To determine The solution of the initial value problem ( D 2 + 0.2 D + 0.26 I ) y = 1.22 e 0.5 x , y ( 0 ) = 3.5 , y ′ ( 0 ) = 0.35 . Explanation Formula used: Method of undetermined coefficients: Consider the linear ordinary differential equation y ″ + a y ′ + b y = r ( x ) . 1. If the term in r ( x ) is of the form k e γ x , then the choice for y p ( x ) is C e γ x . 2. If the term in r ( x ) is of the form k x n ( n = 0 , 1 , … ) , then the choice for y p ( x ) is K n x n + K n − 1 x n − 1 + ⋯ + K 1 x + K 0 . 3. If the term in r ( x ) is of the form k cos ω x or k sin ω x , then the choice for y p ( x ) is K cos ω x + M sin ω x . 4. If the term in r ( x ) is of the form k e α x cos ω x or k e α x sin ω x , then the choice for y p ( x ) is e α x ( K cos ω x + M sin ω x ) . Result used: Basic rule : In this rule, determine the undetermined coefficients of y p by substituting y p and its derivatives in y ″ + a y ′ + b y = r ( x ) . Calculation: The given initial value problem is ( D 2 + 0.2 D + 0.26 I ) y = 1.22 e 0.5 x , y ( 0 ) = 3.5 , y ′ ( 0 ) = 0.35 . Obtain the solution y h of the homogeneous differential equation y ″ + 0.2 y ′ + 0.26 y = 0 as follows. The auxiliary equation of y ″ + 0.2 y ′ + 0.26 y = 0 is m 2 + 0.2 m + 0.26 = 0 . Now obtain the roots of the auxiliary equation m 2 + 0.2 m + 0.26 = 0 as shown below. m = − 0.2 ± ( 0.2 ) 2 − 4 ( 0.26 ) 2 = − 0.2 ± − 1 2 = − 0.2 ± i 2 = − 0.1 ± 0.5 i Here the roots are complex conjugate. Therefore the solution y h is y h = e − 0.1 x ( c 1 cos 0.5 x + c 2 sin 0.5 x ) . Now obtain the particular solution y p as follows. Here the term r ( x ) is 1.22 e 0.5 x which is of the form k e γ x , and hence the choice for y p becomes C e γ x . That is, y p = C e 0.5 x . Differentiate y p = C e 0.5 x with respect to x and obtain the first derivative as shown below. y ′ p = d d x ( C e 0.5 x ) = 0.5 C e 0.5 x Again differentiate y ′ p and obtain the second derivative as follows. y ″ p = d d x ( 0.5 C e 0.5 x ) = ( 0.5 ) ( 0.5 C e 0.5 x ) = 0.25 C e 0.5 x By basic rule, substitute the obtained derivatives in y ″ + 0.2 y ′ + 0.26 y = 1.22 e 0.5 x and determine the undetermined coefficients as shown below. 0.25 C e 0.5 x + 0.2 ( 0.5 C e 0.5 x ) + 0.26 C e 0.5 x = 1.22 e 0.5 x 0.25 C e 0.5 x + 0.1 C e 0.5 x + 0

10th Edition

ISBN: 9781119809210

Author: Kreyszig

Publisher: WILEY

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Chapter 2.7, Problem 17P

To determine

The solution of the initial value problem (D2+0.2D+0.26I)y=1.22e0.5x, y(0)=3.5, y′(0)=0.35.

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Chapter 2 Solutions

ADV.ENG.MATH (LL) W/WILEYPLUS BUNDLE

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