To determine To solve: The boundary value problem x 2 y ″ + 3 x y ′ + y = x 2 , y ( 1 ) = 0 , y ( e ) = 0 . Explanation The given differential equation is x 2 y ″ + 3 x y ′ + y = x 2 . Consider the corresponding homogeneous differential equation x 2 y ″ + 3 x y ′ + y = 0 . The differential equation x 2 y ″ + 3 x y ′ + y = 0 is of the form x 2 y ″ + α x y ′ + β y = 0 with α = 3 and β = 1 . Assume that the differential equation x 2 y ″ + 3 x y ′ + y = 0 has a solution of the form y = x r . Substitute y = x r in x 2 y ″ + 3 x y ′ + y . x 2 y ″ + 3 x y ′ + y = x 2 ( x r ) ′ ′ + 3 x ( x r ) ′ + ( x r ) = x 2 ( r x r − 1 ) ′ + 3 x ( r x r − 1 ) + x r = x 2 ( r ( r − 1 ) x r − 2 ) + 3 r x r + x r = r ( r − 1 ) x r + 3 r x r + x r Simplify further and proceed as follows. x 2 y ″ + 3 x y ′ + y = r ( r − 1 ) x r + 3 r x r + x r = ( r 2 − r ) x r + 3 r x r + x r = ( r 2 − r + 3 r + 1 ) x r = ( r 2 + 2 r + 1 ) x r Solve for r in the expression r 2 + 2 r + 1 . r 2 + 2 r + 1 = 0 ( r + 1 ) 2 = 0 r = − 1 or − 1 Since r has two real equal roots r = − 1 , x − 1 and x − 1 ln x are the solutions of x 2 y ″ + 3 x y ′ + y = 0 . Therefore, a linear combination of x − 1 and x − 1 ln x is the general solution of x 2 y ″ + 3 x y ′ + y = 0 . Hence, the general solution of x 2 y ″ + 3 x y ′ + y = 0 is given by y = c 1 x − 1 + c 2 x − 1 ln x , c 1 and c 2 are arbitrary constants. That is, the complementary function y h of the differential equation x 2 y ″ + 3 x y ′ + y = x 2 is y h = c 1 x − 1 + c 2 x − 1 ln x , c 1 and c 2 are arbitrary constants. Compute the particular integral of the nonhomogeneous differential equation x 2 y ″ + 3 x y ′ + y = x 2 as shown below. The undetermined coefficient set corresponding to the right hand side function x 2 is { 1 , x , x 2 } . Consider the linear combination y = d 1 + d 2 x + d 3 x 2 of 1 , x and x 2 . Compute the first and the second derivatives of the equation y = d 1 + d 2 x + d 3 x 2 . y = d 1 + d 2 x + d 3 x 2 y ′ = d 2 + d 3 ( 2 x ) y ′ = d 2 + 2 d 3 x y ″ = 2 d 3 Substitute for y , y ′ and y ″ in the differential equation x 2 y ″ + 3 x y ′ + y = x 2 . x 2 y ″ + 3 x y ′ + y = x 2 x 2 ( 2 d 3 ) + 3 x ( d 2 + 2 d 3 x ) + ( d 1 + d 2 x + d 3 x 2 ) = x 2 2 d 3 x 2 + 3 d 2 x + 6 d 3 x 2 + d 1 + d 2 x + d 3 x 2 = x 2 ( 2 d 3 + 6 d 3 + d 3 ) x 2 + ( 3 d 2 + d 2 ) x + d 1 = x 2 9 d 3 x 2 + 4 d 2 x + d 1 = x 2 Compare coefficients of x 2 , x and constant terms on both sides of the above equation

Elementary Differential Equations and Boundary Value Problems, Enhanced

11th Edition

ISBN: 9781119381648

Author: Boyce

Publisher: WILEY

See similar textbooks

Concept explainers

Domain

In simple mathematics terms, a domain is defined as the space on which all the possible inputs for the function will lie. It can also be expressed as a set of departures for the required function.

Videos

Question

Chapter 10.1, Problem 12P

To determine

To solve: The boundary value problem x2y″+3xy′+y=x2, y(1)=0, y(e)=0.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 10.1, Problem 11P

Chapter 10.1, Problem 13P

Students have asked these similar questions

Q/Determine the set of points at which - f(z) = 622 2≥ - 4i/z12 i and differentiable analytice is:

sy = f(x) + + + + + + + + + X 3 4 5 7 8 9 The function of shown in the figure is continuous on the closed interval [0, 9] and differentiable on the open interval (0, 9). Which of the following points satisfies conclusions of both the Intermediate Value Theorem and the Mean Value Theorem for f on the closed interval [0, 9] ? (A A B B C D

= Q6 What will be the allowable bearing capacity of sand having p = 37° and ydry 19 kN/m³ for (i) 1.5 m strip foundation (ii) 1.5 m x 1.5 m square footing and (iii)1.5m x 2m rectangular footing. The footings are placed at a depth of 1.5 m below ground level. Assume F, = 2.5. Use Terzaghi's equations. 0 Ne Na Ny 35 57.8 41.4 42.4 40 95.7 81.3 100.4

Chapter 10 Solutions

Elementary Differential Equations and Boundary Value Problems, Enhanced

Ch. 10.1 - In each of Problems 1 through 13, either solve the...Ch. 10.1 - Prob. 2P Ch. 10.1 - In each of Problems 1 through 13, either solve the...Ch. 10.1 - Prob. 4P Ch. 10.1 - In each of Problems 1 through 13, either solve the...Ch. 10.1 - In each of Problems 1 through 13, either solve the...Ch. 10.1 - In each of Problems 1 through 13, either solve the...Ch. 10.1 - Prob. 8P Ch. 10.1 - Prob. 9P Ch. 10.1 - In each of Problems 1 through 13, either solve the...

Knowledge Booster

Learn more about

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