Discussion
Using Euler's method taking step size = 0.1, the approximate value of \(y\) obtained corresponding to \(x = 0.2\) for the initial value problem \(\frac{dy}{dx} = x^{2} + y^{2}\) and \(y(0) = 1\), is
Answer: C
Euler's method gives 1.222.
Euler's method updates \(y_{n+1} = y_{n} + h\,f(x_{n}, y_{n})\) with \(f(x, y) = x^{2} + y^{2}\) and \(h = 0.1\).
\[ \begin{aligned} y_{1} &= 1 + 0.1(0^{2} + 1^{2}) = 1.1 \\ y_{2} &= 1.1 + 0.1(0.1^{2} + 1.1^{2}) \\ &= 1.1 + 0.1(0.01 + 1.21) = 1.222 \end{aligned} \]
So \(y(0.2) \approx 1.222\). The value 1.122 adds the second increment, 0.122, to the starting value 1 instead of to 1.1; 1.110 and 1.322 do not come from the two Euler steps.
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