1. **State the problem:** Given the differential equation $x \frac{dy}{dx} - y = 3$ with the initial condition $x=1$ when $y=-2$, find the relation between $x$ and $y$. 2. **Rewrite the equation:** The equation can be rearranged as $$x \frac{dy}{dx} = y + 3$$ which implies $$\frac{dy}{dx} = \frac{y+3}{x}$$ 3. **Separate variables:** Rewrite as $$\frac{dy}{y+3} = \frac{dx}{x}$$ 4. **Integrate both sides:** $$\int \frac{1}{y+3} dy = \int \frac{1}{x} dx$$ which gives $$\ln|y+3| = \ln|x| + C$$ where $C$ is the constant of integration. 5. **Exponentiate both sides:** $$|y+3| = e^C |x|$$ Let $k = e^C$, so $$|y+3| = k |x|$$ 6. **Use initial condition:** When $x=1$, $y=-2$, substitute to find $k$: $$|-2 + 3| = k |1| \Rightarrow |1| = k \Rightarrow k=1$$ 7. **Final relation:** $$|y+3| = |x|$$ **Answer:** Option (b) $|y + 3| = |x|$