Q16E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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].

Step-by-Step Solution

Answer

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

 

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

Step 2: Find the values of x and y

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

 

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

 

Step 3: 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

Step 4: 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