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Math EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

(a) To verify: The solution to the initial value problem y ″ + 2 y ′ + 2 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = 1 by using the forms d 1 e ( α + i β ) t + d 2 e ( α − i β ) t and c 1 e a t cos β t + c 2 e a t sin β t are equivalent. The given initial value problem is y ″ + 2 y ′ + 2 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = 1 . The associated auxiliary equation r 2 + 2 r + 2 = 0 is factorized using quadratic formula − b ± b 2 − 4 a c 2 a . r = − 2 ± ( 2 ) 2 − 4 ( 1 ) ( 2 ) 2 ( 1 ) = − 2 ± 4 − 8 2 = − 2 ± − 4 2 = − 2 ± 2 i 2 ( − 1 = i ) = − 1 ± i That is, r = − 1 ± 1 i . Here, α = − 1 and β = 1 . Thus, the general solution of the form c 1 e a t cos β t + c 2 e a t sin β t will be y ( t ) = c 1 e − t cos ( t ) + c 2 e − t sin ( t ) . Apply the initial condition y ( 0 ) = 2 in y ( t ) = c 1 e − t cos ( t ) + c 2 e − t sin ( t ) , y ( 0 ) = c 1 e − ( 0 ) cos ( 0 ) + c 2 e − ( 0 ) sin ( 0 ) 2 = c 1 cos ( 0 ) + c 2 sin ( 0 ) 2 = c 1 ( 1 ) + c 2 ( 0 ) c 1 = 2 Differentiate y ( t ) = c 1 e − t cos ( t ) + c 2 e − t sin ( t ) with respect to t as, y ′ ( t ) = c 1 [ − e − t sin ( t ) − e − t cos ( t ) ] + c 2 [ e − t cos ( t ) − e − t sin ( t ) ] = − ( c 1 + c 2 ) e − t sin ( t ) + ( c 2 − c 1 ) e − t cos ( t ) Apply the initial condition y ′ ( 0 ) = 1 in y ′ ( t ) = − ( c 1 + c 2 ) e − t sin ( t ) + ( c 2 − c 1 ) e − t cos ( t ) , y ′ ( 0 ) = − ( c 1 + c 2 ) e − ( 0 ) sin ( 0 ) + ( c 2 − c 1 ) e − ( 0 ) cos ( 0 ) 1 = − ( c 1 + c 2 ) ( 0 ) + ( c 2 − c 1 ) ( 1 ) c 2 − c 1 = 1 Substitute c 1 = 2 in c 2 − c 1 = 1 , c 2 − 2 = 1 c 2 = 3 That is, c 1 = 2 and c 2 = 3 . Thus, the general solution y ( t ) = c 1 e − t cos ( t ) + c 2 e − t sin ( t ) becomes y ( t ) = 2 e − t cos t + 3 e − t sin t . With α = − 1 and β = 1 , the solution of the form d 1 e ( α + i β ) t + d 2 e ( α − i β ) t will be y ( t ) = d 1 e ( − 1 + i ) t + d 2 e ( − 1 − i ) t . Apply the initial condition y ( 0 ) = 2 in y ( t ) = d 1 e ( − 1 + i ) t + d 2 e ( − 1 − i ) t , y ( 0 ) = d 1 e ( − 1 + i ) × 0 + d 2 e ( − 1 − i ) × 0 2 = d 1 e 0 + d 2 e 0 d 1 + d 2 = 2 d 2 = 2 − d 1 Differentiate y ( t ) = d 1 e ( − 1 + i ) t + d 2 e ( − 1 − i ) t with respect to t as, y ′ ( t ) = d 1 ( − 1 + i ) e ( − 1 + i ) t + d 2 ( − 1 − i ) e ( − 1 − i ) t . Apply the initial condition y ′ ( 0 ) = 1 in y ′ ( t ) = d 1 ( − 1 + i ) e ( − 1 + i ) t + d 2 ( − 1 − i ) e ( − 1 − i ) t , y ′ ( t ) = d 1 ( − 1 + i ) e ( − 1 + i ) × 0 + d 2 ( − 1 − i ) e ( − 1 − i ) × 0 1 = d 1 ( − 1 + i ) e 0 + d 2 ( − 1 − i ) e 0 1 = d 1 ( − 1 + i ) + d 2 ( − 1 − i ) d 1 ( − 1 + i ) = 1 − d 2 ( − 1 − i ) Substitute d 2 = 2 − d 1 in d 1 ( − 1 + i ) = 1 − d 2 ( − 1 − i ) , d 1 ( − 1 + i ) = 1 − ( 2 − d 1 ) ( − 1 − i ) − d 1 + i d 1 = 1 − ( − 2 − 2 i + d 1 + i d 1 ) − d 1 + i d 1 = 1 + 2 + 2 i − d 1 − i d 1 2 i d 1 = 3 + 2 i d 1 = 3 + 2 i 2 i Substitute d 1 = 3 + 2 i 2 i in d 2 = 2 − d 1 , d 2 = 2 − 3 + 2 i 2 i = 4 i − 3 − 2 i 2 i = − 3 + 2 i 2 i Simplify d 1 = 3 + 2 i 2 i as, d 1 = 3 + 2 i 2 i × 2 i 2 i = 6 i + 4 i 2 4 i 2 = 6 i − 4 − 4 ( ∵ i 2 = − 1 ) = 4 − 6 i 4 Simplify d 2 = − 3 + 2 i 2 i as, d 2 = − 3 + 2 i 2 i × 2 i 2 i = − 6 i + 4 i 2 4 i 2 = − 6 i − 4 − 4 ( ∵ i 2 = − 1 ) = 4 + 6 i 4 That is, d 1 = 4 − 6 i 4 and d 2 = 4 + 6 i 4 . Note that, d 1 and d 2 are complex conjugate of each other. Thus, the general solution y ( t ) = d 1 e ( − 1 + i ) t + d 2 e ( − 1 − i ) t becomes y ( t ) = ( 4 − 6 i 4 ) e ( − 1 + i ) t + ( 4 + 6 i 4 ) e ( − 1 − i ) t . It is known that, e ( α + i β ) t = e α t ( cos β t + i sin β t ) . Thus, e ( − 1 + i ) t = e − t ( cos t + i sin t ) and e ( − 1 − i ) t = e − t ( cos ( − t ) + i sin ( − t ) ) . The solution y ( t ) = ( 4 − 6 i 4 ) e ( − 1 + i ) t + ( 4 + 6 i 4 ) e ( − 1 − i ) t can be simplified as, y ( t ) = ( 4 − 6 i 4 ) [ e − t ( cos t + i sin t ) ] + ( 4 + 6 i 4 ) [ e − t ( cos ( − t ) + i sin ( − t ) ) ] y ( t ) = ( 4 − 6 i 4 ) [ e − t cos t + i e − t sin t ] + ( 4 + 6 i 4 ) [ e − t cos ( t ) − i e − t sin ( t ) ] y ( t ) = ( 4 + 6 i 4 + 4 − 6 i 4 ) e − t cos ( t ) + i [ 4 − 6 i 4 − 4 + 6 i 4 ] e − t sin ( t ) y ( t ) = 2 e − t cos ( t ) + i ( − 3 i ) e − t sin ( t ) y ( t ) = 2 e − t cos t + 3 e − t sin t Therefore, the solution to the initial value problem y ″ + 2 y ′ + 2 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = 1 by using the forms d 1 e ( α + i β ) t + d 2 e ( α − i β ) t and c 1 e a t cos β t + c 2 e a t sin β t are equivalent. (b) To show: The constants d 1 and d 2 must be complex conjugates in order for the solution to be real. The form y ( t ) = d 1 e ( α + i β ) t + d 2 e ( α − i β ) t is simplified using e ( α + i β ) t = e α t ( cos β t + i sin β t ) as, y ( t ) = d 1 ( e α t ( cos β t + i sin β t ) ) + d 2 ( e α t ( cos ( − β t ) + i sin ( − β t ) ) ) = d 1 ( e α t ( cos β t + i sin β t ) ) + d 2 ( e α t ( cos β t − i sin β t ) ) = d 1 ( e α t cos β t + i e α t sin β t ) + d 2 ( e α t cos β t − i e α t sin β t ) = ( d 1 + d 2 ) e α t cos β t + i ( d 1 − d 2 ) e α t sin β t Given that, d 1 and d 2 are complex constants. Let d 1 = a 1 + i b 1 and d 2 = a 2 + i b 2 . Then, d 1 + d 2 will be, d 1 + d 2 = a 1 + i b 1 + a 2 + i b 2 = a 1 + a 2 + i ( b 1 + b 2 ) Since a 1 + a 2 is already real, the term d 1 + d 2 will be real if the imaginary term b 1 + b 2 is 0. That is, b 1 = − b 2 . Also, i ( d 1 − d 2 ) will be, i ( d 1 − d 2 ) = i ( a 1 + i b 1 − a 2 − i b 2 ) = i a 1 + i 2 b 1 − i a 2 − i 2 b 2 = i a 1 − b 1 − i a 2 + b 2 ( ∵ i 2 = − 1 ) = − b 1 + b 2 + i ( a 1 − a 2 ) Since − b 1 + b 2 is already real, the term i ( d 1 − d 2 ) will be real if the imaginary term a 1 − a 2 is 0. That is, a 1 = a 2 . Therefore, the constants d 1 and d 2 must be complex conjugates for the solution to be real.

(a) To verify: The solution to the initial value problem y ″ + 2 y ′ + 2 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = 1 by using the forms d 1 e ( α + i β ) t + d 2 e ( α − i β ) t and c 1 e a t cos β t + c 2 e a t sin β t are equivalent. The given initial value problem is y ″ + 2 y ′ + 2 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = 1 . The associated auxiliary equation r 2 + 2 r + 2 = 0 is factorized using quadratic formula − b ± b 2 − 4 a c 2 a . r = − 2 ± ( 2 ) 2 − 4 ( 1 ) ( 2 ) 2 ( 1 ) = − 2 ± 4 − 8 2 = − 2 ± − 4 2 = − 2 ± 2 i 2 ( − 1 = i ) = − 1 ± i That is, r = − 1 ± 1 i . Here, α = − 1 and β = 1 . Thus, the general solution of the form c 1 e a t cos β t + c 2 e a t sin β t will be y ( t ) = c 1 e − t cos ( t ) + c 2 e − t sin ( t ) . Apply the initial condition y ( 0 ) = 2 in y ( t ) = c 1 e − t cos ( t ) + c 2 e − t sin ( t ) , y ( 0 ) = c 1 e − ( 0 ) cos ( 0 ) + c 2 e − ( 0 ) sin ( 0 ) 2 = c 1 cos ( 0 ) + c 2 sin ( 0 ) 2 = c 1 ( 1 ) + c 2 ( 0 ) c 1 = 2 Differentiate y ( t ) = c 1 e − t cos ( t ) + c 2 e − t sin ( t ) with respect to t as, y ′ ( t ) = c 1 [ − e − t sin ( t ) − e − t cos ( t ) ] + c 2 [ e − t cos ( t ) − e − t sin ( t ) ] = − ( c 1 + c 2 ) e − t sin ( t ) + ( c 2 − c 1 ) e − t cos ( t ) Apply the initial condition y ′ ( 0 ) = 1 in y ′ ( t ) = − ( c 1 + c 2 ) e − t sin ( t ) + ( c 2 − c 1 ) e − t cos ( t ) , y ′ ( 0 ) = − ( c 1 + c 2 ) e − ( 0 ) sin ( 0 ) + ( c 2 − c 1 ) e − ( 0 ) cos ( 0 ) 1 = − ( c 1 + c 2 ) ( 0 ) + ( c 2 − c 1 ) ( 1 ) c 2 − c 1 = 1 Substitute c 1 = 2 in c 2 − c 1 = 1 , c 2 − 2 = 1 c 2 = 3 That is, c 1 = 2 and c 2 = 3 . Thus, the general solution y ( t ) = c 1 e − t cos ( t ) + c 2 e − t sin ( t ) becomes y ( t ) = 2 e − t cos t + 3 e − t sin t . With α = − 1 and β = 1 , the solution of the form d 1 e ( α + i β ) t + d 2 e ( α − i β ) t will be y ( t ) = d 1 e ( − 1 + i ) t + d 2 e ( − 1 − i ) t . Apply the initial condition y ( 0 ) = 2 in y ( t ) = d 1 e ( − 1 + i ) t + d 2 e ( − 1 − i ) t , y ( 0 ) = d 1 e ( − 1 + i ) × 0 + d 2 e ( − 1 − i ) × 0 2 = d 1 e 0 + d 2 e 0 d 1 + d 2 = 2 d 2 = 2 − d 1 Differentiate y ( t ) = d 1 e ( − 1 + i ) t + d 2 e ( − 1 − i ) t with respect to t as, y ′ ( t ) = d 1 ( − 1 + i ) e ( − 1 + i ) t + d 2 ( − 1 − i ) e ( − 1 − i ) t . Apply the initial condition y ′ ( 0 ) = 1 in y ′ ( t ) = d 1 ( − 1 + i ) e ( − 1 + i ) t + d 2 ( − 1 − i ) e ( − 1 − i ) t , y ′ ( t ) = d 1 ( − 1 + i ) e ( − 1 + i ) × 0 + d 2 ( − 1 − i ) e ( − 1 − i ) × 0 1 = d 1 ( − 1 + i ) e 0 + d 2 ( − 1 − i ) e 0 1 = d 1 ( − 1 + i ) + d 2 ( − 1 − i ) d 1 ( − 1 + i ) = 1 − d 2 ( − 1 − i ) Substitute d 2 = 2 − d 1 in d 1 ( − 1 + i ) = 1 − d 2 ( − 1 − i ) , d 1 ( − 1 + i ) = 1 − ( 2 − d 1 ) ( − 1 − i ) − d 1 + i d 1 = 1 − ( − 2 − 2 i + d 1 + i d 1 ) − d 1 + i d 1 = 1 + 2 + 2 i − d 1 − i d 1 2 i d 1 = 3 + 2 i d 1 = 3 + 2 i 2 i Substitute d 1 = 3 + 2 i 2 i in d 2 = 2 − d 1 , d 2 = 2 − 3 + 2 i 2 i = 4 i − 3 − 2 i 2 i = − 3 + 2 i 2 i Simplify d 1 = 3 + 2 i 2 i as, d 1 = 3 + 2 i 2 i × 2 i 2 i = 6 i + 4 i 2 4 i 2 = 6 i − 4 − 4 ( ∵ i 2 = − 1 ) = 4 − 6 i 4 Simplify d 2 = − 3 + 2 i 2 i as, d 2 = − 3 + 2 i 2 i × 2 i 2 i = − 6 i + 4 i 2 4 i 2 = − 6 i − 4 − 4 ( ∵ i 2 = − 1 ) = 4 + 6 i 4 That is, d 1 = 4 − 6 i 4 and d 2 = 4 + 6 i 4 . Note that, d 1 and d 2 are complex conjugate of each other. Thus, the general solution y ( t ) = d 1 e ( − 1 + i ) t + d 2 e ( − 1 − i ) t becomes y ( t ) = ( 4 − 6 i 4 ) e ( − 1 + i ) t + ( 4 + 6 i 4 ) e ( − 1 − i ) t . It is known that, e ( α + i β ) t = e α t ( cos β t + i sin β t ) . Thus, e ( − 1 + i ) t = e − t ( cos t + i sin t ) and e ( − 1 − i ) t = e − t ( cos ( − t ) + i sin ( − t ) ) . The solution y ( t ) = ( 4 − 6 i 4 ) e ( − 1 + i ) t + ( 4 + 6 i 4 ) e ( − 1 − i ) t can be simplified as, y ( t ) = ( 4 − 6 i 4 ) [ e − t ( cos t + i sin t ) ] + ( 4 + 6 i 4 ) [ e − t ( cos ( − t ) + i sin ( − t ) ) ] y ( t ) = ( 4 − 6 i 4 ) [ e − t cos t + i e − t sin t ] + ( 4 + 6 i 4 ) [ e − t cos ( t ) − i e − t sin ( t ) ] y ( t ) = ( 4 + 6 i 4 + 4 − 6 i 4 ) e − t cos ( t ) + i [ 4 − 6 i 4 − 4 + 6 i 4 ] e − t sin ( t ) y ( t ) = 2 e − t cos ( t ) + i ( − 3 i ) e − t sin ( t ) y ( t ) = 2 e − t cos t + 3 e − t sin t Therefore, the solution to the initial value problem y ″ + 2 y ′ + 2 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = 1 by using the forms d 1 e ( α + i β ) t + d 2 e ( α − i β ) t and c 1 e a t cos β t + c 2 e a t sin β t are equivalent. (b) To show: The constants d 1 and d 2 must be complex conjugates in order for the solution to be real. The form y ( t ) = d 1 e ( α + i β ) t + d 2 e ( α − i β ) t is simplified using e ( α + i β ) t = e α t ( cos β t + i sin β t ) as, y ( t ) = d 1 ( e α t ( cos β t + i sin β t ) ) + d 2 ( e α t ( cos ( − β t ) + i sin ( − β t ) ) ) = d 1 ( e α t ( cos β t + i sin β t ) ) + d 2 ( e α t ( cos β t − i sin β t ) ) = d 1 ( e α t cos β t + i e α t sin β t ) + d 2 ( e α t cos β t − i e α t sin β t ) = ( d 1 + d 2 ) e α t cos β t + i ( d 1 − d 2 ) e α t sin β t Given that, d 1 and d 2 are complex constants. Let d 1 = a 1 + i b 1 and d 2 = a 2 + i b 2 . Then, d 1 + d 2 will be, d 1 + d 2 = a 1 + i b 1 + a 2 + i b 2 = a 1 + a 2 + i ( b 1 + b 2 ) Since a 1 + a 2 is already real, the term d 1 + d 2 will be real if the imaginary term b 1 + b 2 is 0. That is, b 1 = − b 2 . Also, i ( d 1 − d 2 ) will be, i ( d 1 − d 2 ) = i ( a 1 + i b 1 − a 2 − i b 2 ) = i a 1 + i 2 b 1 − i a 2 − i 2 b 2 = i a 1 − b 1 − i a 2 + b 2 ( ∵ i 2 = − 1 ) = − b 1 + b 2 + i ( a 1 − a 2 ) Since − b 1 + b 2 is already real, the term i ( d 1 − d 2 ) will be real if the imaginary term a 1 − a 2 is 0. That is, a 1 = a 2 . Therefore, the constants d 1 and d 2 must be complex conjugates for the solution to be real.

Solution Summary: The author explains the auxiliary equation r2+2r+0 is factorized using a quadratic formula.

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

9th Edition

ISBN: 9780135963777

Author: Nagle

Publisher: PEARSON CO

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Chapter 4.3, Problem 35E

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Chapter 4 Solutions

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

Ch. 4.1 - Verify that for b = 0 and Fext(t) = 0, equation...Ch. 4.1 - If Fext(t) = 0, equation (3) becomes my + by + ky...Ch. 4.1 - Show that if Fext(t) = 0, m = 1, k = 9, and b = 6,...Ch. 4.1 - Verify that y = sin 3t + 2 cos 3t is a solution to...Ch. 4.1 - Prob. 5E Ch. 4.1 - An external force F(t) = 2 cos 2t is applied to a...Ch. 4.1 - Prob. 7E Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - Prob. 10E

Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - Prob. 3E Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - Prob. 5E Ch. 4.2 - Prob. 6E Ch. 4.2 - Prob. 7E Ch. 4.2 - Prob. 8E Ch. 4.2 - Prob. 9E Ch. 4.2 - Prob. 10E Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 15E Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 17E Ch. 4.2 - Prob. 18E Ch. 4.2 - Prob. 19E Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 21E Ch. 4.2 - Prob. 22E Ch. 4.2 - Prob. 23E Ch. 4.2 - Prob. 24E Ch. 4.2 - Prob. 25E Ch. 4.2 - Prob. 26E Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - Prob. 31E Ch. 4.2 - Prob. 32E Ch. 4.2 - Prob. 33E Ch. 4.2 - Prob. 34E Ch. 4.2 - Prob. 35E Ch. 4.2 - Using the definition in Problem 35, prove that if...Ch. 4.2 - Prob. 37E Ch. 4.2 - Prob. 38E Ch. 4.2 - Prob. 39E Ch. 4.2 - Prob. 40E Ch. 4.2 - In Problems 3741, find three linearly independent...Ch. 4.2 - Prob. 42E Ch. 4.2 - Prob. 43E Ch. 4.2 - Solve the initial value problem: y 2y y + 2y =...Ch. 4.2 - Prob. 45E Ch. 4.2 - Prob. 46E Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - Prob. 5E Ch. 4.3 - Prob. 6E Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 920, find a general solution. 9. y 8y...Ch. 4.3 - In Problems 920, find a general solution. 10. y +...Ch. 4.3 - Prob. 11E Ch. 4.3 - Prob. 12E Ch. 4.3 - In Problems 920, find a general solution. 13. y ...Ch. 4.3 - Prob. 14E Ch. 4.3 - Prob. 15E Ch. 4.3 - Prob. 16E Ch. 4.3 - In Problems 920, find a general solution. 17. y y...Ch. 4.3 - In Problems 920, find a general solution. 18. 2y +...Ch. 4.3 - Prob. 19E Ch. 4.3 - In Problems 920, find a general solution. 20. y y...Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 22E Ch. 4.3 - Prob. 23E Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 25E Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 27E Ch. 4.3 - Prob. 28E Ch. 4.3 - Prob. 29E Ch. 4.3 - Prob. 30E Ch. 4.3 - Prob. 31E Ch. 4.3 - Prob. 32E Ch. 4.3 - Prob. 33E Ch. 4.3 - Prob. 34E Ch. 4.3 - Swinging Door. The motion of a swinging door with...Ch. 4.3 - Prob. 36E Ch. 4.3 - Prob. 37E Ch. 4.3 - Prove the sum of angles formula for the sine...Ch. 4.4 - In Problems 18, decide whether or not the method...Ch. 4.4 - Prob. 2E Ch. 4.4 - Prob. 3E Ch. 4.4 - Prob. 4E Ch. 4.4 - Prob. 5E Ch. 4.4 - Prob. 6E Ch. 4.4 - Prob. 7E Ch. 4.4 - Prob. 8E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 16E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 18E Ch. 4.4 - Prob. 19E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 23E Ch. 4.4 - Prob. 24E Ch. 4.4 - Prob. 25E Ch. 4.4 - Prob. 26E Ch. 4.4 - Prob. 27E Ch. 4.4 - Prob. 28E Ch. 4.4 - Prob. 29E Ch. 4.4 - Prob. 30E Ch. 4.4 - Prob. 31E Ch. 4.4 - Prob. 32E Ch. 4.4 - Prob. 33E Ch. 4.4 - Prob. 34E Ch. 4.4 - Prob. 35E Ch. 4.4 - In Problems 3336, use the method of undetermined...Ch. 4.5 - Given that y1(t) = cos t is a solution to y y + y...Ch. 4.5 - Prob. 2E Ch. 4.5 - Prob. 3E Ch. 4.5 - Prob. 4E Ch. 4.5 - Prob. 5E Ch. 4.5 - Prob. 6E Ch. 4.5 - Prob. 7E Ch. 4.5 - In Problems 38, a nonhomogeneous equation and a...Ch. 4.5 - Prob. 9E Ch. 4.5 - Prob. 10E Ch. 4.5 - Prob. 11E Ch. 4.5 - Prob. 12E Ch. 4.5 - Prob. 13E Ch. 4.5 - In Problems 916 decide whether the method of...Ch. 4.5 - Prob. 15E Ch. 4.5 - Prob. 16E Ch. 4.5 - Prob. 17E Ch. 4.5 - Prob. 18E Ch. 4.5 - Prob. 19E Ch. 4.5 - In Problems 1722, find a general solution to the...Ch. 4.5 - Prob. 21E Ch. 4.5 - Prob. 22E Ch. 4.5 - Prob. 23E Ch. 4.5 - Prob. 24E Ch. 4.5 - Prob. 25E Ch. 4.5 - Prob. 26E Ch. 4.5 - Prob. 27E Ch. 4.5 - Prob. 28E Ch. 4.5 - Prob. 29E Ch. 4.5 - Prob. 30E Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - Prob. 33E Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - Prob. 35E Ch. 4.5 - Prob. 36E Ch. 4.5 - Prob. 37E Ch. 4.5 - Prob. 38E Ch. 4.5 - In Problems 3740, find a particular solution to...Ch. 4.5 - Prob. 40E Ch. 4.5 - Prob. 41E Ch. 4.5 - Prob. 42E Ch. 4.5 - Prob. 43E Ch. 4.5 - Prob. 44E Ch. 4.5 - Prob. 46E Ch. 4.5 - Prob. 47E Ch. 4.5 - Prob. 48E Ch. 4.6 - Prob. 1E Ch. 4.6 - In Exercises 12, rectangles have been drawn to...Ch. 4.6 - Prob. 3E Ch. 4.6 - Prob. 4E Ch. 4.6 - Prob. 5E Ch. 4.6 - Prob. 6E Ch. 4.6 - Prob. 7E Ch. 4.6 - Prob. 8E Ch. 4.6 - Prob. 9E Ch. 4.6 - Prob. 10E Ch. 4.6 - Prob. 11E Ch. 4.6 - Prob. 12E Ch. 4.6 - Prob. 13E Ch. 4.6 - Prob. 14E Ch. 4.6 - Prob. 15E Ch. 4.6 - In Exercises 1318, use a calculator or a computer...Ch. 4.6 - Prob. 17E Ch. 4.6 - In Exercises 1318, use a calculator or a computer...Ch. 4.6 - For f(x) = 3x + 2, evaluate the Riemann sums: (a)...Ch. 4.6 - For g(x) = 4x 1, evaluate the Riemann sums: (a)...Ch. 4.6 - Prob. 21E Ch. 4.6 - Use Table 5.9 to evaluate the Riemann sums: Table...Ch. 4.6 - The graph of f(t) is in Figure 5.40. Which of the...Ch. 4.6 - Prob. 24E Ch. 4.6 - Prob. 25E Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - Prob. 3E Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - Prob. 5E Ch. 4.7 - In Problems 5 through 8, determine whether Theorem...Ch. 4.7 - Prob. 7E Ch. 4.7 - In Problems 5 through 8, determine whether Theorem...Ch. 4.7 - Prob. 9E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - Prob. 11E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - Prob. 13E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - In Problems 15 through 18, find a general solution...Ch. 4.7 - In Problems 15 through 18, find a general solution...Ch. 4.7 - Prob. 17E Ch. 4.7 - Prob. 18E Ch. 4.7 - Prob. 19E Ch. 4.7 - Prob. 20E Ch. 4.7 - Prob. 21E Ch. 4.7 - Prob. 22E Ch. 4.7 - Prob. 23E Ch. 4.7 - Solve the following CauchyEuler equations by using...Ch. 4.7 - Prob. 25E Ch. 4.7 - Prob. 26E Ch. 4.7 - Prob. 27E Ch. 4.7 - Prob. 28E Ch. 4.7 - Prove that if y1 and y2 are linearly independent...Ch. 4.7 - Prob. 30E Ch. 4.7 - Prob. 31E Ch. 4.7 - Prob. 32E Ch. 4.7 - Prob. 33E Ch. 4.7 - Prob. 34E Ch. 4.7 - Prob. 35E Ch. 4.7 - Prob. 36E Ch. 4.7 - Prob. 37E Ch. 4.7 - Prob. 38E Ch. 4.7 - Prob. 39E Ch. 4.7 - Prob. 40E Ch. 4.7 - Prob. 41E Ch. 4.7 - In Problems 41 through 44, a differential equation...Ch. 4.7 - Prob. 43E Ch. 4.7 - Prob. 44E Ch. 4.7 - Prob. 45E Ch. 4.7 - Prob. 46E Ch. 4.7 - Prob. 47E Ch. 4.7 - Prob. 48E Ch. 4.7 - Prob. 49E Ch. 4.7 - Prob. 50E Ch. 4.7 - Prob. 51E Ch. 4.7 - Prob. 52E Ch. 4.8 - Prob. 1E Ch. 4.8 - Prob. 2E Ch. 4.8 - Try to predict the qualitative features of the...Ch. 4.8 - Prob. 4E Ch. 4.8 - Prob. 5E Ch. 4.8 - Prob. 6E Ch. 4.8 - Prob. 7E Ch. 4.8 - Prob. 8E Ch. 4.8 - Prob. 9E Ch. 4.8 - Prob. 10E Ch. 4.8 - Prob. 11E Ch. 4.8 - Use reduction of order to derive the solution...Ch. 4.8 - Prob. 13E Ch. 4.8 - Prob. 14E Ch. 4.8 - Prob. 15E Ch. 4.8 - Prob. 16E Ch. 4.8 - Prob. 17E Ch. 4.9 - Prob. 1E Ch. 4.9 - All problems refer to the massspring configuration...Ch. 4.9 - Prob. 3E Ch. 4.9 - Prob. 4E Ch. 4.9 - Prob. 5E Ch. 4.9 - Prob. 6E Ch. 4.9 - Prob. 7E Ch. 4.9 - Prob. 8E Ch. 4.9 - All problems refer to the massspring configuration...Ch. 4.9 - Prob. 10E Ch. 4.9 - Prob. 11E Ch. 4.9 - Prob. 12E Ch. 4.9 - Prob. 13E Ch. 4.9 - Prob. 14E Ch. 4.9 - Prob. 15E Ch. 4.9 - Prob. 16E Ch. 4.9 - Prob. 17E Ch. 4.9 - Prob. 18E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 2E Ch. 4.10 - Prob. 3E Ch. 4.10 - Prob. 4E Ch. 4.10 - Prob. 5E Ch. 4.10 - Prob. 6E Ch. 4.10 - Prob. 7E Ch. 4.10 - Prob. 8E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 10E Ch. 4.10 - Prob. 11E Ch. 4.10 - Prob. 12E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 14E Ch. 4.10 - Prob. 15E Ch. 4 - In Problems 128, find a general solution to the...Ch. 4 - Prob. 2RP Ch. 4 - Prob. 3RP Ch. 4 - Prob. 4RP Ch. 4 - Prob. 5RP Ch. 4 - Prob. 6RP Ch. 4 - Prob. 7RP Ch. 4 - Prob. 8RP Ch. 4 - Prob. 9RP Ch. 4 - Prob. 10RP Ch. 4 - Prob. 11RP Ch. 4 - Prob. 12RP Ch. 4 - Prob. 13RP Ch. 4 - Prob. 14RP Ch. 4 - Prob. 15RP Ch. 4 - Prob. 16RP Ch. 4 - Prob. 17RP Ch. 4 - In Problems 128, find a general solution to the...Ch. 4 - Prob. 19RP Ch. 4 - Prob. 20RP Ch. 4 - Prob. 21RP Ch. 4 - Prob. 22RP Ch. 4 - Prob. 23RP Ch. 4 - Prob. 24RP Ch. 4 - Prob. 25RP Ch. 4 - Prob. 26RP Ch. 4 - Prob. 27RP Ch. 4 - Prob. 28RP Ch. 4 - Prob. 29RP Ch. 4 - Prob. 30RP Ch. 4 - Prob. 31RP Ch. 4 - Prob. 32RP Ch. 4 - Prob. 33RP Ch. 4 - Prob. 34RP Ch. 4 - Prob. 35RP Ch. 4 - Prob. 36RP Ch. 4 - Prob. 37RP Ch. 4 - Prob. 38RP Ch. 4 - A 32-lb weight is attached to a vertical spring,...Ch. 4 - Prob. 1TWE Ch. 4 - Prob. 2TWE Ch. 4 - Prob. 3TWE Ch. 4 - For students with a background in linear algebra:...

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