(a) Explanation Procedure used: 1. Make the characteristic equation by substituting y ″ = r 2 , y ′ = r . The characteristic equation is a r 2 + b r + c = 0 . 2. Find the roots of the characteristic equation. 3. If roots are real: Case-1 If r = r 1 = r 2 , then the general solution is y ( t ) = c 1 e r t + c 2 t e r t Case-2 If r 1 ≠ r 2 , then the general solution is y ( t ) = c 1 e r 1 t + c 2 e r 2 t . 4. If roots are imaginary in the form of r 1 , 2 = α ± β i , then the general solution is y ( t ) = e a t ( c 1 cos β t + c 2 sin β t ) . Calculation: Consider the non linear equation. ( y ″ ) 2 − 5 y ″ y + 4 y 2 = 0 The factoring the nonlinear equation. ( y ″ ) 2 − 5 y ″ y + 4 y 2 = 0 ( y ″ ) 2 − 4 y ″ y − y ″ y + 4 y 2 = 0 ( y ″ − 4 y ) ( y ″ − y ) = 0 Consider the one solution by factoring ( y ″ − y ) = 0 . The characteristic equation is r 2 − 1 = 0 . The roots of the above equation are, r 2 − 1 = 0 r 2 = 1 r = 1 , − 1 Here, roots are not equal (b) To determine To find: The solution of ( y ″ ) 2 − 2 y ″ y + y 2 = 0 using factoring.

5th Edition

ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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Chapter 4.2, Problem 42E

(a)

To determine

To solve: The differential equation (y″)2−5y″y+4y2=0 using factoring and whether the Principle of Superposition valid.

(b)

To determine

To find: The solution of (y″)2−2y″y+y2=0 using factoring.

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Chapter 4.2, Problem 41E

Chapter 4.2, Problem 43E

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Chapter 4 Solutions

Introductory Differential Equations

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The differential equation ( y ″ ) 2 − 5 y ″ y + 4 y 2 = 0 using factoring and whether the Principle of Superposition valid.

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To solve: The differential equation (y″)2−5y″y+4y2=0 using factoring and whether the Principle of Superposition valid.

To find: The solution of (y″)2−2y″y+y2=0 using factoring.

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Chapter 4 Solutions