Explanation The given differential equation is y ″ + 2 y ′ + 5 y = { 1 , 0 ≤ t ≤ π 2 0 , t > π 2 , y ( 0 ) = 0 , y ′ ( 0 ) = 0 . When 0 ≤ t ≤ π 2 , the differential equation is y ″ + 2 y ′ + 5 y = 1 . The characteristic equation by replacing y ″ with r 2 , y ′ with r and y with 1 in the differential equation: r 2 + 2 r + 5 = 0 r = − 2 ± ( 2 ) 2 − 4 ( 1 ) ( 5 ) 2 ( 1 ) r = − 2 ± 4 − 20 2 r = − 2 ± i 4 2 r = − 1 ± i 2 Note that, when the complex root, then the general solution is y = e α t ( c 1 cos ( β t ) + c 2 sin ( β t ) ) . Thus, the complementary solution is y = e − t ( c 1 cos ( 2 t ) + c 2 sin ( 2 t ) ) . Obtain the particular solution by undetermined coefficients as follows. Let Y = A . Obtain the first derivative of Y = A as follows. Y ′ = d d t ( A ) = 0 Obtain the second derivative of Y = A as follows. Y ″ = d d t ( Y ′ ) = d d t ( 0 ) = 0 Substitute Y ″ = 0 , Y ′ = 0 and Y = A in Y ″ + Y = t . 0 + 2 ( 0 ) + 5 A = 1 ⇒ A = 1 5 Therefore, the general solution is y = e − t ( c 1 cos ( 2 t ) + c 2 sin ( 2 t ) ) + 1 5 . Obtain the first derivative of y as follows. y ′ ( t ) = d d t ( e − t ( c 1 cos ( 2 t ) + c 2 sin ( 2 t ) ) + 1 5 ) y ′ ( t ) = e − t [ − 2 c 1 sin ( 2 t ) + 2 c 2 cos ( 2 t ) ] + ( c 1 cos ( 2 t ) + c 2 sin ( 2 t ) ) ( − e − t ) + 0 = e − t [ − 2 c 1 sin ( 2 t ) + 2 c 2 cos ( 2 t ) ] + ( c 1 cos ( 2 t ) + c 2 sin ( 2 t ) ) ( − e − t ) Initial condition y ( 0 ) = 0 : Substitute 0 for t and 0 for y in y = e − t ( c 1 cos ( 2 t ) + c 2 sin ( 2 t ) ) + 1 5 . y ( 0 ) = e 0 ( c 1 cos ( 0 ) + c 2 sin ( 0 ) ) + 1 5 0 = c 1 + 0 + 1 5 − 1 5 = c 1 Initial condition y ′ ( 0 ) = 1 : Substitute 0 for t and 1 for y ′ in y ′ ( t ) . y ′ ( 0 ) = e 0 [ − 2 c 1 sin ( 0 ) + 2 c 2 cos ( 0 ) ] + ( c 1 cos ( 0 ) + c 2 sin ( 0 ) ) ( − e 0 ) 0 = 0 + 2 c 2 − c 1 2 c 2 = ( − 1 5 ) c 2 = − 1 10 Thus, the solution of the initial value problem is y = e − t ( − 1 5 cos ( 2 t ) − 1 10 sin ( 2 t ) ) + 1 5 _ . When t > π 2 , the differential equation is y ″ + 2 y ′ + 5 y = 0 . The general solution is y = e − t ( c 1 cos ( 2 t ) + c 2 sin ( 2 t ) )

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

See similar textbooks

Question

Chapter 3.5, Problem 32P

To determine

The solution of the differential equation y″+2y′+5y={1,0≤t≤π20,t>π2 with the initial conditions y(0)=0 and y′(0)=0.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 3.5, Problem 31P

Chapter 3.5, Problem 33P

Students have asked these similar questions

A sample of 1,000 people was asked how many cups of coffee they drink in the morning. You are given the following sample information. Cups of Coffee Frequency 200 0 1 300 2 350 3 150 1000 Total Frequencies The expected number of cups of coffee that each person drinks in the morning is O 1.0 1.45 1.65 1.5

Suppose that the cumulative distribution function of the random variable X is x 6) = (c) P(-1

For the Big-M tableau (of a maximization LP and row0 at bottom and M=1000), Z Ꮖ 1 x2 x3 81 82 83 e4 a4 RHS 0 7 0 0 1 0 4 3 -3 20 0 -4.5 0 0 0 1 -8 -2.5 2.5 6 0 7 0 1 0 0 8 3 -3 4 0 -1 50 1 0 0 0-2 -1 1 4 0000 0 30 970 200 If the original value of c₁ is increased by 60, what is the updated value of c₁ (meaning keeping the same set for BV. -10? Having made that change, what is the new optimal value for ž?

Chapter 3 Solutions

Elementary Differential Equations

Knowledge Booster

The solution of the differential equation y ″ + 2 y ′ + 5 y = { 1 , 0 ≤ t ≤ π 2 0 , t &gt; π 2 with the initial conditions y ( 0 ) = 0 and y ′ ( 0 ) = 0 .

The solution of the differential equation y″+2y′+5y={1,0≤t≤π20,t>π2 with the initial conditions y(0)=0 and y′(0)=0.

Want to see the full answer?

Chapter 3 Solutions

The solution of the differential equation y ″ + 2 y ′ + 5 y = { 1 , 0 ≤ t ≤ π 2 0 , t > π 2 with the initial conditions y ( 0 ) = 0 and y ′ ( 0 ) = 0 .