Explanation Given: The boundary value problem is x 2 y ″ + 3 x y ′ + 3 y = 0 with the initial conditions as y ( 1 ) = 5 and y ( 2 ) = 0 . Calculation: The given equation can be rewritten as follows: x 2 y ″ + 3 x y ′ + 3 y = 0 y ″ + 3 y ′ x + 3 y x 2 = 0 ( Divide by x 2 ) Compare the obtained equation y ″ + 3 y ′ x + 3 y x 2 = 0 with equation y ″ + P ( x ) y ′ + Q ( x ) y = f ( x ) to obtain the value of P ( x ) , Q ( x ) and f ( x ) . P ( x ) = 3 x Q ( x ) = 3 x 2 f ( x ) = 0 The value of interior points are obtained as follows: x n − 1 = a + ( n − 1 ) h ...... ( 1 ) It is given that the value of a is 1 and the value of b is 2. For n = 8 the value of h is obtained as follows: h = b − a n = 2 − 1 8 = 1 8 = 0.125 The finite difference equation for an approximation of the differential equation is as follows: ( 1 + h 2 P i ) y i + 1 + ( − 2 + h 2 Q i ) y i + ( 1 − h 2 P i ) y i − 1 = h 2 f i ...... ( 2 ) Substitute 0.125 for h , 3 x for P , 3 x 2 for Q and 0 for f ( x ) in equation (1). ( 1 + 0.125 2 ( 3 x i ) ) y i + 1 + ( − 2 + ( 0.125 ) 2 ( 3 x i 2 ) ) y i + ( 1 − 0.125 2 ( 3 x i ) ) y i − 1 = 0 ( 1 + 0.1875 x i ) y i + 1 + ( − 2 + 0.0469 x i 2 ) y i + ( 1 − 0.1875 x i ) y i − 1 = 0 ...... ( 3 ) To obtain the value of interior point, x 1 substitute 2 for n in equation (1). x 1 = a + ( 2 − 1 ) ( 1 8 ) = 1 + 1 8 = 9 8 To obtain the value of interior point x 2 , substitute 3 for n in equation (1). x 2 = a + ( 3 − 1 ) 1 8 = 1 + ( 2 ) 1 8 = 1 + 1 4 = 5 4 To obtain the value of interior point x 3 , substitute 4 for n in equation (1). x 3 = a + ( 4 − 1 ) 1 8 = 1 + ( 3 ) 1 8 = 1 + 3 8 = 11 8 To obtain the value of interior point x 4 , substitute 5 for n in equation (1). x 4 = a + ( 5 − 1 ) 1 8 = 1 + ( 4 ) 1 8 = 1 + 1 2 = 3 2 To obtain the value of interior point x 5 , substitute 6 for n in equation (1). x 5 = a + ( 6 − 1 ) 1 8 = 1 + ( 5 ) 1 8 = 1 + 5 8 = 13 8 To obtain the value of interior point x 6 , substitute 7 for n in equation (1). x 6 = a + ( 7 − 1 ) 1 8 = 1 + ( 6 ) 1 8 = 1 + 3 4 = 7 4 To obtain the value of interior point x 7 , substitute 8 for n in equation (1). x 7 = a + ( 8 − 1 ) 1 8 = 1 + ( 7 ) 1 8 = 1 + 7 8 = 15 8 To obtain the value of interior point x 8 , substitute 9 for n in equation (1). x 8 = a + ( 9 − 1 ) 1 8 = 1 + ( 8 ) 1 8 = 1 + 1 = 2 Now substitute 1 for i in equation (3). ( 1 + 0.1875 x 1 ) y 2 + ( − 2 + 0.0469 x 1 2 ) y 1 + ( 1 − 0.1875 x 1 ) y 0 = 0 ( 1 + 0.1875 ( 9 / 8 ) ) y 2 + ( − 2 + 0.0469 ( 9 / 8 ) 2 ) y 1 + ( 1 − 0.1875 ( 9 / 8 ) ) y 0 = 0 ( x 1 = 9 8 ) 1.66667 y 2 − 1.96294 y 1 + 0.83334 y 0 = 0 ...... ( 4 ) Substitute 2 for i in equation (3). ( 1 + 0.1875 x 2 ) y 3 + ( − 2 + 0.0469 x 2 2 ) y 2 + ( 1 − 0.1875 x 2 ) y 1 = 0 ( 1 + 0.1875 ( 5 / 4 ) ) y 3 + ( − 2 + 0.0469 ( 5 / 4 ) 2 ) y 2 + ( 1 − 0.1875 ( 5 / 4 ) ) y 1 = 0 ( x 2 = 5 4 ) 1.15 y 3 − 1.9379 y 2 + 0.916667 y 1 = 0

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

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ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 9.5, Problem 7E

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The approximate solution of the boundary value problem x2y″+3xy′+3y=0, y(1)=5 and y(2)=0 for n=8.

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The approximate solution of the boundary value problem x 2 y ″ + 3 x y ′ + 3 y = 0 , y ( 1 ) = 5 and y ( 2 ) = 0 for n = 8 .

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The approximate solution of the boundary value problem x2y″+3xy′+3y=0, y(1)=5 and y(2)=0 for n=8.

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