Explanation Procedure used: Consider the differential equation a t 2 y ″ ( t ) + b t y ′ ( t ) + c y = 0 . The associated auxiliary equation is a r 2 + ( b − a ) r + c = 0 . Solving the auxiliary equation gives the roots r 1 and r 2 . (a) If the auxiliary equation has distinct real roots r 1 and r 2 , then y ( t ) = c 1 t r 1 + c 2 t r 2 is a general solution. (b) If the auxiliary equation has repeated root r , then y ( t ) = c 1 t r + c 2 t r ln t is a general solution. (b) If the auxiliary equation has complex root r = α ± i β , then y ( t ) = c 1 t α cos ( β ln t ) + c 2 t α sin ( β ln t ) is a general solution. Calculation: The given differential equation is, ( t + 1 ) 2 y ″ ( t ) + 10 ( t + 1 ) y ′ ( t ) + 14 y ( t ) = 0 . Let u = t + 1 . Then, t = u − 1 and d u = d t . The first derivative will be, y ′ = d y d t = d y d u × d u d t = d y d u ( ∵ d u = d t ) The second derivative will be, y ″ = d 2 y d t 2 = d d t ( d y d t ) = d d t ( d y d u ) ( ∵ d y d t = d y d u ) = d 2 y d u 2 ( ∵ d u = d t ) The differential equation ( t + 1 ) 2 y ″ ( t ) + 10 ( t + 1 ) y ′ ( t ) + 14 y ( t ) = 0 becomes, u 2 d 2 y d u 2 + 10 u d y d u + 14 y ( u ) = 0

Fundamentals of Differential Equations (9th Edition)

9th Edition

ISBN: 9780321977069

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

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Chapter 4.7, Problem 22E

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The general solution to the Cauchy-Euler equation (t+1)2y″(t)+10(t+1)y′(t)+14y(t)=0 for t>−1.

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Chapter 4.7, Problem 21E

Chapter 4.7, Problem 23E

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The general solution to the Cauchy-Euler equation ( t + 1 ) 2 y ″ ( t ) + 10 ( t + 1 ) y ′ ( t ) + 14 y ( t ) = 0 for t > − 1 .

Solution Summary: The author explains the general solution to the Cauchy-Euler equation (t+1)2y′′.

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The general solution to the Cauchy-Euler equation (t+1)2y″(t)+10(t+1)y′(t)+14y(t)=0 for t>−1.

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The general solution to the Cauchy-Euler equation ( t + 1 ) 2 y ″ ( t ) + 10 ( t + 1 ) y ′ ( t ) + 14 y ( t ) = 0 for t &gt; − 1 .

Solution Summary: The author explains the general solution to the Cauchy-Euler equation (t+1)2y′′.

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The general solution to the Cauchy-Euler equation (t+1)2y″(t)+10(t+1)y′(t)+14y(t)=0 for t>−1.

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Chapter 4 Solutions

The general solution to the Cauchy-Euler equation ( t + 1 ) 2 y ″ ( t ) + 10 ( t + 1 ) y ′ ( t ) + 14 y ( t ) = 0 for t > − 1 .