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Expert-verified Found in: Page 140 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # 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 1: Find the values of $${{\bf{k}}_{\bf{i}}}{\bf{.i = 1,2,3,4}}$$

Using the improved 4th order Runge-Kutta subroutine.

Since $${\bf{f(x,y) = cos}}\;{\bf{5y - x}}$$ 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 ### Want to see more solutions like these? 