## Deriving the Recurrence Relation and Particular Solution of a Differential EquationGiven the differential equation:\[ y'' - 2y' + y = 0 \]with initial conditions $ y(0) = 0 $ and $ y'(0) = 5 $, follow the steps below to derive a recurrence relation, determine initial values, and find the particular solution in terms of elementary functions.1. **Derive a recurrence relation for $ c_n $ in terms of $ c_0 $ or $ c_1 $:** The recurrence relation is $ c_{n+1} = \boxed{\frac{2nc_n}{n + 1}} $ for $ n \ge 1 $. (Type an expression using $ n $, $ c_n $, and $ c_{n-1} $ as the variables.)2. **Apply initial conditions to find the values of $ c_0 $ and $ c_1 $:** The constants are $ c_0 = \boxed{0} $ and $ c_1 = \boxed{5} $. (Type integers or fractions.)3. **Determine the particular solution $ c_n $ in terms of $ n $:** The explicit formula for the coefficients is $ c_n = \boxed{\frac{5 \cdot 2^{n-1}}{n!}} $ for $ n \ge 1 $. 4. **Identify the particular solution in terms of elementary functions:** The particular solution in terms of elementary functions is $ y(x) = \boxed{5xe^x} $.

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First derive a recurrence relation giving c₁ for n ≥2 in terms of co or c₁ (or both). Then apply the given initial conditions to find the values of co and c₁. Next determine cn (in terms of n) and, finally, identify the particular solution in terms of familiar elementary functions. y'' - 2y'+y=0; y(0) = 0, y'(0) = 5 The recurrence relation is n+1 = for n ≥ 1. (Type an expression using n, cn, and C₁-1 as the variables.) and C₁ The constants are co = (Type integers or fractions.) The explicit formula for the coefficients is cn = for n ≥ 1. The particular solution in terms of elementary functions is y(x) = CH

First derive a recurrence relation giving c₁ for n ≥2 in terms of co or c₁ (or both). Then apply the given initial conditions to find the values of co and c₁. Next determine cn (in terms of n) and, finally, identify the particular solution in terms of familiar elementary functions. y'' - 2y'+y=0; y(0) = 0, y'(0) = 5 The recurrence relation is n+1 = for n ≥ 1. (Type an expression using n, cn, and C₁-1 as the variables.) and C₁ The constants are co = (Type integers or fractions.) The explicit formula for the coefficients is cn = for n ≥ 1. The particular solution in terms of elementary functions is y(x) = CH

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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Question

## Deriving the Recurrence Relation and Particular Solution of a Differential Equation

Given the differential equation:

\[ y'' - 2y' + y = 0 \]
with initial conditions $ y(0) = 0 $ and $ y'(0) = 5 $, follow the steps below to derive a recurrence relation, determine initial values, and find the particular solution in terms of elementary functions.

1. **Derive a recurrence relation for $ c_n $ in terms of $ c_0 $ or $ c_1 $:**

The recurrence relation is $ c_{n+1} = \boxed{\frac{2nc_n}{n + 1}} $ for $ n \ge 1 $.
(Type an expression using $ n $, $ c_n $, and $ c_{n-1} $ as the variables.)

2. **Apply initial conditions to find the values of $ c_0 $ and $ c_1 $:**

The constants are $ c_0 = \boxed{0} $ and $ c_1 = \boxed{5} $.
(Type integers or fractions.)

3. **Determine the particular solution $ c_n $ in terms of $ n $:**

The explicit formula for the coefficients is $ c_n = \boxed{\frac{5 \cdot 2^{n-1}}{n!}} $ for $ n \ge 1 $.

4. **Identify the particular solution in terms of elementary functions:**

The particular solution in terms of elementary functions is $ y(x) = \boxed{5xe^x} $.

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Step 1: To find differential equation by method of power series

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Step 2: recurrence relation

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Step 3: coefficients

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Step 4: general solution of given differential equation

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Step 5: using initial conditions , we can find particular solution

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