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Mobile App Chapter 3: Q15E (page 140)
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\(\left[ {{\bf{0,3}}} \right]\).
Short Answer
\({{\rm{x}}_{\rm{n}}}\) | \({{\rm{y}}_{\rm{n}}}\) |
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.
Step by step solution
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
\(\begin{array}{l}{{\bf{k}}_{\bf{1}}}{\bf{ = h}}{\rm{f}}{\bf{(x,y) = 0}}{\bf{.1(cos5y - x)}}\\{{\bf{k}}_{\bf{2}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.1}}\left( {{\bf{cos}}\left( {{\bf{5y + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{ - (x + 0}}{\bf{.05)}}} \right)\\{{\bf{k}}_{\bf{3}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.1}}\left( {{\bf{cos}}\left( {{\bf{5y + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{ - (x + 0}}{\bf{.05)}}} \right)\\{{\bf{k}}_{\bf{4}}}{\bf{ = hf}}\left( {{\bf{x + h,y + }}{{\bf{k}}_{\bf{3}}}} \right){\bf{ = 0}}{\bf{.1(y + }}{{\bf{k}}_{\bf{3}}}{\bf{)(cos(5(y + }}{{\bf{k}}_{\bf{3}}}{\bf{) - (x + 0}}{\bf{.05))}}\\{{\bf{k}}_{\bf{1}}}{\bf{ = h(x,y) = 0}}{\bf{.1}}\\{{\bf{k}}_{\bf{2}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.091891}}\\{{\bf{k}}_{\bf{3}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.092373}}\\{{\bf{k}}_{\bf{4}}}{\bf{ = hf}}\left( {{\bf{x + h,y + }}{{\bf{k}}_{\bf{3}}}} \right){\bf{ = 0}}{\bf{.079522}}\end{array}\)
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.
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 |
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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