Explanation Procedure used: Method of undetermined coefficients: A particular solution to the differential equation a y ″ + b y ′ + c y = P m ( t ) e α t cos β t + Q n ( t ) e α t sin β t , β ≠ 0 , where P m ( t ) and Q n ( t ) is a polynomial of degree m and n respectively, is of the form y p ( t ) = t s ( A k t k + ⋯ + A 1 t + A 0 ) e α t cos β t + t s ( B k t k + ⋯ + B 1 t + B 0 ) e α t sin β t , where k is the larger of m and n : (i) s = 0 , if α + i β is not a root of the associated auxiliary equation; (ii) s = 1 , if α + i β is a root of the associated auxiliary equation. Calculation: The given initial value problem is, y ″ − 2 y ′ + 10 y = 6 cos 3 t − sin 3 t , y ( 0 ) = 2 , y ′ ( 0 ) = − 8 . The corresponding auxiliary equation is r 2 − 2 r + 10 = 0 . Factorize r 2 − 2 r + 10 = 0 using quadratic formula − b ± b 2 − 4 a c 2 a as, r = − ( − 2 ) ± ( − 2 ) 2 − 4 ( 1 ) ( 10 ) 2 ( 1 ) = 2 ± 4 − 40 2 = 2 ± − 36 2 = 2 ± 6 i 2 ( ∵ i = − 1 ) = 1 ± 3 i That is, the roots are r 1 = 1 + 3 i and r 2 = 1 − 3 i . Here, α = 1 and β = 3 . Thus, the general solution for the homogenous equation is, y ( t ) = c 1 e t cos 3 t + c 2 e t sin 3 t . The RHS of the differential equation y ″ − 2 y ′ + 10 y = 6 cos 3 t − sin 3 t is 6 cos 3 t − sin 3 t . Comparing 6 cos 3 t − sin 3 t with the form P m ( t ) e α t cos β t + Q n ( t ) e α t sin β t , it is observed that m = 0 , n = 0 , α = 0 and β = 3 . Since α + i β = 0 + 3 i is a not a root of the associated auxiliary equation, s = 0 . Note that, k is the largest of m and n , which is 0. Thus, the particular solution form y p ( t ) = t s ( A k t k + ⋯ + A 1 t + A 0 ) e α t cos β t + t s ( B k t k + ⋯ + B 1 t + B 0 ) e α t sin β t for s = 0 , k = 0 , α = 0 and β = 3 will be, y p ( t ) = t 0 ( A 0 ) e 0 cos 3 t + t 0 ( B 0 ) e 0 sin 3 t = A 0 cos 3 t + B 0 sin 3 t Obtain the first and second order derivatives of y p ( t ) = A 0 cos 3 t + B 0 sin 3 t as follows: Differentiate y p ( t ) = A 0 cos 3 t + B 0 sin 3 t with respect to t as, y ′ p ( t ) = − 3 A 0 sin 3 t + 3 B 0 cos 3 t . Differentiate y ′ p ( t ) = − 3 A 0 sin 3 t + 3 B 0 cos 3 t with respect to t as, y ″ p ( t ) = − 9 A 0 cos 3 t − 9 B 0 sin 3 t . Thus, y ″ − 2 y ′ + 10 y for y p will be, y ″ p − 2 y ′ p + 10 y p = [ − 9 A 0 cos 3 t − 9 B 0 sin 3 t − 2 ( − 3 A 0 sin 3 t + 3 B 0 cos 3 t ) + 10 ( A 0 cos 3 t + B 0 sin 3 t ) ] = − 9 A 0 cos 3 t − 9 B 0 sin 3 t + 6 A 0 sin 3 t − 6 B 0 cos 3 t + 10 A 0 cos 3 t + 10 B 0 sin 3 t = ( − 9 A 0 − 6 B 0 + 10 A 0 ) cos 3 t + ( − 9 B 0 + 6 A 0 + 10 B 0 ) sin 3 t = ( A 0 − 6 B 0 ) cos 3 t + ( B 0 + 6 A 0 ) sin 3 t Setting ( A 0 − 6 B 0 ) cos 3 t + ( B 0 + 6 A 0 ) sin 3 t = 6 cos 3 t − sin 3 t and comparing the coefficients result in A 0 − 6 B 0 = 6 and B 0 + 6 A 0 = − 1

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ISBN: 9780135963777

Author: Nagle

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Chapter 4, Problem 31RP

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The solution to the initial value problem y″−2y′+10y=6cos3t−sin3t,y(0)=2,y′(0)=−8.

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The solution to the initial value problem y ″ − 2 y ′ + 10 y = 6 cos 3 t − sin 3 t , y ( 0 ) = 2 , y ′ ( 0 ) = − 8 .

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The solution to the initial value problem y″−2y′+10y=6cos3t−sin3t,y(0)=2,y′(0)=−8.

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