Explanation Given: d v d u = u ( u 2 + 1 ) v 3 + ( u + 5 2 ) v 2 , v ( 0 ) = 0.1 , absolute tolerance, ∈ = 0.0001 Approach: The recursive formulas for Runge-Kutta method are u n + 1 = u n + h , v n + 1 = v n + 1 6 ( k 1 + 2 k 2 + 2 k 3 + k 4 ) , Where k 1 = h f ( u n + v n ) , k 2 = h f ( u n + h 2 , v n + k 1 2 ) , k 3 = h f ( u n + h 2 , v n + k 2 2 ) , k 4 = h f ( u n + h , v n + k 3 ) Calculation: The inputs are u 0 = 2 , v 0 = 0.1 , c = 3 , ∈ = 0.0001 f ( u , v ) = u ( u 2 + 1 ) v 3 + ( u + 5 2 ) v 2 Consider, z = v 0 = 0.1 The initial value for N in this algorithm is N = 1 , so h = ( c − u 0 N ) = ( 3 − 2 1 ) = 1 ∴ k 1 = h f ( u 0 , v 0 ) = 1 f ( 2 , 0.1 ) = 0.049 , k 2 = h f ( u 0 + h 2 , v 0 + k 1 2 ) = 1 f ( 2.5 , 0.1245 ) = 0.088356 k 3 = h f ( u 0 + h 2 , v 0 + k 2 2 ) = 1 f ( 2.5 , 0.144178 ) = 0.120795 k 4 = h f ( u 0 + h , v 0 + k 3 ) = 1 f ( 2

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ISBN: 9780135902738

Author: Nagle

Publisher: PEARSON

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Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

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Chapter 3.7, Problem 19E

To determine

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The solution of the given initial value problem v=3.

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Chapter 3.7, Problem 18E

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