Explanation Given: The differential equation is given by, y ″ ( θ ) + 2 y ′ ( θ ) + 2 y ( θ ) = e − θ cos θ ... ( 1 ) Approach: 1. Find the characteristic equation by substitutions y ″ = r 2 , y ′ = r . 2. Find the roots of the characteristic equation. 3. If roots are real. Case-1 The general solution if r = r 1 = r 2 is, y ( t ) = c 1 e r t + c 2 t e r t Case-2 The general solution if r 1 ≠ r 2 is, y ( t ) = c 1 e r 1 t + c 2 e r 2 t 4. The general solution if roots are imaginary in form of r 1 , 2 = α ± β i is, y ( t ) = e a t ( c 1 cos β t + c 2 sin β t ) For each function in f i ( t ) in f ( t ) , determine the associated set of functions F i . 5. F i can be obtained as, The associate function if f i ( t ) = t m is, F = { 1 , t , t 2 , ... , t m } The associate function if f i ( t ) = t m e α t is, F = { e α t , e α t t , e α t t 2 , ... , e α t t m } The associate function if f i ( t ) = t m e α t cos β t or f i ( t ) = t m e α t sin β t is, F = { e α t cos β t , e α t t cos β t , e α t t 2 cos β t , ... , e α t t m cos β t , e α t sin β t , e α t t sin β t , e α t t 2 sin β t , ... , e α t t m sin β t } 6. The general solution if roots are imaginary in form of r 1 , 2 = α ± β i is, y ( t ) = e a t ( c 1 cos β t + c 2 sin β t ) The particular solution can be obtained by taking the linear combination of all functions in the associated sets. If any repeated function is present then it should appear only once in the particular solution. The general solution of the differential equation is given by, y = y c + y p ... ( 3 ) Here, y c is the complementary function and y p is the particular solution. Calculation: Consider the given differential equation. y ″ ( θ ) + 2 y ′ ( θ ) + 2 y ( θ ) = e − θ cos θ The auxiliary equation is written as follows r 2 + 2 r + 2 = 0 Solve the above equation for r . r 2 + 2 r + 2 = 0 r 2 + 2 r + 1 + 1 = 0 ( r + 1 ) 2 + 1 = 0 ( r + 1 ) 2 = − 1 Further simplify the above equation as, ( r + 1 ) = ± i r = − 1 ± i From the above equation the roots of the equation are, r 1 = − 1 + i r 2 = − 1 − i The complementary solution for the equation ( 1 ) using the roots of auxiliary equation is, y c = c 1 e − θ cos θ + c 2 e − θ sin θ Let the particular solution is, y p ( θ ) = θ e − θ ( A cos θ + B sin θ ) y p ( θ ) = e − θ ( A θ cos θ + B θ sin θ ) ... ( 4 ) Differentiate the above equation. y ′ p ( θ ) = d d θ e − θ ( A θ cos θ + B θ sin θ ) y ′ p ( θ ) = [ e − θ ( A ( − θ sin θ + cos θ ) + B ( θ cos θ + sin θ ) ) + ( A θ cos θ + B θ sin θ ) ( − e − θ ) ] y ′ p ( θ ) = e − θ [ ( − A − B ) θ sin θ + ( B − A ) θ cos θ + A cos θ + B sin θ ] Again differentiate the above equation.

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 4.5, Problem 21E

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The general solution to the given differential equation.

y″(θ)+2y′(θ)+2y(θ)=e−θcosθ

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Chapter 4.5, Problem 20E

Chapter 4.5, Problem 22E

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Fundamentals of Differential Equations and Boundary Value Problems

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The general solution to the given differential equation. y ″ ( θ ) + 2 y ′ ( θ ) + 2 y ( θ ) = e − θ cos θ

Solution Summary: The author explains how to find the general solution to the given differential equation.

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To find:

The general solution to the given differential equation.

y″(θ)+2y′(θ)+2y(θ)=e−θcosθ

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Chapter 4 Solutions

The general solution to the given differential equation. y ″ ( θ ) + 2 y ′ ( θ ) + 2 y ( θ ) = e − θ cos θ

Solution Summary: The author explains how to find the general solution to the given differential equation.

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To find: The general solution to the given differential equation. y″(θ)+2y′(θ)+2y(θ)=e−θcosθ

Want to see the full answer?

Chapter 4 Solutions

To find:

The general solution to the given differential equation.

y″(θ)+2y′(θ)+2y(θ)=e−θcosθ