1. **State the problem:** Solve the system of differential equations using Laplace transform: $$2 \frac{dx}{dt} + \frac{dy}{dt} = 5 e^t$$ $$\frac{dy}{dt} - 3 \frac{dx}{dt} = 5$$ with initial conditions $x(0) = 0$ and $y(0) = 0$. 2. **Recall Laplace transform properties:** - $\mathcal{L}\{\frac{df}{dt}\} = sF(s) - f(0)$ where $F(s) = \mathcal{L}\{f(t)\}$. - Initial conditions simplify transforms since $x(0) = y(0) = 0$. 3. **Apply Laplace transform to both equations:** For the first equation: $$2 \mathcal{L}\{\frac{dx}{dt}\} + \mathcal{L}\{\frac{dy}{dt}\} = 5 \mathcal{L}\{e^t\}$$ Using linearity and initial conditions: $$2 (sX(s) - 0) + (sY(s) - 0) = 5 \cdot \frac{1}{s-1}$$ Simplify: $$2 s X(s) + s Y(s) = \frac{5}{s-1}$$ For the second equation: $$\mathcal{L}\{\frac{dy}{dt}\} - 3 \mathcal{L}\{\frac{dx}{dt}\} = 5 \mathcal{L}\{1\}$$ $$s Y(s) - 3 s X(s) = \frac{5}{s}$$ 4. **Write the system in terms of $X(s)$ and $Y(s)$:** $$2 s X(s) + s Y(s) = \frac{5}{s-1}$$ $$-3 s X(s) + s Y(s) = \frac{5}{s}$$ 5. **Solve the system:** Subtract second from first to eliminate $Y(s)$: $$\left(2 s X(s) + s Y(s)\right) - \left(-3 s X(s) + s Y(s)\right) = \frac{5}{s-1} - \frac{5}{s}$$ Simplify left side: $$2 s X(s) + s Y(s) + 3 s X(s) - s Y(s) = 5 s X(s)$$ Right side: $$\frac{5}{s-1} - \frac{5}{s} = 5 \left(\frac{1}{s-1} - \frac{1}{s}\right) = 5 \frac{s - (s-1)}{s(s-1)} = \frac{5}{s(s-1)}$$ So: $$5 s X(s) = \frac{5}{s(s-1)}$$ Divide both sides by $5 s$: $$X(s) = \frac{5}{s(s-1)} \cdot \frac{1}{5 s} = \frac{1}{s^2 (s-1)}$$ 6. **Find $Y(s)$ using one of the original equations:** From first equation: $$2 s X(s) + s Y(s) = \frac{5}{s-1}$$ Solve for $Y(s)$: $$s Y(s) = \frac{5}{s-1} - 2 s X(s)$$ $$Y(s) = \frac{5}{s(s-1)} - 2 X(s)$$ Substitute $X(s)$: $$Y(s) = \frac{5}{s(s-1)} - 2 \cdot \frac{1}{s^2 (s-1)} = \frac{5}{s(s-1)} - \frac{2}{s^2 (s-1)}$$ 7. **Perform partial fraction decomposition to invert Laplace transforms:** For $X(s) = \frac{1}{s^2 (s-1)}$: Assume: $$\frac{1}{s^2 (s-1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s-1}$$ Multiply both sides by $s^2 (s-1)$: $$1 = A s (s-1) + B (s-1) + C s^2$$ Set $s=0$: $$1 = B (-1) \Rightarrow B = -1$$ Set $s=1$: $$1 = A \cdot 1 \cdot 0 + B \cdot 0 + C \cdot 1^2 = C \Rightarrow C = 1$$ Set $s=2$: $$1 = A \cdot 2 \cdot 1 + B \cdot 1 + C \cdot 4 = 2A + B + 4C$$ Substitute $B=-1$, $C=1$: $$1 = 2A -1 + 4 \Rightarrow 2A + 3 = 1 \Rightarrow 2A = -2 \Rightarrow A = -1$$ So: $$X(s) = -\frac{1}{s} - \frac{1}{s^2} + \frac{1}{s-1}$$ For $Y(s) = \frac{5}{s(s-1)} - \frac{2}{s^2 (s-1)}$: Rewrite as: $$Y(s) = \frac{5}{s(s-1)} - 2 X(s) = \frac{5}{s(s-1)} - 2 \left(-\frac{1}{s} - \frac{1}{s^2} + \frac{1}{s-1}\right)$$ Simplify: $$Y(s) = \frac{5}{s(s-1)} + \frac{2}{s} + \frac{2}{s^2} - \frac{2}{s-1}$$ Partial fraction for $\frac{5}{s(s-1)}$: $$\frac{5}{s(s-1)} = \frac{D}{s} + \frac{E}{s-1}$$ Multiply both sides by $s(s-1)$: $$5 = D (s-1) + E s$$ Set $s=0$: $$5 = D (-1) \Rightarrow D = -5$$ Set $s=1$: $$5 = E \cdot 1 \Rightarrow E = 5$$ So: $$Y(s) = -\frac{5}{s} + \frac{5}{s-1} + \frac{2}{s} + \frac{2}{s^2} - \frac{2}{s-1} = \left(-5 + 2\right) \frac{1}{s} + \frac{2}{s^2} + \left(5 - 2\right) \frac{1}{s-1}$$ Simplify coefficients: $$Y(s) = -\frac{3}{s} + \frac{2}{s^2} + \frac{3}{s-1}$$ 8. **Find inverse Laplace transforms:** Recall: - $\mathcal{L}^{-1}\{\frac{1}{s}\} = 1$ - $\mathcal{L}^{-1}\{\frac{1}{s^2}\} = t$ - $\mathcal{L}^{-1}\{\frac{1}{s-a}\} = e^{a t}$ So: $$x(t) = -1 - t + e^t$$ $$y(t) = -3 + 2 t + 3 e^t$$ 9. **Check initial conditions:** At $t=0$: $$x(0) = -1 - 0 + 1 = 0$$ $$y(0) = -3 + 0 + 3 = 0$$ Initial conditions satisfied. **Final answer:** $$\boxed{x(t) = e^t - t - 1, \quad y(t) = 3 e^t + 2 t - 3}$$