For $\frac{d y}{d x} = x + y^{2}$ , given that $y = 1$ at $x = 0$ ; value of $y$ at $x = 0.2$ , using Runge Kutta fourth order method (h = 0.2) will be _____ (upto 4 decimal places)
1.2735

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Solution

The correct option is A 1.2735
Given, $x_{0} = 0$ $y_{0} = 1$ $h = 0.2$
$f (x, y) = x + y^{2}$
$y_{1} = y_{(0.2)}$
$k_{1} = h f (x_{0}, y_{0}) = 0.2 f (0, 1) = 0.2 \times 1 = 0.2$
$k_{2} = h f (x_{0} + \frac{h}{2}, y_{0} + \frac{k_{1}}{2})$
$k_{2} = 0.2 f (0.1, 1.1) = 0.262$

$k_{3} = h f (x_{0} + \frac{h}{2}, y_{0} + \frac{k_{2}}{2})$
$= (0.2) f (0.1, 1.131) = 0.2758$
$k_{4} = h f (x_{0} + h, y_{0} + k_{3})$
$= 0.2 f (0.2, 1.2758) = 0.3655$
$k = \frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4}) = \frac{1}{6} (0.2 + 2 (0.262) + 2 (0.2758) + 0.3655)$
$= 0.2735$
$y_{1} = y_{(0.2)} = y_{0} + k = 1 + 0.2735 = 1.2735$

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