Discussion
For a given initial value problem
\[ y' = 1 + xy,\ y(0) = 2, \]
the value of \(y(0.1)\) by Euler's method with \(n = 0.1\) is :
Answer: A
Euler's method gives \(y(0.1) = 2.1\).
The printed \(n = 0.1\) is the step size \(h\). Euler's method uses \(y_{1} = y_{0} + h f(x_{0}, y_{0})\) with \(f(x, y) = 1 + xy\):
\[ y_{1} = 2 + 0.1(1 + 0 \cdot 2) = 2 + 0.1 = 2.1 \]
The slope at the start is only 1, because the \(xy\) term vanishes at \(x = 0\). 2.2 and 2.4 would need slopes of 2 and 4, and 2.0 ignores the step altogether.
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