Explanation Procedure used: “Find an approximate solution of a boundary-value problem.” (1) Compare given differential equation with y ″ + P ( x ) y ′ + Q ( x ) y = f ( x ) to identify the functions P ( x ) , Q ( x ) and f ( x ) ; and the boundary conditions with y ( a ) = α , y ( b ) = β to identify the boundary [ a , b ] . Also, identify the number of steps n . (2) Compute step size of the approximation using the formula h = b − a n and use the value to identify the interior mesh points x i = a + ( i + 1 ) h for i = 0 , 1 , 2 , … , n . (3) At each interior mesh point x i with i = 1 , 2 , … , n − 1 as obtained in Step 2, define p i = P ( x i ) , q i = Q ( x i ) and f i = f ( x i ) . (4) Formulate the system of difference equations for the interior mesh points corresponding to i = 1 , 2 , … , n − 1 as ( 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 using the values obtained in Step 3. (5) Solve the system using given boundary value conditions to obtain the functional values at each interior mesh point. Calculation: Given differential equation is y ″ + 2 y ′ + y = 5 x with boundary values as y ( 0 ) = 0 and y ( 1 ) = 0 . Compare the equation with general form to identify the required functions. P ( x ) = 2 Q ( x ) = 1 f ( x ) = 5 x The boundary which is to be partitioned into n = 5 interior mesh points and in which the solution is to be approximated is [ 0 , 1 ] . Compute the value of h as shown below: h = 1 − 0 5 = 0.2 Find the interior mesh points and the functional values p i , q i and f i corresponding to each point x i . The value of x 1 is calculated as follows: x 1 = 0 + ( 1 ⋅ 0.2 ) = 0.2 The value of p 1 is calculated as follows: p 1 = P ( 0.2 ) = 2 The value of q 1 is calculated as follows: q 1 = Q ( 0.2 ) = 1 The value of f 1 is calculated as follows: f 1 = f ( 0.2 ) = 5 ⋅ ( 0.2 ) = 1 This gives that when i = 1 , then p 1 = 2 , q 1 = 1 , f 1 = 1 . The difference equation is as follows: ( 1 + 0.2 2 × 2 ) y 1 + 1 + { − 2 + ( 0.2 ) 2 × 1 } y 1 + ( 1 − 0.2 2 × 2 ) y 1 − 1 = ( 0.2 ) 2 × 1 1.2 y 2 − 1 . 9 6 y 1 + 0.8 y 0 = 0.04 The value of x 2 is calculated as follows: x 2 = 0 + ( 2 ⋅ 0.2 ) = 0.4 The value of p 2 is calculated as follows: p 2 = P ( 0.4 ) = 2 The value of q 2 is calculated as follows: q 2 = Q ( 0.4 ) = 1 The value of f 2 is calculated as follows: f 2 = f ( 0.4 ) = 5 ⋅ ( 0.4 ) = 2 This gives that when i = 2 , then p 2 = 2 , q 2 = 1 , f 2 = 2 . The difference equation is as follows: ( 1 + 0.2 2 × 2 ) y 2 + 1 + { − 2 + ( 0.2 ) 2 × 1 } y 2 + ( 1 − 0.2 2 × 2 ) y 2 − 1 = ( 0.2 ) 2 × 2 1.2 y 3 − 1 . 9 6 y 2 + 0.8 y 1 = 0.08 The value of x 3 is calculated as follows: x 3 = 0 + 3 × 0.2 = 0.6 The value of p 3 is calculated as follows: p 3 = P ( 0.6 ) = 2 The value of q 3 is calculated as follows: q 3 = Q ( 0.6 ) = 1 The value of f 3 is calculated as follows: f 3 = f ( 0.6 ) = 5 ⋅ ( 0.6 ) = 3 This gives that when i = 3 , then p 3 = 2 , q 3 = 1 , f 3 = 3 . The difference equation is as follows: ( 1 + 0.2 2 × 2 ) y 3 + 1 + { − 2 + ( 0.2 ) 2 × 1 } y 3 + ( 1 − 0.2 2 × 2 ) y 3 − 1 = ( 0.2 ) 2 × 3 1.2 y 4 − 1 . 9 6 y 3 + 0.8 y 2 = 0.12 The value of x 4 is calculated as follows: x 4 = 0 + 4 × 0.2 = 0.8 The value of p 4 is calculated as follows: p 4 = P ( 0.8 ) = 2 The value of q 4 is calculated as follows: q 4 = Q ( 0

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

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ISBN: 9781337515573

Author: ZILL

Publisher: Cengage

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