**Differential Equations - Problem Solving****Problem Statement:** Two solutions to the differential equation \( y'' + 11y' + 28y = 0 \) are \( y_1 = e^{-7t} \) and \( y_2 = e^{-4t} \).a) Find the Wronskian.\[ W = \_\_\_\_\_\_ \]b) Find the solution that satisfies the initial conditions \( y(0) = 0 \), \( y'(0) = -18 \).\[ y = \_\_\_\_\_\_ \]---**Solution Approach:**1. **Wronskian Calculation:** The Wronskian \( W \) of two solutions \( y_1(t) \) and \( y_2(t) \) is given by: \[ W = y_1(t) y_2'(t) - y_2(t) y_1'(t). \] - For \( y_1(t) = e^{-7t} \): \[ y_1'(t) = -7e^{-7t}. \] - For \( y_2(t) = e^{-4t} \): \[ y_2'(t) = -4e^{-4t}. \] Substituting into the Wronskian formula: \[ W = e^{-7t} (-4e^{-4t}) - e^{-4t} (-7e^{-7t}). \]2. **Finding the Particular Solution:** We can express the general solution of the differential equation as: \[ y(t) = C_1 y_1(t) + C_2 y_2(t) = C_1 e^{-7t} + C_2 e^{-4t}. \] Using the initial conditions \( y(0) = 0 \) and \( y'(0) = -18 \): - At \( t = 0 \): \[ y(0) = C_1 e^{0} + C_2 e^{0} = C_1 + C_2 = 0. \] Therefore, \( C_1 + C_2 = 0 \). - The derivative of \( y \) is: \[ y'(t) = -7C

Name: **Differential Equations - Problem Solving****Problem Statement:** Tw
Uploaded: 2024-03-20T06:46:55.000Z
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Description: **Differential Equations - Problem Solving****Problem Statement:** Two solutions to the differential equation \( y'' + 11y' + 28y = 0 \) are \( y_1 = e^{-7t} \) and \( y_2 = e^{-4t} \).a) Find the Wronskian.\[ W = \_\_\_\_\_\_ \]b) Find the solution