To determine The solution of the given initial value problem; sketch the graph of the solution and describe its behaviour for increasing t . Explanation The given differential equation with initial condition is 9 y ″ − 12 y ′ + 4 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = − 1 . Obtain the characteristic equation by replacing y ″ with r 2 , y ′ with r and y with 1 in the differential equation: 9 r 2 − 12 r + 4 = 0 . The roots of the characteristic equation is as follows. 9 r 2 − 12 r + 4 = 0 r 1 , 2 = − ( − 12 ) ± ( − 12 ) 2 − 4 ( 9 ) ( 4 ) 2 ( 9 ) ( ∵ x 1 , 2 = − b ± b 2 − 4 a c 2 a ) r 1 , 2 = 12 ± 144 − 144 18 r 1 , 2 = 12 ± 0 18 Simplify further, r 1 , 2 = 12 18 r 1 , 2 = 2 3 r = 2 3 Note that, the general equation is y ( t ) = c 1 e r t + c 2 t e r t where r is the double root of the characteristic equation. Substitute 2 3 for r in y ( t ) = c 1 e r t + c 2 t e r t . y ( t ) = c 1 e 2 3 t + c 2 t e 2 3 t Therefore, the general solution of the given differential equation is y ( t ) = c 1 e 2 3 t + c 2 t e 2 3 t _ . Obtain the derivative of the general solution as follows. y ′ = d d t ( c 1 e 2 3 t + c 2 t e 2 3 t ) = c 1 2 3 e 2 3 t + c 2 d d t ( t e 2 3 t ) = c 1 2 3 e 2 3 t + c 2 [ t d d t ( e 2 3 t ) + e 2 3 t d d t ( t ) ] = c 1 2 3 e 2 3 t + c 2 [ t ( 2 3 e 2 3 t ) + e 2 3 t ] = c 1 2 3 e 2 3 t + c 2 2 3 t e 2 3 t + c 2 e 2 3 t Thus, the derivative of the general solution is y ′ ( t ) = c 1 2 3 e 2 3 t + c 2 2 3 t e 2 3 t + c 2 e 2 3 t

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 3.4, Problem 11P

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The solution of the given initial value problem; sketch the graph of the solution and describe its behaviour for increasing t.

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