By experimenting with the fourth-order Runge-Kutta subroutine, find the maximum value over the interval \(\left[ {{\bf{1,2}}} \right]\)of the solution to the initial value problem\({\bf{y' = }}\frac{{{\bf{1}}{\bf{.8}}}}{{{{\bf{x}}^{\bf{4}}}}}{\bf{ - }}{{\bf{y}}^{\bf{2}}}{\bf{,y(1) = - 1}}\) . Where does this maximum occur? Give your answers to two decimal places.

Since \({\bf{f(x,y) = }}\frac{{{\bf{1}}{\bf{.8}}}}{{{{\bf{x}}^{\bf{4}}}}}{\bf{ - }}{{\bf{y}}^{\bf{2}}}{\bf{,y(1) = - 1}}\) and \({\bf{x = }}{{\bf{x}}_{\bf{0}}}{\bf{ = 1,y = }}{{\bf{y}}_{\bf{o}}}{\bf{ = - 1}}\) and h = 0.005, M = 200

\(\begin{array}{l}{{\bf{k}}_{\bf{1}}}{\bf{ = h(x,y) = 0}}{\bf{.004}}\\{{\bf{k}}_{\bf{2}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.00393054}}\\{{\bf{k}}_{\bf{3}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.00393019}}\\{{\bf{k}}_{\bf{4}}}{\bf{ = hf}}\left( {{\bf{x + h,y + }}{{\bf{k}}_{\bf{3}}}} \right){\bf{ = 0}}{\bf{.00386145}}\end{array}\)

\(\begin{array}{c}{\bf{x = 1 + 0}}{\bf{.005 = 1}}{\bf{.005}}\\{\bf{y = - 1 + }}\frac{{\bf{1}}}{{\bf{6}}}\left( {{{\bf{k}}_{\bf{1}}}{\bf{ + 2}}{{\bf{k}}_{\bf{2}}}{\bf{ + 2}}{{\bf{k}}_{\bf{3}}}{\bf{ + }}{{\bf{k}}_{\bf{4}}}} \right)\\{\bf{ = - 1 + }}\frac{{\bf{1}}}{{\bf{6}}}{\bf{(0}}{\bf{.004 + 2}}\left( {{\bf{0}}{\bf{.003931}}} \right){\bf{ + 2}}\left( {{\bf{0}}{\bf{.00393}}} \right){\bf{ + }}\left( {{\bf{0}}{\bf{.00386}}} \right){\bf{)}}\\{\bf{ = - 0}}{\bf{.996}}\end{array}\)

Apply the same procedure to the values of x and y.

x = 1.185, y = -0.929

\({\bf{x}} \in \left[ {{\bf{1}}{\bf{.190,1}}{\bf{.215}}} \right]{\bf{,y = 0}}{\bf{.928}}\)

x = 1.216, y = -0.929

This means that the solution of the given initial value problem crosses the x-axis in between x= 1.190 and 1.215.i.e 1.20.

Hence the solution is x = 1.20