Consider the first order initial value problem y' = y + 2x − x2, y(0) = 1, (0 ≤ x < ∞)
Given y1 = y + 2x – x2 ⇒ f(x,y) = y + 2x – x2
From Runge-kutta second order method we have:
\({{\rm{y}}_1} = {{\rm{y}}_0} + \frac{1}{2}\left( {{{\rm{K}}_1} + {{\rm{K}}_2}} \right)\)
where: K1 = hf (x0, y0) and
K2 = hf(x0 + h, y0 + K1)
Given x0 = 0, y0 = 1, h = 0.1
Then K1 = 0.1 (y0 + 2x0 - x20 = 0.1 × 1 = 0.1
\( {{\rm{K}}_2}{\rm{\;}} = {\rm{\;}}0.1{\rm{\;f}}\left( {0{\rm{\;}} + {\rm{\;}}0.1,{\rm{\;}}1{\rm{\;}} + {\rm{\;}}0.1} \right) \)
\(= {\rm{\;}}\left( {0.1} \right){\rm{\;f\;}}\left( {0.1,{\rm{\;}}1.1} \right)\)
\(= {\rm{}}\left( {0.1} \right){\rm{}}\left( {1.1{\rm{}} + {\rm{}}2\left( {0.1} \right){\rm{}}-{\rm{}}0.01} \right){\rm{}} = {\rm{}}0.129\)
\( \therefore {\rm{}}{{\rm{y}}_1}{\rm{}} = {\rm{y}}\left( {0.1} \right){\rm{}} = {\rm{\;}}{{\rm{y}}_0} + \frac{1}{2}\left( {0.1 + 0.129} \right) \)
\(= 1.1145\) ---(1)
But the exact solution is given as:
\({\rm{y}}\left( {\rm{x}} \right){\rm{\;}} = {\rm{\;}}{{\rm{x}}^2}{\rm{\;}} + {\rm{\;}}{{\rm{e}}^{\rm{x}}}\)
\({\rm{y}}\left( {0.1} \right){\rm{}} = {\rm{}}{\left( {0.1} \right)^2}{\rm{\;}} + {\rm{}}{{\rm{e}}^{0.1}}{\rm{}} = {\rm{}}1.1152\) ---(2)
\({\rm{\% difference}} = \frac{{1.1152 - 1.1145}}{{1.1152}} × 100 \)
\(= 0.0627{\rm{\% }} \)
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