Ivp And System 04E0A4

1. **Problem (a):** Solve the initial value problem $$y'' + 3y' - 4y = 6e^{2x} - 3$$ with initial conditions $$y(1.5) = 4$$ and $$y'(1.5) = 5$$. 2. **Step 1: Solve the homogeneous equation** The associated homogeneous equation is: $$y'' + 3y' - 4y = 0$$ The characteristic equation is: $$r^2 + 3r - 4 = 0$$ 3. **Step 2: Solve the characteristic equation** Using the quadratic formula: $$r = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm \sqrt{25}}{2}$$ So, $$r_1 = \frac{-3 + 5}{2} = 1, \quad r_2 = \frac{-3 - 5}{2} = -4$$ 4. **Step 3: Write the general solution of the homogeneous equation** $$y_h = C_1 e^{x} + C_2 e^{-4x}$$ 5. **Step 4: Find a particular solution** The nonhomogeneous term is $$6e^{2x} - 3$$, which is a sum of an exponential and a constant. Try a particular solution of the form: $$y_p = Ae^{2x} + B$$ 6. **Step 5: Compute derivatives of $$y_p$$** $$y_p' = 2Ae^{2x}$$ $$y_p'' = 4Ae^{2x}$$ 7. **Step 6: Substitute $$y_p$$ and its derivatives into the differential equation** $$y_p'' + 3y_p' - 4y_p = 6e^{2x} - 3$$ Substitute: $$4Ae^{2x} + 3(2Ae^{2x}) - 4(Ae^{2x} + B) = 6e^{2x} - 3$$ Simplify: $$4Ae^{2x} + 6Ae^{2x} - 4Ae^{2x} - 4B = 6e^{2x} - 3$$ $$ (4A + 6A - 4A) e^{2x} - 4B = 6e^{2x} - 3$$ $$6Ae^{2x} - 4B = 6e^{2x} - 3$$ 8. **Step 7: Equate coefficients** For $$e^{2x}$$ terms: $$6A = 6 \implies A = 1$$ For constant terms: $$-4B = -3 \implies B = \frac{3}{4}$$ 9. **Step 8: Write the particular solution** $$y_p = e^{2x} + \frac{3}{4}$$ 10. **Step 9: Write the general solution** $$y = y_h + y_p = C_1 e^{x} + C_2 e^{-4x} + e^{2x} + \frac{3}{4}$$ 11. **Step 10: Use initial conditions to find $$C_1$$ and $$C_2$$** Calculate $$y'(x)$$: $$y' = C_1 e^{x} - 4C_2 e^{-4x} + 2e^{2x}$$ Apply initial conditions at $$x=1.5$$: $$y(1.5) = C_1 e^{1.5} + C_2 e^{-6} + e^{3} + \frac{3}{4} = 4$$ $$y'(1.5) = C_1 e^{1.5} - 4C_2 e^{-6} + 2 e^{3} = 5$$ 12. **Step 11: Solve the system for $$C_1$$ and $$C_2$$** Let $$a = e^{1.5}$$, $$b = e^{-6}$$, $$c = e^{3}$$. Equations: $$C_1 a + C_2 b = 4 - c - \frac{3}{4}$$ $$C_1 a - 4 C_2 b = 5 - 2 c$$ Add the two equations: $$2 C_1 a - 3 C_2 b = (4 - c - \frac{3}{4}) + (5 - 2 c)$$ But better to solve by substitution or elimination: From first: $$C_1 a = 4 - c - \frac{3}{4} - C_2 b$$ Substitute into second: $$(4 - c - \frac{3}{4} - C_2 b) - 4 C_2 b = 5 - 2 c$$ $$4 - c - \frac{3}{4} - 5 C_2 b = 5 - 2 c$$ $$- 5 C_2 b = 5 - 2 c - 4 + c + \frac{3}{4}$$ $$- 5 C_2 b = 1 - c + \frac{3}{4} = \frac{7}{4} - c$$ $$C_2 = - \frac{\frac{7}{4} - c}{5 b}$$ Then $$C_1 = \frac{4 - c - \frac{3}{4} - C_2 b}{a}$$ 13. **Final answer for (a):** $$y = C_1 e^{x} + C_2 e^{-4x} + e^{2x} + \frac{3}{4}$$ with $$C_1, C_2$$ as above. --- 14. **Problem (b):** Find the general solution of the system $$\begin{cases} y_1' = y_1 - 2 y_2 + \cos t \\ y_2' = -2 y_1 + y_2 - \sin t \end{cases}$$ 15. **Step 1: Write system in matrix form** $$\mathbf{y}' = A \mathbf{y} + \mathbf{f}(t)$$ where $$\mathbf{y} = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix}, \quad A = \begin{bmatrix} 1 & -2 \\ -2 & 1 \end{bmatrix}, \quad \mathbf{f}(t) = \begin{bmatrix} \cos t \\ -\sin t \end{bmatrix}$$ 16. **Step 2: Solve the homogeneous system** $$\mathbf{y}' = A \mathbf{y}$$ Find eigenvalues $$\lambda$$ from $$\det(A - \lambda I) = 0$$ $$\det \begin{bmatrix} 1 - \lambda & -2 \\ -2 & 1 - \lambda \end{bmatrix} = (1 - \lambda)^2 - 4 = 0$$ $$ (1 - \lambda)^2 = 4$$ $$1 - \lambda = \pm 2$$ So, $$\lambda_1 = 1 - 2 = -1, \quad \lambda_2 = 1 + 2 = 3$$ 17. **Step 3: Find eigenvectors** For $$\lambda_1 = -1$$: $$(A + I) \mathbf{v} = 0$$ $$\begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = 0$$ From first row: $$2 v_1 - 2 v_2 = 0 \implies v_1 = v_2$$ Eigenvector: $$\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$ For $$\lambda_2 = 3$$: $$(A - 3I) \mathbf{v} = 0$$ $$\begin{bmatrix} -2 & -2 \\ -2 & -2 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = 0$$ From first row: $$-2 v_1 - 2 v_2 = 0 \implies v_1 = -v_2$$ Eigenvector: $$\mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$$ 18. **Step 4: Write homogeneous solution** $$\mathbf{y}_h = C_1 e^{-t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + C_2 e^{3t} \begin{bmatrix} 1 \\ -1 \end{bmatrix}$$ 19. **Step 5: Find particular solution** Try a particular solution of the form: $$\mathbf{y}_p = \begin{bmatrix} A \cos t + B \sin t \\ C \cos t + D \sin t \end{bmatrix}$$ 20. **Step 6: Compute $$\mathbf{y}_p'$$** $$\mathbf{y}_p' = \begin{bmatrix} -A \sin t + B \cos t \\ -C \sin t + D \cos t \end{bmatrix}$$ 21. **Step 7: Substitute into the system** $$\mathbf{y}_p' = A \mathbf{y}_p + \mathbf{f}(t)$$ Write components: First equation: $$-A \sin t + B \cos t = 1 \cdot (A \cos t + B \sin t) - 2 (C \cos t + D \sin t) + \cos t$$ Second equation: $$-C \sin t + D \cos t = -2 (A \cos t + B \sin t) + 1 (C \cos t + D \sin t) - \sin t$$ 22. **Step 8: Group coefficients of $$\cos t$$ and $$\sin t$$** First equation: $$-A \sin t + B \cos t = A \cos t + B \sin t - 2C \cos t - 2D \sin t + \cos t$$ Group $$\cos t$$: Left: $$B \cos t$$ Right: $$A \cos t - 2C \cos t + \cos t = (A - 2C + 1) \cos t$$ Group $$\sin t$$: Left: $$-A \sin t$$ Right: $$B \sin t - 2D \sin t = (B - 2D) \sin t$$ Equate coefficients: $$B = A - 2C + 1$$ $$-A = B - 2D$$ Second equation: $$-C \sin t + D \cos t = -2 A \cos t - 2 B \sin t + C \cos t + D \sin t - \sin t$$ Group $$\cos t$$: Left: $$D \cos t$$ Right: $$-2 A \cos t + C \cos t = (-2 A + C) \cos t$$ Group $$\sin t$$: Left: $$-C \sin t$$ Right: $$-2 B \sin t + D \sin t - \sin t = (-2 B + D - 1) \sin t$$ Equate coefficients: $$D = -2 A + C$$ $$-C = -2 B + D - 1$$ 23. **Step 9: Solve the system of equations** Equations: (1) $$B = A - 2C + 1$$ (2) $$-A = B - 2D$$ (3) $$D = -2 A + C$$ (4) $$-C = -2 B + D - 1$$ Substitute (1) and (3) into (2) and (4): From (2): $$-A = (A - 2C + 1) - 2(-2 A + C) = A - 2C + 1 + 4 A - 2 C = 5 A - 4 C + 1$$ So: $$-A = 5 A - 4 C + 1 \implies -A - 5 A + 4 C = 1 \implies -6 A + 4 C = 1$$ From (4): $$-C = -2 (A - 2C + 1) + (-2 A + C) - 1 = -2 A + 4 C - 2 - 2 A + C - 1 = -4 A + 5 C - 3$$ So: $$-C = -4 A + 5 C - 3 \implies -C - 5 C = -4 A - 3 \implies -6 C = -4 A - 3$$ Rewrite: $$-6 A + 4 C = 1 \quad (i)$$ $$-6 C = -4 A - 3 \implies 6 C = 4 A + 3 \quad (ii)$$ From (ii): $$C = \frac{4 A + 3}{6}$$ Substitute into (i): $$-6 A + 4 \cdot \frac{4 A + 3}{6} = 1$$ Multiply both sides by 6: $$-36 A + 4 (4 A + 3) = 6$$ $$-36 A + 16 A + 12 = 6$$ $$-20 A + 12 = 6$$ $$-20 A = -6$$ $$A = \frac{3}{10}$$ Then, $$C = \frac{4 \cdot \frac{3}{10} + 3}{6} = \frac{\frac{12}{10} + 3}{6} = \frac{\frac{12}{10} + \frac{30}{10}}{6} = \frac{\frac{42}{10}}{6} = \frac{42}{60} = \frac{7}{10}$$ Now find $$B$$ and $$D$$: $$B = A - 2 C + 1 = \frac{3}{10} - 2 \cdot \frac{7}{10} + 1 = \frac{3}{10} - \frac{14}{10} + 1 = -\frac{11}{10} + 1 = -\frac{1}{10}$$ $$D = -2 A + C = -2 \cdot \frac{3}{10} + \frac{7}{10} = -\frac{6}{10} + \frac{7}{10} = \frac{1}{10}$$ 24. **Step 10: Write the particular solution** $$\mathbf{y}_p = \begin{bmatrix} \frac{3}{10} \cos t - \frac{1}{10} \sin t \\ \frac{7}{10} \cos t + \frac{1}{10} \sin t \end{bmatrix}$$ 25. **Step 11: Write the general solution** $$\mathbf{y} = C_1 e^{-t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + C_2 e^{3t} \begin{bmatrix} 1 \\ -1 \end{bmatrix} + \begin{bmatrix} \frac{3}{10} \cos t - \frac{1}{10} \sin t \\ \frac{7}{10} \cos t + \frac{1}{10} \sin t \end{bmatrix}$$