Numerical Solution of ODEs Questions and Answers

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.

Answer: C

The second-order Runge-Kutta choice is \(a = 0.25, b = 0.75, \alpha = 2/3, \beta = 2/3\).

Matching the Taylor expansion of \(y(x_{n})\) to second order requires

\[ \begin{aligned} a + b &= 1 \\ b\alpha &= \tfrac{1}{2},\quad b\beta = \tfrac{1}{2} \end{aligned} \]

For \(a = 0.25, b = 0.75, \alpha = \beta = 2/3\): \(a + b = 1\) and \(b\alpha = b\beta = 0.5\). This is Ralston's method.

\(a = b = 1\) gives \(a + b = 2\). \(a = b = 0.5, \alpha = \beta = 2\) gives \(b\alpha = 1\). \(a = b = 0.5, \alpha = 1, \beta = 2\) gives \(b\beta = 1\). Each fails a condition.