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Q15E

Expert-verified

Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 140

Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

Use the fourth-order Runge–Kutta subroutine with h = 0.1 to approximate the solution to\({\bf{y' = cos}}\;{\bf{5y - x,y(0) = 0}}\),at the points x = 0, 0.1, 0.2, . . ., 3.0. Use your answers to make a rough sketch of the solution on\(\left[ {{\bf{0,3}}} \right]\).

\({{\rm{x}}_{\rm{n}}}\)	\({{\rm{y}}_{\rm{n}}}\)
0.5	0.21462
1.0	0.13890
1.5	-0.02668
2.0	-0.81879
2.5	-1.69491
3.0	-2.99510

The rough sketch is given below.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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

Using the improved 4^th order Runge-Kutta subroutine.

Since \({\bf{f(x,y) = cos}}\;{\bf{5y - x}}\) and \({\bf{x = }}{{\bf{x}}_{\bf{0}}}{\bf{ = 0,y = }}{{\bf{y}}_{\bf{o}}}{\bf{ = 0}}\) and h = 0.1, M = 30

\(\begin{array}{l}{{\bf{k}}_{\bf{1}}}{\bf{ = h}}{\rm{f}}{\bf{(x,y) = 0}}{\bf{.1(cos5y - x)}}\\{{\bf{k}}_{\bf{2}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.1}}\left( {{\bf{cos}}\left( {{\bf{5y + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{ - (x + 0}}{\bf{.05)}}} \right)\\{{\bf{k}}_{\bf{3}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.1}}\left( {{\bf{cos}}\left( {{\bf{5y + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{ - (x + 0}}{\bf{.05)}}} \right)\\{{\bf{k}}_{\bf{4}}}{\bf{ = hf}}\left( {{\bf{x + h,y + }}{{\bf{k}}_{\bf{3}}}} \right){\bf{ = 0}}{\bf{.1(y + }}{{\bf{k}}_{\bf{3}}}{\bf{)(cos(5(y + }}{{\bf{k}}_{\bf{3}}}{\bf{) - (x + 0}}{\bf{.05))}}\\{{\bf{k}}_{\bf{1}}}{\bf{ = h(x,y) = 0}}{\bf{.1}}\\{{\bf{k}}_{\bf{2}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.091891}}\\{{\bf{k}}_{\bf{3}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.092373}}\\{{\bf{k}}_{\bf{4}}}{\bf{ = hf}}\left( {{\bf{x + h,y + }}{{\bf{k}}_{\bf{3}}}} \right){\bf{ = 0}}{\bf{.079522}}\end{array}\)

Step 2: Find the values of x and y

\(\begin{array}{c}{\bf{x = 0 + 0}}{\bf{.1 = 0}}{\bf{.1}}\\{\bf{y = 0 + }}\frac{{\bf{1}}}{{\bf{6}}}\left( {{{\bf{k}}_{\bf{1}}}{\bf{ + 2}}{{\bf{k}}_{\bf{2}}}{\bf{ + 2}}{{\bf{k}}_{\bf{3}}}{\bf{ + }}{{\bf{k}}_{\bf{4}}}} \right)\\{\bf{ = 0}}{\bf{.091342}}\end{array}\)

The solution of the given IVP at x=0.1 is approx. 0.091.

Step 3: Evaluate the other values of x and y

x	y	x	y	x	y
0.2	0.157	1.2	0.087	2.2	-1.173
0.3	0.195	1.3	0.055	2.3	-1.300
0.4	0.212	1.4	0.019	2.4	-1.454
0.5	0.215	1.5	-0.027	2.5	-1.695
0.6	0.208	1.6	-0.086	2.6	-2.037
0.7	0.196	1.7	-0.170	2.7	-2.309
0.8	0.180	1.8	-0.306	2.8	-2.501
0.9	0.161	1.9	-0.535	2.9	-2.698
1	0.139	2	-0.819	3	-2.995
1.1	0.114	2.1	-0.029

Step 4: Plot the Graph

Hence the solution is

\({{\rm{x}}_{\rm{n}}}\)	\({{\rm{y}}_{\rm{n}}}\)
0.5	0.21462
1.0	0.13890
1.5	-0.02668
2.0	-0.81879
2.5	-1.69491
3.0	-2.99510

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