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Numerical methods for differential equations questions

Replace derivatives by differences and a differential equation becomes arithmetic. Step along the solution one h at a time, and Simpson's rule measures the area you pass over.

7 original questions · 25 marks · the numerical methods for differential equations notes · Further Pure 1

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• 2 marks · write · warm-up

  Write down the forward and central difference approximations to dy/dx at the point x_n, and say which is more accurate.

  Worked answer

  Forward: (y_n+1 − y_n)/h. Central: (y_n+1 − y_n−1)/(2h). The central difference is the more accurate for the same step length, because it straddles the point rather than looking only ahead. B1 for the two differences, B1 for naming the central one with a reason.

• 3 marks · warm-up

  For dy/dx = x² + y with y = 1 at x = 0 and h = 0.1, use the forward difference to estimate y at x = 0.2.

  Worked answer

  First step: y₁ = 1 + 0.1(0 + 1) = 1.1. Second step: y₂ = 1.1 + 0.1(0.01 + 1.1) = 1.211. M1 for the Euler step, A1 for y₁, A1 for y₂. Each step reuses the previous value, so an early rounding error propagates through everything after it.

• 3 marks · write

  Write down the central difference approximation to the second derivative, and rearrange it to give y_n+1.

  Worked answer

  The approximation is (y_n+1 − 2y_n + y_n−1)/h². Rearranged: y_n+1 = 2y_n − y_n−1 + h² times the second derivative at x_n. B1 for the approximation, M1 for rearranging, A1 for the result. Two previous values are needed, That is the reason a second order problem always supplies two starting conditions.

• 4 marks · use

  Use Simpson's rule with four strips to estimate ∫ e^−x² dx from 0 to 1, to 4 decimal places.

  Worked answer

  h = 0.25 and the ordinates are 1, 0.939413, 0.778801, 0.569783, 0.367879. The weighted sum with pattern 1, 4, 2, 4, 1 is 8.962265, and multiplying by h/3 gives 0.7469. B1 for h = 0.25, M1 for the ordinates, M1 for the 1, 4, 2, 4, 1 weighting, A1 for 0.7469. The true value is 0.7468, so four strips are accurate to three decimal places on a function with no elementary antiderivative.

• 2 marks · explain

  Explain why Simpson's rule requires an even number of strips.

  Worked answer

  The rule fits a parabola through each consecutive group of three ordinates, which uses up two strips at a time. An odd number leaves one strip with no partner and no parabola to fit, so the rule cannot be applied as it stands. B1 for a parabola through each group of three ordinates, B1 for strips being used two at a time.

• 4 marks

  For y'' = 2x − y with y = 1 at x = 0 and y = 1.05 at x = 0.1, use the central second-difference approximation with h = 0.1 to estimate y at x = 0.2 and at x = 0.3.

  Worked answer

  Rearranged, y_n+1 = 2y_n − y_n−1 + h²(2x_n − y_n).
  At n = 1, with x₁ = 0.1: y₂ = 2(1.05) − 1 + 0.01(0.2 − 1.05) = 1.1 − 0.0085 = 1.09150.
  At n = 2, with x₂ = 0.2: y₃ = 2(1.0915) − 1.05 + 0.01(0.4 − 1.0915) = 1.133 − 0.006915 = 1.12609.
  M1 for the recurrence, M1 for the first step, A1 for y at 0.2, A1 for y at 0.3.
  The recurrence needs the two values before it, so a second order problem always supplies two starting conditions. Keep five decimal places through the working; rounding y₂ to 1.09 shifts y₃ in the third place.

• 7 marks · stretch

  For dy/dx = x² + y with y = 1 at x = 0, use the improved Euler method with h = 0.1 to estimate y at x = 0.2, giving your answer to five decimal places. The exact solution is y = 3e^x − x² − 2x − 2. Compare the accuracy of your estimate with the value 1.211 obtained from the forward difference method with the same step length.

  Worked answer

  The improved Euler step is y_n+1 = y_n + ½(k₁ + k₂), with k₁ = hf(x_n, y_n) and k₂ = hf(x_n + h, y_n + k₁). The second increment is evaluated at the predicted point, not at the current one.
  First step, from (0, 1):
  k₁ = 0.1(0² + 1) = 0.1.
  k₂ = 0.1(0.1² + 1.1) = 0.1(1.11) = 0.111.
  y₁ = 1 + ½(0.1 + 0.111) = 1.1055.
  Second step, from (0.1, 1.1055):
  k₁ = 0.1(0.01 + 1.1055) = 0.11155.
  k₂ = 0.1(0.04 + 1.21705) = 0.125705.
  y₂ = 1.1055 + ½(0.11155 + 0.125705) = 1.1055 + 0.1186275 = 1.22413 to five decimal places.
  Comparison. The exact value is 3e^0.2 − 0.04 − 0.4 − 2 = 1.22421. The improved Euler estimate is out by about 0.00008; the forward difference value 1.211 is out by about 0.0132, roughly 160 times worse.
  M1 for k₁ and k₂ with k₂ at the predicted point, A1 for k₂, A1 for y₁, M1 for the second step, A1 for 1.22413, B1 for the exact value, B1 for the comparison of errors.
  The forward difference uses the gradient at the left end of each interval and so lags a curve that is bending upwards. Averaging the gradients at the two ends removes most of that lag, at the cost of one extra evaluation of the gradient per step.

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