Skip to main content

Math Advanced Math Fundamentals of Differential Equations (9th Edition)

(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.

Fundamentals of Differential Equations (9th Edition)

Fundamentals of Differential Equations (9th Edition)

9th Edition

ISBN: 9780321977069

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

See similar textbooks

Concept explainers

Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

Videos

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

Blurred answer

Chapter 4.3, Problem 35E

Chapter 4.3, Problem 37E

Students have asked these similar questions

EXAMPLE 3 Find S X √√2-2x2 dx. SOLUTION Let u = 2 - 2x². Then du = Χ dx = 2- 2x² = 信 du dx, so x dx = du and u-1/2 du (2√u) + C + C (in terms of x).

Let g(z) = z-i z+i' (a) Evaluate g(i) and g(1). (b) Evaluate the limits lim g(z), and lim g(z). 2-12 (c) Find the image of the real axis under g. (d) Find the image of the upper half plane {z: Iz > 0} under the function g.

k (i) Evaluate k=7 k=0 [Hint: geometric series + De Moivre] (ii) Find an upper bound for the expression 1 +2x+2 where z lies on the circle || z|| = R with R > 10. [Hint: Use Cauchy-Schwarz]

Chapter 4 Solutions

Fundamentals of Differential Equations (9th Edition)

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:...

Knowledge Booster

Background pattern image

$Advanced Math$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions

SEE MORE QUESTIONS

Recommended textbooks for you

Discrete Mathematics and Its Applications ( 8th I...
Math
ISBN:9781259676512
Author:Kenneth H Rosen
Publisher:McGraw-Hill Education
Mathematics for Elementary Teachers with Activiti...
Math
ISBN:9780134392790
Author:Beckmann, Sybilla
Publisher:PEARSON
Calculus Volume 1
Math
ISBN:9781938168024
Author:Strang, Gilbert
Publisher:OpenStax College
Thinking Mathematically (7th Edition)
Math
ISBN:9780134683713
Author:Robert F. Blitzer
Publisher:PEARSON
Discrete Mathematics With Applications
Math
ISBN:9781337694193
Author:EPP, Susanna S.
Publisher:Cengage Learning,
Pathways To Math Literacy (looseleaf)
Math
ISBN:9781259985607
Author:David Sobecki Professor, Brian A. Mercer
Publisher:McGraw-Hill Education

Text book image

Discrete Mathematics and Its Applications ( 8th I...

Math

ISBN:9781259676512

Author:Kenneth H Rosen

Publisher:McGraw-Hill Education

Text book image

Mathematics for Elementary Teachers with Activiti...

Math

ISBN:9780134392790

Author:Beckmann, Sybilla

Publisher:PEARSON

Text book image

Calculus Volume 1

Math

ISBN:9781938168024

Author:Strang, Gilbert

Publisher:OpenStax College

Text book image

Thinking Mathematically (7th Edition)

Math

ISBN:9780134683713

Author:Robert F. Blitzer

Publisher:PEARSON

Text book image

Discrete Mathematics With Applications

Math

ISBN:9781337694193

Author:EPP, Susanna S.

Publisher:Cengage Learning,

Text book image

Pathways To Math Literacy (looseleaf)

Math

ISBN:9781259985607

Author:David Sobecki Professor, Brian A. Mercer

Publisher:McGraw-Hill Education

SEE MORE TEXTBOOKS

01 - What Is A Differential Equation in Calculus? Learn to Solve Ordinary Differential Equations.; Author: Math and Science;https://www.youtube.com/watch?v=K80YEHQpx9g;License: Standard YouTube License, CC-BY

Higher Order Differential Equation with constant coefficient (GATE) (Part 1) l GATE 2018; Author: GATE Lectures by Dishank;https://www.youtube.com/watch?v=ODxP7BbqAjA;License: Standard YouTube License, CC-BY

Solution of Differential Equations and Initial Value Problems; Author: Jefril Amboy;https://www.youtube.com/watch?v=Q68sk7XS-dc;License: Standard YouTube License, CC-BY