Consider the first order initial value problem $y^{'} = y + 2 x - x^{2}, y (0) = 1, (0 \leq x < \infty)$ with exact solutions $y (x) = x^{2} + e^{x}$ . For x =0.1 , the percentage difference between the exact solution and the solution obtained using a single iteration of the second-order Runge-Kutta with step-size h = 0.1 is
0.063

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Solution

The correct option is A 0.063
Given, $y^{'} = f (x, y) = y + 2 x - x^{2}, y (0) = 1 (0 \leq x < \infty), h = 0.1$
Now by using runge-kutta method of $2^{n d}$ order
$y_{1} = y_{0} + \frac{1}{2} [k_{1} = k_{2}]$
where,
$k_{1} = h f (x_{0}, y_{0})$
$k_{2} = h f (x_{0} + h, y_{0} + k_{1})$
Here, $x_{0} = 0, y_{0} = 1, h = 0.1$
$∴ k_{1} = 0.1 [1 + 2 (0) - 0^{2}] = 0.1$
$k_{2} = 0.1 [(1 + 0.1) + 2 (0.1) - (0.1)^{2}]$
= 0.129
$∴ y_{1} = 1 + \frac{1}{2} [0.1 + 0.129] = 1 + 0.1145$
= 1.1145
Now, the exact solution,
$y (x) = x^{2} + e^{x}$
$\Rightarrow y (0.1) = (0.1)^{2} + e^{0.1} = 0.01 + 1.1052$
= 1.1152
$∴$ percentage difference
$= \frac{1.1152 - 1.1145}{1.1152} \times 100$ %
= 0.063%

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