Q11E
Question
The solution to the initial value problem\({\bf{y' = }}\frac{{\bf{2}}}{{{{\bf{x}}^{\bf{4}}}}}{\bf{ - }}{{\bf{y}}^{\bf{2}}}{\bf{,y(1) = - 0}}{\bf{.414}}\), crosses the x-axis at a point in the interval \(\left[ {{\bf{1,2}}} \right]\).By experimenting with the fourth-order Runge–Kutta subroutine, determine this point to two decimal places
Step-by-Step Solution
Verifiedx = 1.41
Since \({\bf{y' = }}\frac{{\bf{2}}}{{{{\bf{x}}^{\bf{4}}}}}{\bf{ - }}{{\bf{y}}^{\bf{2}}}{\bf{,y(1) = - 0}}{\bf{.414}}\) and \({\bf{x = }}{{\bf{x}}_{\bf{0}}}{\bf{ = 0,y = }}{{\bf{y}}_{\bf{o}}}{\bf{ = - 0}}{\bf{.414}}\) and h = 0.005, M = 200
\(\begin{array}{l}{{\bf{k}}_{\bf{1}}}{\bf{ = h}}{\rm{f}}{\bf{(x,y) = 0}}{\bf{.005(1 - 0) = 0}}{\bf{.914302}}\\{{\bf{k}}_{\bf{2}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.00989131}}\\{{\bf{k}}_{\bf{3}}}{\bf{ = hf}}\left( {{\bf{x + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,y + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.00906399}}\\{{\bf{k}}_{\bf{4}}}{\bf{ = hf}}\left( {{\bf{x + h,y + }}{{\bf{k}}_{\bf{3}}}} \right){\bf{ = 0}}{\bf{.00898261}}\end{array}\)
\(\begin{array}{c}{\bf{x = 1 + 0}}{\bf{.005 = 1}}{\bf{.005}}\\{\bf{y = - 0}}{\bf{.414 + }}\frac{{\bf{1}}}{{\bf{6}}}\left( {{{\bf{k}}_{\bf{1}}}{\bf{ + 2}}{{\bf{k}}_{\bf{2}}}{\bf{ + 2}}{{\bf{k}}_{\bf{3}}}{\bf{ + }}{{\bf{k}}_{\bf{4}}}} \right)\\{\bf{ = - 0}}{\bf{.414 + }}\frac{{\bf{1}}}{{\bf{6}}}{\bf{(0}}{\bf{.914302 + 2}}\left( {{\bf{0}}{\bf{.0099}}} \right){\bf{ + 2}}\left( {{\bf{0}}{\bf{.0091}}} \right){\bf{ + }}\left( {{\bf{0}}{\bf{.00898}}} \right){\bf{)}}\\{\bf{ = - 0}}{\bf{.405}}\end{array}\)
Apply the same procedure to the values of x and y.
x = 1.410, y = -0.002
x = 1.415, y = 0.001
This means that the solution of the given initial value problem crosses the x-axis in between x= 1.410 and 1.415.i.e 1.41.
Hence the solution is x = 1.41