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Chapter 3: Q16E (page 140)

Use the fourth-order Runge–Kutta subroutine with h= 0.1 to approximate the solution to\({\bf{y' = 3cos(y - 5x),y(0) = 0}}\) , at the points x= 0, 0.1, 0.2, . . ., 4.0. Use your answers to make a rough sketch of the solution on [0, 4].

Short Answer

Expert verified

\({{\rm{x}}_{\rm{n}}}\)

\({{\rm{y}}_{\rm{n}}}\)

0.5

1.177

1.0

0.376

1.5

1.359

2.0

2.667

2.5

2.007

3.0

2.723

3.5

4.112

4

3.721

The rough sketch is given below.

Step by step solution

01

Find the values of \({{\bf{k}}_{\bf{i}}}{\bf{.i = 1,2,3,4}}\)

Using the improved 4th order Runge-Kutta subroutine.

Since \({\bf{f(x,y) = 3cos(y - 5x)}}\) and \({\bf{x = }}{{\bf{x}}_{\bf{0}}}{\bf{ = 0,y = }}{{\bf{y}}_{\bf{o}}}{\bf{ = 0}}\) and h = 0.1, M = 40

\(\begin{array}{l}{{\bf{k}}_{\bf{1}}}{\bf{ = h}}{\rm{f}}{\bf{(x,y) = (0}}{\bf{.1)3(cosy - 5x)}}\\{{\bf{k}}_{\bf{2}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{ = (0}}{\bf{.1)3}}\left( {{\bf{cos}}\left( {{\bf{y + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{ - 5(x + 0}}{\bf{.05)}}} \right)\\{{\bf{k}}_{\bf{3}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{ = (0}}{\bf{.1)3}}\left( {{\bf{cos}}\left( {{\bf{y + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{ - 5(x + 0}}{\bf{.05)}}} \right)\\{{\bf{k}}_{\bf{4}}}{\bf{ = hf}}\left( {{\bf{x + h,y + }}{{\bf{k}}_{\bf{3}}}} \right){\bf{ = (0}}{\bf{.1)3cos(y + }}{{\bf{k}}_{\bf{3}}}{\bf{) - 5(x + 0}}{\bf{.1))}}\\{{\bf{k}}_{\bf{1}}}{\bf{ = h(x,y) = 0}}{\bf{.3}}\\{{\bf{k}}_{\bf{2}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.298501}}\\{{\bf{k}}_{\bf{3}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.298479}}\\{{\bf{k}}_{\bf{4}}}{\bf{ = hf}}\left( {{\bf{x + h,y + }}{{\bf{k}}_{\bf{3}}}} \right){\bf{ = 0}}{\bf{.293929}}\end{array}\)

02

Find the values of x and y

\(\begin{array}{c}{\bf{x = 0 + 0}}{\bf{.1 = 0}}{\bf{.1}}\\{\bf{y = 0 + }}\frac{{\bf{1}}}{{\bf{6}}}\left( {{{\bf{k}}_{\bf{1}}}{\bf{ + 2}}{{\bf{k}}_{\bf{2}}}{\bf{ + 2}}{{\bf{k}}_{\bf{3}}}{\bf{ + }}{{\bf{k}}_{\bf{4}}}} \right)\\{\bf{ = 0}}{\bf{.297}}\end{array}\)

The solution of the given IVP at x=0.1 is approx. 0.297.

03

Evaluate the other values of x and y

x

y

x

y

x

y x y

0.2

0.583

1.2

0.576

2.2

2.723 3.2 3.317

0.3

0.841

1.3

0.801

2.3

2.520 3.3 3.609

0.4

1.049

1.4

1.069

2.4

2.228 3.4 3.880

0.5

1.177

1.5

1.359

2.5

2.007 3.5 4.112

0.6

1.180

1.6

1.658

2.6

1.946 3.6 4.278

0.7

1.023

1.7

1.954

2.7

2.028 3.7 4.339

0.8

0.742

1.8

2.233

2.8

2.207 3.8 4.251

0.9

0.486

1.9

2.479

2.9

2.447 3.9 4.009

1

0.376

2

2.667

3

2.723 4 3.721

1.1

0.421

2.1

2.764

3.1

3.017

04

Plot the Graph

Hence the solution is

\({{\rm{x}}_{\rm{n}}}\)

\({{\rm{y}}_{\rm{n}}}\)

0.5

1.177

1.0

0.376

1.5

1.359

2.0

2.667

2.5

2.007

3.0

2.723

3.5

4.112

4

3.721

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Most popular questions from this chapter

Rocket Flight. A model rocket having initial mass mo kg is launched vertically from the ground. The rocket expels gas at a constant rate of a kg/sec and at a constant velocity of b m/sec relative to the rocket. Assume that the magnitude of the gravitational force is proportional to the mass with proportionality constant g. Because the mass is not constant, Newton’s second law leads to the equation (mo - αt) dv/dt - αβ = -g(m0 – αt), where v = dx/dt is the velocity of the rocket, x is its height above the ground, and m0 - αt is the mass of the rocket at t sec after launch. If the initial velocity is zero, solve the above equation to determine the velocity of the rocket and its height above ground for 0≤t<m0/α.

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