1. **Problem statement:** Given the continuous LTI system described by the differential equation $$\frac{d^2y(t)}{dt^2} + 6 \frac{dy(t)}{dt} + 8 y(t) = x(t),$$ we need to find: (i) The impulse response $h(t)$. (ii) The output $y(t)$ when the input is $x(t) = e^{-2t} u(t)$, where $u(t)$ is the unit step function. 2. **Impulse response calculation:** The impulse response $h(t)$ is the output when the input $x(t)$ is the Dirac delta function $\delta(t)$. The differential equation becomes: $$\frac{d^2h(t)}{dt^2} + 6 \frac{dh(t)}{dt} + 8 h(t) = \delta(t).$$ 3. **Solve the homogeneous equation:** $$\frac{d^2h(t)}{dt^2} + 6 \frac{dh(t)}{dt} + 8 h(t) = 0.$$ The characteristic equation is: $$r^2 + 6r + 8 = 0.$$ Solve for $r$: $$r = \frac{-6 \pm \sqrt{36 - 32}}{2} = \frac{-6 \pm 2}{2}.$$ So, $$r_1 = -2, \quad r_2 = -4.$$ The homogeneous solution is: $$h_h(t) = C_1 e^{-2t} + C_2 e^{-4t}.$$ 4. **Find the impulse response using Laplace transform:** Taking Laplace transform of the differential equation: $$s^2 H(s) + 6 s H(s) + 8 H(s) = 1,$$ since $\mathcal{L}\{\delta(t)\} = 1$. Factor out $H(s)$: $$H(s)(s^2 + 6s + 8) = 1,$$ so $$H(s) = \frac{1}{(s+2)(s+4)}.$$ 5. **Partial fraction decomposition:** $$\frac{1}{(s+2)(s+4)} = \frac{A}{s+2} + \frac{B}{s+4}.$$ Multiply both sides by $(s+2)(s+4)$: $$1 = A(s+4) + B(s+2).$$ Set $s = -2$: $$1 = A(2) \Rightarrow A = \frac{1}{2}.$$ Set $s = -4$: $$1 = B(-2) \Rightarrow B = -\frac{1}{2}.$$ 6. **Inverse Laplace transform:** $$h(t) = \frac{1}{2} e^{-2t} u(t) - \frac{1}{2} e^{-4t} u(t) = \frac{1}{2} (e^{-2t} - e^{-4t}) u(t).$$ 7. **Output for input $x(t) = e^{-2t} u(t)$:** Use convolution: $$y(t) = h(t) * x(t) = \int_0^t h(\tau) x(t-\tau) d\tau.$$ Substitute $h(\tau)$ and $x(t-\tau)$: $$y(t) = \int_0^t \frac{1}{2} (e^{-2\tau} - e^{-4\tau}) e^{-2(t-\tau)} d\tau = \frac{1}{2} e^{-2t} \int_0^t (e^{-2\tau} - e^{-4\tau}) e^{2\tau} d\tau.$$ Simplify inside the integral: $$e^{-2\tau} e^{2\tau} = 1, \quad e^{-4\tau} e^{2\tau} = e^{-2\tau}.$$ So, $$y(t) = \frac{1}{2} e^{-2t} \int_0^t (1 - e^{-2\tau}) d\tau = \frac{1}{2} e^{-2t} \left[ \tau + \frac{1}{2} e^{-2\tau} \right]_0^t = \frac{1}{2} e^{-2t} \left(t - \frac{1}{2} + \frac{1}{2} e^{-2t} \right).$$ 8. **Final output expression:** $$y(t) = \frac{1}{2} e^{-2t} t - \frac{1}{4} e^{-2t} + \frac{1}{4} e^{-4t}.$$ **Summary:** - Impulse response: $$h(t) = \frac{1}{2} (e^{-2t} - e^{-4t}) u(t).$$ - Output for $x(t) = e^{-2t} u(t)$: $$y(t) = \frac{1}{2} t e^{-2t} - \frac{1}{4} e^{-2t} + \frac{1}{4} e^{-4t}.$$