1. **State the problem:** Solve the system of differential equations: $$\begin{cases} x' = -4x \\ y' = -6x - y \end{cases}$$ with initial conditions $x(0) = -1$ and $y(0) = 0$. 2. **Solve for $x(t)$:** The first equation is $x' = -4x$, a separable ODE. Using the formula for first-order linear ODEs, the solution is: $$x(t) = Ce^{-4t}$$ Apply initial condition $x(0) = -1$: $$-1 = Ce^{0} = C \implies C = -1$$ So, $$x(t) = -e^{-4t}$$ 3. **Solve for $y(t)$:** Substitute $x(t)$ into the second equation: $$y' = -6x - y = -6(-e^{-4t}) - y = 6e^{-4t} - y$$ Rewrite as: $$y' + y = 6e^{-4t}$$ 4. **Use integrating factor method:** The integrating factor is: $$\mu(t) = e^{\int 1 dt} = e^{t}$$ Multiply both sides by $e^{t}$: $$e^{t}y' + e^{t}y = 6e^{t}e^{-4t} = 6e^{-3t}$$ This simplifies to: $$(e^{t}y)' = 6e^{-3t}$$ 5. **Integrate both sides:** $$e^{t}y = \int 6e^{-3t} dt = 6 \int e^{-3t} dt = 6 \left(-\frac{1}{3} e^{-3t}\right) + C = -2e^{-3t} + C$$ 6. **Solve for $y(t)$:** $$y(t) = e^{-t}(-2e^{-3t} + C) = -2e^{-4t} + Ce^{-t}$$ 7. **Apply initial condition $y(0) = 0$:** $$0 = -2e^{0} + C e^{0} = -2 + C \implies C = 2$$ 8. **Final solution:** $$x(t) = -e^{-4t}$$ $$y(t) = -2e^{-4t} + 2e^{-t}$$ This completes the solution to the system.