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Q15E
ExpertverifiedUse the fourthorder 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.
Using the improved 4^{th} order RungeKutta 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
\(\begin{array}{l}{{\bf{k}}_{\bf{1}}}{\bf{ = h}}{\rm{f}}{\bf{(x,y) = 0}}{\bf{.1(cos5y  x)}}\\{{\bf{k}}_{\bf{2}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.1}}\left( {{\bf{cos}}\left( {{\bf{5y + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{  (x + 0}}{\bf{.05)}}} \right)\\{{\bf{k}}_{\bf{3}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.1}}\left( {{\bf{cos}}\left( {{\bf{5y + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{  (x + 0}}{\bf{.05)}}} \right)\\{{\bf{k}}_{\bf{4}}}{\bf{ = hf}}\left( {{\bf{x + h,y + }}{{\bf{k}}_{\bf{3}}}} \right){\bf{ = 0}}{\bf{.1(y + }}{{\bf{k}}_{\bf{3}}}{\bf{)(cos(5(y + }}{{\bf{k}}_{\bf{3}}}{\bf{)  (x + 0}}{\bf{.05))}}\\{{\bf{k}}_{\bf{1}}}{\bf{ = h(x,y) = 0}}{\bf{.1}}\\{{\bf{k}}_{\bf{2}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.091891}}\\{{\bf{k}}_{\bf{3}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.092373}}\\{{\bf{k}}_{\bf{4}}}{\bf{ = hf}}\left( {{\bf{x + h,y + }}{{\bf{k}}_{\bf{3}}}} \right){\bf{ = 0}}{\bf{.079522}}\end{array}\)
\(\begin{array}{c}{\bf{x = 0 + 0}}{\bf{.1 = 0}}{\bf{.1}}\\{\bf{y = 0 + }}\frac{{\bf{1}}}{{\bf{6}}}\left( {{{\bf{k}}_{\bf{1}}}{\bf{ + 2}}{{\bf{k}}_{\bf{2}}}{\bf{ + 2}}{{\bf{k}}_{\bf{3}}}{\bf{ + }}{{\bf{k}}_{\bf{4}}}} \right)\\{\bf{ = 0}}{\bf{.091342}}\end{array}\)
The solution of the given IVP at x=0.1 is approx. 0.091.
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 


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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