would you like further details or explanations?

5 Related Questions:

1. How do you handle absolute values when solving equations?
2. What are the properties of logarithmic functions that lead to absolute values in solutions?
3. How would the solution change if the initial condition was different?
4. What is the general solution to a first-order linear differential equation?
5. How do you interpret the solution graphically?

Tip:

Always check the sign of your variables when solving logarithmic or exponential equations, as this often leads to the appearance of absolute values in the solution.