Let's solve **both problems** step by step, using the **method of undetermined coefficients** and the given initial conditions. --- ## **Problem 4** **Equation:** \( 6y'' + y' - y = 25e^{2x} + 7e^{-2x} \), \( y(0) = 1 \), \( y'(0) = -1 \) ### **Step 1: Solve the Homogeneous Equation** Homogeneous equation: \( 6y'' + y' - y = 0 \) The characteristic equation: \( 6r^2 + r - 1 = 0 \) Solve for \( r \): \[ r = \frac{-1 \pm \sqrt{1^2 - 4(6)(-1)}}{2 \cdot 6} = \frac{-1 \pm \sqrt{1 + 24}}{12} = \frac{-1 \pm 5}{12} \] So, - \( r_1 = \frac{4}{12} = \frac{1}{3} \) - \( r_2 = \frac{-6}{12} = -\frac{1}{2} \) So the general solution to the homogeneous part is: \[ y_h(x) = C_1 e^{x/3} + C_2 e^{-x/2} \] --- ### **Step 2: Find a Particular Solution \(y_p(x)\)** Since the right side is \(25e^{2x} + 7e^{-2x}\), try: \[ y_p(x) = A e^{2x} + B e^{-2x} \] Compute derivatives: - \( y_p' = 2A e^{2x} - 2B e^{-2x} \) - \( y_p'' = 4A e^{2x} + 4B e^{-2x} \) Substitute into the original equation: \[ 6[4A e^{2x} + 4B e^{-2x}] + [2A e^{2x} - 2B e^{-2x}] - [A e^{2x} + B e^{-2x}] \] \[ = (24A + 2A - A) e^{2x} + (24B - 2B - B) e^{-2x} \] \[ = (25A) e^{2x} + (21B) e^{-2x} \] Set equal to the original right side: \( 25A e^{2x} + 21B e^{-2x} = 25e^{2x} + 7e^{-2x} \) Matching coefficients: - \( 25A = 25 \implies A = 1 \) - \( 21B = 7 \implies B = \frac{1}{3} \) So, \[ y_p(x) = e^{2x} + \frac{1}{3} e^{-2x} \] --- ### **Step 3: General Solution** \[ y(x) = C_1 e^{x/3} + C_2 e^{-x/2} + e^{2x} + \frac{1}{3} e^{-2x} \] --- ### **Step 4: Apply Initial Conditions** Find \( y(0) = 1 \): \[ y(0) = C_1 + C_2 + 1 + \frac{1}{3} = 1 \] \[ C_1 + C_2 = 1 - 1 - \frac{1}{3} = -\frac{1}{3} \] \[ C_1 + C_2 = -\frac{1}{3} \] Find \( y'(x) \): \[ y'(x) = \frac{1}{3} C_1 e^{x/3} - \frac{1}{2} C_2 e^{-x/2} + 2e^{2x} - \frac{2}{3} e^{-2x} \] At \( x = 0 \): \[ y'(0) = \frac{1}{3} C_1 - \frac{1}{2} C_2 + 2 - \frac{2}{3} \] \[ = \frac{1}{3} C_1 - \frac{1}{2} C_2 + \frac{4}{3} \] But \( y'(0) = -1 \): \[ \frac{1}{3}C_1 - \frac{1}{2}C_2 + \frac{4}{3} = -1 \] \[ \frac{1}{3}C_1 - \frac{1}{2}C_2 = -1 - \frac{4}{3} = -\frac{7}{3} \] --- ### **Step 5: Solve for \( C_1, C_2 \)** You have: \[ C_1 + C_2 = -\frac{1}{3} \tag{1} \] \[ \frac{1}{3}C_1 - \frac{1}{2}C_2 = -\frac{7}{3} \tag{2} \] Multiply (1) by \( \frac{1}{3} \): \[ \frac{1}{3}C_1 + \frac{1}{3}C_2 = -\frac{1}{9} \] Subtract from (2): \[ \left(\frac{1}{3}C_1 - \frac{1}{2}C_2\right) - \left(\frac{1}{3}C_1 + \frac{1}{3}C_2\right) = -\frac{7}{3} + \frac{1}{9} \] \[ \left(-\frac{1}{2}C_2 - \frac{1}{3}C_2\right) = -\frac{21}{9} + \frac{1}{9} = -\frac{20}{9} \] \[ -\frac{5}{6}C_2 = -\frac{20}{9} \] \[ C_2 = \frac{20}{9} \cdot \frac{6}{5} = \frac{24}{9} = \frac{8}{3} \] Plug into (1): \[ C_1 + \frac{8}{3} = -\frac{1}{3} \implies C_1 = -\frac{1}{3} - \frac{8}{3} = -3 \] --- ### **Final Answer for Problem 4:** \[ \boxed{ y(x) = -3 e^{x/3} + \frac{8}{3} e^{-x/2} + e^{2x} + \frac{1}{3} e^{-2x} } \] --- ## **Problem 5** **Equation:** \( y'' - 4y' + 4y = 25\cos(x) \), \( y(0) = 5, \quad y'(0) = -1 \) ### **Step 1: Homogeneous Solution** Characteristic equation: \[ r^2 - 4r + 4 = 0 \implies (r-2)^2 = 0 \implies r = 2 \text{ (double root)} \] Homogeneous solution: \[ y_h(x) = (C_1 + C_2 x) e^{2x} \] --- ### **Step 2: Particular Solution** Try \( y_p(x) = A \cos(x) + B \sin(x) \): \[ y_p' = -A\sin(x) + B\cos(x) \] \[ y_p'' = -A\cos(x) - B\sin(x) \] Substitute into the ODE: \[ (-A\cos(x) - B\sin(x)) - 4(-A\sin(x) + B\cos(x)) + 4(A\cos(x) + B\sin(x)) = 25\cos(x) \] Expand: \[ [-A\cos(x) - B\sin(x)] + [4A\sin(x) - 4B\cos(x)] + [4A\cos(x) + 4B\sin(x)] = 25\cos(x) \] Group \(\cos(x)\) and \(\sin(x)\) terms: - \(\cos(x)\): \(-A - 4B + 4A = (3A - 4B)\) - \(\sin(x)\): \(-B + 4A + 4B = (4A + 3B)\) So, \[ (3A - 4B) \cos(x) + (4A + 3B) \sin(x) = 25\cos(x) \] Set coefficients equal: \[ 3A - 4B = 25 \tag{1} \] \[ 4A + 3B = 0 \tag{2} \] From (2): \( 4A + 3B = 0 \implies B = -\frac{4}{3}A \) Plug into (1): \[ 3A - 4\left(-\frac{4}{3}A\right) = 25 \implies 3A + \frac{16}{3}A = 25 \] \[ \frac{9A + 16A}{3} = 25 \implies \frac{25A}{3} = 25 \implies A = 3 \] Then, \[ B = -\frac{4}{3} \times 3 = -4 \] So, \[ y_p(x) = 3\cos(x) - 4\sin(x) \] --- ### **Step 3: General Solution** \[ y(x) = (C_1 + C_2 x) e^{2x} + 3\cos(x) - 4\sin(x) \] --- ### **Step 4: Apply Initial Conditions** \( y(0) = 5 \): \[ y(0) = (C_1 + 0) e^{0} + 3 \cdot 1 - 4 \cdot 0 = C_1 + 3 = 5 \implies C_1 = 2 \] \( y'(x) = \) First, \[ y'(x) = \left[C_2 e^{2x} + (C_1 + C_2 x) 2e^{2x}\right] - 3\sin(x) - 4\cos(x) \] \[ = C_2 e^{2x} + 2C_1 e^{2x} + 2C_2 x e^{2x} - 3\sin(x) - 4\cos(x) \] At \( x = 0 \): \[ y'(0) = C_2 \cdot 1 + 2C_1 \cdot 1 + 0 - 0 - 4 \cdot 1 = C_2 + 2C_1 - 4 \] Set to \( y'(0) = -1 \): \[ C_2 + 2C_1 - 4 = -1 \] But \( C_1 = 2 \): \[ C_2 + 4 - 4 = -1 \implies C_2 = -1 \] --- ### **Final Answer for Problem 5:** \[ \boxed{ y(x) = (2 - x) e^{2x} + 3\cos(x) - 4\sin(x) } \] --- ### **Summary of Final Answers** #### **Problem 4:** \[ \boxed{ y(x) = -3 e^{x/3} + \frac{8}{3} e^{-x/2} + e^{2x} + \frac{1}{3} e^{-2x} } \] #### **Problem 5:** \[ \boxed{ y(x) = (2 - x) e^{2x} + 3\cos(x) - 4\sin(x) } \] If you need further explanation on any step, let me know!