Explanation Formula used: The Euler’s formula for approximation is, y n + 1 = y n + h f ( x n , y n ) , n = 0 , 1 , 2.... Calculation: Consider the given initial value problem. y ′ = − 1 2 y , y ( 0 ) = 3 Start with x 0 = 0 and y 0 = 3 h = b − a n Substitute 2 for b and 0 for a in above equation. h = 2 − 0 n n = 2 h … ( 1 ) Substitute − 1 2 y n for f ( x n , y n ) in Euler’s formula. y n + 1 = y n + h ( − 1 2 y n ) = y n ( 1 − h 2 ) … ( 2 ) Substitute 0 for n in equation ( 2 ) . y 0 + 1 = y 0 ( 1 − h 2 ) y 1 = 3 ( 1 − h 2 ) Substitute 1 for n in equation ( 2 )

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Chapter 3.6, Problem 2E

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The solution of initial value problem y′=−12y, y(0)=3 at x=2 with step size h is 3(1−h2)2h using Euler’s method.

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Chapter 3.6, Problem 1E

Chapter 3.6, Problem 3E

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Chapter 3 Solutions

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The solution of initial value problem y ′ = − 1 2 y , y ( 0 ) = 3 at x = 2 with step size h is 3 ( 1 − h 2 ) 2 h using Euler’s method.

To show: The solution of initial value problem y′=−12y, y(0)=3 at x=2 with step size h is 3(1−h2)2h using Euler’s method.

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Chapter 3 Solutions

To show:

The solution of initial value problem y′=−12y, y(0)=3 at x=2 with step size h is 3(1−h2)2h using Euler’s method.