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Part CCSIR NET December 2024two-conditions-on-the-second-order-terms-fix-b-and-c

Two conditions on the second order terms fix b and c

Let f:[0,1]×R→Rf : [0, 1] \times \mathbb{R} \to \mathbb{R} be a function such that f and its partial derivatives of orders less than or equal to 3 are continuous and bounded. Let y:[0,1]→Ry : [0, 1] \to \mathbb{R} be the solution of dy/dt =f(t,y(t)),0≤t≤1,y(0)=y0= f(t, y(t)), 0 \le t \le 1, y(0) = y_{0}, where y0∈Ry_{0} \in \mathbb{R}. For h > 0, denote t_j = jh. Let Y_j be an approximation of y(t_j), defined by Y_(j+1) = Y_j + ahf(t_j, Y_j) + bhf(t_(j+1), Y_j + chf(tj,Yj)),0<(j+1)h<1,Y0=y0(t_j, Y_j)), 0 < (j+1)h < 1, Y_{0} = y_{0}, where a,b,c∈Ra, b, c \in \mathbb{R}. If there exists an M > 0 such that |y(t_(j+1)) − y(t_j) − ahf(t_j, y(t_j)) − bhf(t_(j+1), y(t_j) + chf(t_j, y(t_j)))| ≤ Mh3^{3} for every h > 0, and every j with 0 ≤ (j+1)h < 1, then

  1. A.a + b = 1✓ Correct
  2. B.a + b = c✓ Correct
  3. C.a + b + c = 0
  4. D.a2+b2=c2a^{2} + b^{2} = c^{2}

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests numerical convergence.

See pricing

50 are analysed free — try those first.

The trap it tests

Numerical convergence

A scheme assumed to converge, or to converge at its advertised rate.

Drill statements like this

Related counterexample: A convergent method is stable for any step size

More on this topic

The chapter behind this: Numerical methods for ODE — free to read

From Numerical Analysis › Numerical ODE: Euler, Runge–Kutta

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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