1. **State the problem:** Solve the initial-value problem $$\frac{dy}{dx} = \frac{xy}{x+3}, \quad y(-2) = 3,$$ assuming $x > -3$. 2. **Rewrite the differential equation:** Separate variables to isolate $y$ and $x$ terms: $$\frac{dy}{y} = \frac{x}{x+3} dx.$$ 3. **Integrate both sides:** $$\int \frac{1}{y} dy = \int \frac{x}{x+3} dx.$$ 4. **Simplify the right integral:** Use polynomial division: $$\frac{x}{x+3} = 1 - \frac{3}{x+3}.$$ 5. **Rewrite the integral:** $$\int \frac{x}{x+3} dx = \int \left(1 - \frac{3}{x+3}\right) dx = \int 1 dx - 3 \int \frac{1}{x+3} dx.$$ 6. **Integrate:** $$\int 1 dx = x,$$ $$\int \frac{1}{x+3} dx = \ln|x+3|.$$ 7. **Combine results:** $$\int \frac{x}{x+3} dx = x - 3 \ln|x+3| + C,$$ where $C$ is the constant of integration. 8. **Integrate left side:** $$\int \frac{1}{y} dy = \ln|y|.$$ 9. **Equate integrals:** $$\ln|y| = x - 3 \ln|x+3| + C.$$ 10. **Simplify using $x > -3$ assumption:** Since $x+3 > 0$, $|x+3| = x+3$. 11. **Exponentiate both sides:** $$|y| = e^{x - 3 \ln(x+3) + C} = e^C e^x (x+3)^{-3}.$$ 12. **Replace $e^C$ with $A$:** $$y = A e^x (x+3)^{-3}.$$ 13. **Use initial condition $y(-2) = 3$ to find $A$:** $$3 = A e^{-2} (1)^{-3} = A e^{-2} \implies A = 3 e^{2}.$$ 14. **Final solution:** $$\boxed{y = 3 e^{2+x} (x+3)^{-3}}.$n This is the exact formula for $y$ satisfying the initial-value problem with $x > -3$.