Explanation The given differential equation is t 2 y ″ + 5 t y ′ + 13 y = 0 , t > 0 . Compare t 2 y ″ + 5 t y ′ + 13 y = 0 with general form t 2 y ″ + α t y ′ + β y = 0 . Here, α = 5 and β = 13 . Substitute α = 5 and β = 13 in d 2 y d t 2 + ( α − 1 ) d y d t + β y = 0 . d 2 y d t 2 + ( 5 − 1 ) d y d t + 13 y = 0 d 2 y d t 2 + 4 d y d t + 13 y = 0 y ″ + 4 y ′ + 13 y = 0 Obtain the characteristic equation by replacing y ″ with r 2 , y ′ with r and y with 1 in the differential equation: r 2 + 4 r + 13 = 0 . The roots of the characteristic equation are as follows. r 2 + 4 r + 13 = 0 r 1 , 2 = − 4 ± ( 4 ) 2 − 4 ( 1 ) ( 13 ) 2 ( 1 ) ( ∵ x 1 , 2 = − b ± b 2 − 4 a c 2 a ) r 1 , 2 = − 4 ± 16 − 52 2 r 1 , 2 = − 4 ± − 36 2 Simplify further as, r 1 , 2 = − 4 ± i 6 2 r 1 , 2 = 2 ( − 2 ± i 3 ) 2 r = − 2 ± i 3 ( ∵ r = a + b i ) Here, roots are complex

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Differential Equations

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Chapter 3.4, Problem 45P

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The solution of the given differential equation.

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Elementary Differential Equations

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