a. Explanation Given: A second order equation t 2 y ″ ( t ) + 2 t y ′ ( t ) − 12 y ( t ) = 0 and its solution is in the form y ( t ) = t p . Calculation: Take derivative on both sides, y ′ ( t ) = p t p − 1 Again take derivative on both sides, y ″ ( t ) = p ( p − 1 ) t p − 2 Substitute the values of y ″ ( t ) , y ′ ( t ) , y ( t ) in the second order equation, t 2 y ″ ( t ) + 2 t y ′ ( t ) − 12 y ( t ) = 0 t< b. To determine To find: The general solution of the equation t 2 y ″ ( t ) + 2 t y ′ ( t ) − 12 y ( t ) = 0 .

Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

3rd Edition

ISBN: 9780134996103

Author: William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz

Publisher: PEARSON

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Chapter 9, Problem 33RE

To determine

To find: The two values of p that gives the solution of the given differential equation.

To determine

To find: The general solution of the equation t2y″(t)+2ty′(t)−12y(t)=0 .

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Chapter 9, Problem 32RE

Chapter 10.1, Problem 1QC

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Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

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The two values of p that gives the solution of the given differential equation.

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