To determine The solution of the initial value problem ( x 2 D 2 + x D − I ) y = 16 x 3 , y ( 1 ) = − 1 , y ′ ( 1 ) = 1 and graph it. Explanation Procedure used: “Suppose that, the non-homogeneous linear ordinary differential equation of the form y ″ + p ( x ) y ′ + q ( x ) y = r ( x ) and it is continuous on the interval I . Then, the particular solution y p on I is y p ( x ) = − y 1 ∫ y 2 r W d x + y 2 ∫ y 1 r W d x . Here, y 1 , y 2 form a basis of solutions of the corresponding ordinary differential equation. The value of W is obtained by Wronskian of y 1 , y 2 . That is, W = | y 1 y 2 y ′ 1 y ′ 2 | ” Also, the solution of the ordinary differential equation y ″ + p ( x ) y ′ + q ( x ) y = r ( x ) is the sum of general solution y h and particular solution y p . That is, y ( x ) = y h ( x ) + y p ( x ) . Result used: The general solution of Euler-Cauchy equation: “Suppose that, the linear ordinary differential equation of the form x 2 y ″ + a x y ′ + b y = 0 . In order to obtain the auxiliary equation, substitute y = x m , y ′ = m x m − 1 and y ″ = m ( m − 1 ) x m − 2 . The auxiliary equation is m 2 + ( a − 1 ) m + b = 0 . Then, then the general solution of the equation is of the form: Case 1: If the roots are real distinct m 1 and m 2 , then the general solution is, y h ( x ) = c 1 x m 1 + c 2 x m 2 . Case 2: If the roots are real double roots m 1 and m 1 , then the general solution is, y h ( x ) = ( c 1 + c 2 ln x ) x m 1 . Case 3: If the roots are complex a ± i b , then the general solution is, y h ( x ) = x a ( A cos ( b ln x ) + B sin ( b ln x ) ) . Calculation: Given that, the non-homogeneous linear ordinary differential equation is ( x 2 D 2 + x D − I ) y = 16 x 3 . It can be written as, x 2 y ″ + x y ′ − y = 16 x 3 . Divide the both sides of the equation x 2 y ″ + x y ′ − y = 16 x 3 by x 2 as, y ″ + 1 x y ′ − 1 x 2 y = 16 x . Compare the equation y ″ + 1 x y ′ − 1 x 2 y = 16 x with the equation y ″ + p ( x ) y ′ + q ( x ) y = r ( x ) . Here, it is observed that the function p ( x ) = 1 x , q ( x ) = − 1 x 2 and r ( x ) = 16 x . First compute the general solution y h of the non-homogeneous linear ordinary differential equation as follows. Here, the non-homogeneous linear ordinary differential equation is in form of x 2 y ″ + a x y ′ + b y = 0 . The auxiliary equation of x 2 y ″ + x y ′ − y = 16 x 3 is m 2 + ( 1 − 1 ) m − 1 = 0 . m 2 + ( 1 − 1 ) m − 1 = 0 m 2 − 1 = 0 m 2 = 1 m = ± 1 Thus, the roots of the auxiliary equation m 2 + ( 1 − 1 ) m − 9 = 0 are ± 1 . Here, the roots are real distinct. So, the general solution y h of the non-homogeneous linear ordinary differential equation is, y h ( x ) = A x + B x − 1 . Here, it is observed that the function y 1 ( x ) = x and the function y 2 ( x ) = x − 1 . Thus, the value of y ′ 1 ( x ) = 1 and the value of y ′ 2 ( x ) = − x − 2 . In order to obtain the value of y p , first find the Wronskian value as follows. W = | y 1 y 2 y ′ 1 y ′ 2 | = | x x − 1 1 − x − 2 | = − x 1 − 2 − 1 x − 1 = − x − 1 − x − 1 = − 2 x − 1 Thus, the value of W is − 2 x − 1

10th Edition

ISBN: 9781119809210

Author: Kreyszig

Publisher: WILEY

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Chapter 2, Problem 21RQ

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The solution of the initial value problem (x2D2+xD−I)y=16x3,y(1)=−1,y′(1)=1 and graph it.

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Chapter 2, Problem 20RQ

Chapter 2, Problem 22RQ

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