The solution to the initial value problem \(\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ = }}{{\bf{y}}^{\bf{2}}}{\bf{ - 2}}{{\bf{e}}^{\bf{x}}}{\bf{y + }}{{\bf{e}}^{{\bf{2x}}}}{\bf{ + }}{{\bf{e}}^{\bf{x}}}{\bf{,y(0) = 3}}\)has a vertical asymptote (“blows up”) at some point in the interval\(\left[ {{\bf{0,2}}} \right]\). By experimenting with the fourth-order Runge–Kutta subroutine, determine this point to two decimal places.

When the solution of given IVP “blows up” using the improved 4^th order Runge-Kutta subroutine.

Since \({\bf{f(x,y) = }}{{\bf{y}}^{\bf{2}}}{\bf{ - 2}}{{\bf{e}}^{\bf{x}}}{\bf{y + }}{{\bf{e}}^{{\bf{2x}}}}{\bf{ + }}{{\bf{e}}^{\bf{x}}}\) and \({\bf{x = }}{{\bf{x}}_{\bf{0}}}{\bf{ = 0,y = }}{{\bf{y}}_{\bf{o}}}{\bf{ = 3}}\) and h = 0.005, M = 400

\(\begin{array}{l}{{\bf{k}}_{\bf{1}}}{\bf{ = h}}{\rm{f}}{\bf{(x,y) = 0}}{\bf{.005(}}{{\bf{y}}^{\bf{2}}}{\bf{ - 2}}{{\bf{e}}^{\bf{x}}}{\bf{y + }}{{\bf{e}}^{{\bf{2x}}}}{\bf{ + }}{{\bf{e}}^{\bf{x}}}{\bf{)}}\\{{\bf{k}}_{\bf{2}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.005}}\left( {{{\left( {{\bf{y + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right)}^{\bf{2}}}} \right){\bf{ - 2}}{{\bf{e}}^{{\bf{x + 0}}{\bf{.0025}}}}\left( {{\bf{y + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{ + }}{{\bf{e}}^{{\bf{2(x + 0}}{\bf{.0025)}}}}{\bf{ + }}{{\bf{e}}^{{\bf{x + 0}}{\bf{.0025}}}}\\{{\bf{k}}_{\bf{3}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.005}}\left( {{{\left( {{\bf{y + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right)}^{\bf{2}}}} \right){\bf{ - 2}}{{\bf{e}}^{{\bf{x + 0}}{\bf{.0025}}}}\left( {{\bf{y + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{ + }}{{\bf{e}}^{{\bf{2(x + 0}}{\bf{.0025)}}}}{\bf{ + }}{{\bf{e}}^{{\bf{x + 0}}{\bf{.0025}}}}\\{{\bf{k}}_{\bf{4}}}{\bf{ = hf}}\left( {{\bf{x + h,y + }}{{\bf{k}}_{\bf{3}}}} \right){\bf{ = 0}}{\bf{.005}}\left( {{{{\bf{(y + }}{{\bf{k}}_{\bf{3}}}{\bf{)}}}^{\bf{2}}}{\bf{ - 2}}{{\bf{e}}^{{\bf{x + 0}}{\bf{.0025}}}}{\bf{ + }}{{\bf{e}}^{{\bf{2(x + 0}}{\bf{.0025)}}}}{\bf{ + }}{{\bf{e}}^{{\bf{x + 0}}{\bf{.0025}}}}} \right)\\{{\bf{k}}_{\bf{1}}}{\bf{ = h(x,y) = 0}}{\bf{.025}}\\{{\bf{k}}_{\bf{2}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.02522}}\\{{\bf{k}}_{\bf{3}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.02522}}\\{{\bf{k}}_{\bf{4}}}{\bf{ = hf}}\left( {{\bf{x + h,y + }}{{\bf{k}}_{\bf{3}}}} \right){\bf{ = 0}}{\bf{.02543}}\end{array}\)

\(\begin{array}{c}{\bf{x = 1 + 0}}{\bf{.005 = 1}}{\bf{.005}}\\{\bf{y = 3 + }}\frac{{\bf{1}}}{{\bf{6}}}\left( {{{\bf{k}}_{\bf{1}}}{\bf{ + 2}}{{\bf{k}}_{\bf{2}}}{\bf{ + 2}}{{\bf{k}}_{\bf{3}}}{\bf{ + }}{{\bf{k}}_{\bf{4}}}} \right)\\{\bf{ = 3}}{\bf{.025}}\end{array}\)

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

\(\begin{array}{l}{\bf{x = 0}}{\bf{.505}},{\bf{y}} \approx 2020103\\{\bf{x = 0}}{\bf{.510,y}} \approx 9550268\end{array}\)

This means that the solution of the given initial value problem “blows up” on the interval \(\left[ {{\bf{0,2}}} \right]\) at x=0.51.

Hence the solution is x = 0.51