Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis
Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis
5th Edition
ISBN: 9780321816252
Author: C. Henry Edwards, David E. Penney, David Calvis
Publisher: PEARSON
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Chapter 2.6, Problem 1P
Program Plan Intro

Program Description: Purpose of the problem is to construct a table for the approximation solution and the actual solution of y=y up to five decimal places by the use of Runge-Kutta’s method.

Summary Introduction:

Purpose will use Runge-Kutta’s method to construct the table of the approximation solution and the actual solution y=y(x) is the solution of a differential equation dydx=f(x,y) with y(x0)=y0 where approximate value can be calculated with the formula yi+1=yi+kh and the value of k can be calculated as k=16(k1+2k2+2k3+k4) where the value of k1=f(xi,yi) , k2=f(xi+12h,yi+12hk1) , k3=f(xi+12h,yi+12hk2) and k4=f(xi+1,yi+hk3) .

Expert Solution & Answer
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Explanation of Solution

Given information:

In the interval of [0,0.5] the function have 2 subintervals as h=0.25 .

The differential equation is dydx=y and y(x)=2ex with y(0)=2 .

Calculation:

The initial value problem dydx=f(x,y) , y(x0)=y0 can be calculated with RungeKutta’s method with the size of h .

The differential equation is dydx=y and y(x)=2ex with y(0)=2 .

The value of k1 is f(xi,yi) .

Therefore, the value of k1 can be expressed as,

  k1=yi

Substitute 0 for i in the predicted formula to obtain the formula of u1 as,

  k1=y0 ........ (1)

Substitute 2 for y0 in equation (1) to obtain the value of k1 as,

  k1=2

The value of k2 is f(xi+12h,yi+12hk1) .

Therefore, the value of k2 can be expressed as,

  k2=(yi+12hk1)

Substitute 0 for i in the predicted formula to obtain the formula of k2 as,

  k2=(y0+12hk1) ........ (2)

Substitute 2 for y0 , 0.25 for h and 2 for k1 in equation (2) to obtain the value of k2 as,

  k2=(2+12( 0.25)( 2))k2=1.75

The value of k3 is f(xi+12h,yi+12hk2) .

Therefore, the value of k3 can be expressed as,

  k3=(yi+12hk2)

Substitute 0 for i in the predicted formula to obtain the formula of k3 as,

  k3=(y0+12hk2) ........ (3)

Substitute 2 for y0 , 0.25 for h and 1.75 for k2 in equation (3) to obtain the value of k3 as,

  k3=(2+12( 0.25)( 1.75))k3=1.78125

The value of k4 is f(xi+1,yi+hk3) .

Therefore, the value of k4 can be expressed as,

  k4=(yi+hk3)

Substitute 0 for i in the predicted formula to obtain the formula of k4 as,

  k4=(y0+hk3) ........ (4)

Substitute 2 for y0 , 0.25 for h and 1.78125 for k3 in equation (4) to obtain the value of k4 as,

  k4=(2+( 0.25)( 1.78125))k4=1.55469

The value of k can be calculated as,

  k=16(2+2( 1.75)+2( 1.78125)+( 1.55469))k=1.76953

The approximate value can be calculated as,

  yi+1=yi+khy1=2+(1.76953)(0.25)y1=1.55762

Now actual value can be calculated by substituting the given values of xi in the y(x)=2ex .

Substitute 0.25 for x in the equation y(x)=2ex to obtain the actual solution of y for x=0.25 .

  y(0.25)=2e( 0.25)y(0.25)=1.55760

Similarly, further values can also be calculated using the above steps.

    k1k2k3k4xiapprox value yiActual value     y(xi)
    21.751.781251.554690.251.557621.55760
    1.557621.362921.387251.21080.51.213091.21306

Therefore, the above table shows all the values of approximation value yi and actual value yi .

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Chapter 2 Solutions

Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis

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