To determine The solution of the given IVP. Explanation Given: The given IVP is ( D 4 − 26 D 2 + 25 I ) y = 50 ( x + 1 ) 2 , y ( 0 ) = 12.16 , D y ( 0 ) = − 6 , D 2 y ( 0 ) = 34 , D 3 y ( 0 ) = − 130 . Procedure used: Let y ( n ) + a n − 1 ( x ) y ( n − 1 ) + ⋯ + a 1 ( x ) y ′ + a 0 ( x ) y = r ( x ) be the n th order homogeneous ODE with constant coefficients, where y ( n ) = d n y d x n is the first term and r ( x ) ≠ 0 . A general solution of the above equation on an open interval I of the x -axis is of the form r ( x ) ≠ 0 Obtain the characteristic equation by substituting y = e λ x in the ODE as follows. y ( x ) = y h ( x ) + y p ( x ) Here y h ( x ) is a general solution of the corresponding homogeneous ODE. And y p ( x ) is any solution of the given equation on I containing no arbitrary constants. Calculation: Consider the given IVP, ( D 4 − 26 D 2 + 25 I ) y = 50 ( x + 1 ) 2 , y ( 0 ) = 12.16 , D y ( 0 ) = − 6 , D 2 y ( 0 ) = 34 , D 3 y ( 0 ) = − 130 y i v − 26 y ″ + 25 y = 50 ( x + 1 ) 2 , y ( 0 ) = 12.16 , y ′ ( 0 ) = − 6 , y ″ ( 0 ) = 34 , y ′ ′ ′ ( 0 ) = − 130 Consider the corresponding homogeneous equation of the given ODE, y i v − 26 y ″ + 25 y = 0 Obtain the characteristic equation as follows. λ 4 − 26 λ 2 + 25 = 0 Compute the roots of the characteristic equation as follows. λ 4 − 26 λ 2 + 25 = 0 λ 2 = − ( − 26 ) ± ( − 26 ) 2 − 4 ( 1 ) ( 25 ) 2 λ 2 = 26 ± 576 2 λ 2 = 26 ± 24 2 Simplify further as follows. λ 2 = 25 and λ 2 = 1 λ 1 = − 5 , λ 2 = 5 , λ 3 = − 1 , λ 4 = 1 The homogeneous solution of the given ODE is given as follows. y h = c 1 e − 5 x + c 2 e 5 x + c 3 e − x + c 4 e x Let the particular solution is of the form y p = A x 2 + B x + C . Obtain the first three derivatives of the above equation as follows. y ′ p = 2 A x + B y ″ p = 2 A y ′ ′ ′ p = 0 y i v = 0 Substitute the above derivatives in the given ODE as follows. 0 − 26 ( 2 A ) + 25 ( A x 2 + B x + C ) = 50 ( x + 1 ) 2 − 52 A + 25 A x 2 + 25 B x + 25 C = 50 x 2 + 100 x + 50 25 A x 2 + 25 B x + ( 25 C − 52 A ) = 50 x 2 + 100 x + 50 Equate like terms on both sides as follows. 25 A = 50 , 25 B = 100 and 25 C − 52 A = 50 A = 2 , B = 4 and C = 50 + 52 ( 2 ) 25 = 6.16 The particular solution of the given ODE is given as follows. y p = 2 x 2 + 4 x + 6.16 The general solution of the given ODE is as follows, y = y h + y p = c 1 e − 5 x + c 2 e 5 x + c 3 e − x + c 4 e x + 2 x 2 + 4 x + 6.16 Obtain the first 3 derivatives of the general solution obtained as follows. y = c 1 e − 5 x + c 2 e 5 x + c 3 e − x + c 4 e x + 2 x 2 + 4 x + 6.16 y ′ = − 5 c 1 e − 5 x + 5 c 2 e 5 x − c 3 e − x + c 4 e x + 4 x + 4 y ″ = 25 c 1 e − 5 x + 25 c 2 e 5 x + c 3 e − x + c 4 e x + 4 y ′ ′ ′ = − 125 c 1 e − 5 x + 125 c 2 e 5 x − c 3 e − x + c 4 e x Apply the given initial values on the general solution and its derivatives obtain as follows

Custom Kreyszig: Advanced Engineering Mathematics

10th Edition

ISBN: 9781119166856

Author: Kreyszig

Publisher: JOHN WILEY+SONS INC.CUSTOM

See similar textbooks

Concept explainers

Ratios

A ratio is a comparison between two numbers of the same kind. It represents how many times one number contains another. It also represents how small or large one number is compared to the other.

Trigonometric Ratios

Trigonometric ratios give values of trigonometric functions. It always deals with triangles that have one angle measuring 90 degrees. These triangles are right-angled. We take the ratio of sides of these triangles.

Videos

Question

Chapter 3, Problem 18RQ

To determine

The solution of the given IVP.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 3, Problem 17RQ

Chapter 3, Problem 19RQ

Students have asked these similar questions

Question 1 20 pts Test data on the bending strength of construction wood poles of various diameter are presented below assuming the same length. Kip- 1000 lbf. Using the following data with 2nd order Newton polynomial interpolation, we want to determine the strength of the material for x=4.9 in. Which data point will be used as x? After you found x0, enter the value of x-xo in the solution. Answer shall be in one decimal place. Distance (in) Strength (kips) 100 3.6 1.1 5.6 3.6 5.6 200 300 400 500

Test data on the bending strength of construction wood poles of various diameter are presented below assuming the same length. Kip- 1000 lbf. Using the following data with 2nd order Newton polynomial interpolation, we want to determine the strength of the material for x=4.3 in. Which data point will be used as x0? After you found x0, enter the value of x-xo in the solution. Answer shall be in one decimal place. Distance (in) Strength (kips) 100 2.7 1 6.8 0.6 5.7 200 300 400 500

2/2. prove that if G is Euler then so is L (G).

Chapter 3 Solutions

Custom Kreyszig: Advanced Engineering Mathematics

Ch. 3.1 - 1–6 BASES: TYPICAL EXAMPLES To get a feel for...Ch. 3.1 - 1–6 BASES: TYPICAL EXAMPLES To get a feel for...Ch. 3.1 - 1–6 BASES: TYPICAL EXAMPLES To get a feel for...Ch. 3.1 - 1–6 BASES: TYPICAL EXAMPLES To get a feel for...Ch. 3.1 - Prob. 5P Ch. 3.1 - Prob. 6P Ch. 3.1 - Prob. 8P Ch. 3.1 - Prob. 9P Ch. 3.1 - Prob. 10P Ch. 3.1 - Prob. 11P

Knowledge Booster

$Advanced Math$

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.