**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

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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

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Answered: 7t Two solutions to y''+ 11y' + 28y = 0…

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7t Two solutions to y''+ 11y' + 28y = 0 are yi = e , 42 = e-4t. a) Find the Wronskian. W = b) Find the solution satisfying the initial conditions y(0) = 0, y'(0) = - 18

Advanced Engineering Mathematics

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ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Fundamentals of Trigonometric Identities

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**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

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