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Part BCSIR NET December 2023

Numerical ODE: Euler, Runge–Kutta

Using Euler's method with step size 0.05, the approximate value of the solution of dy/dx =3x+2y+1,y(1)=1= \sqrt{3x + 2y + 1}, y(1) = 1, at x = 1.1 (rounded to two decimal places) is

  1. A.1.50
  2. B.1.65
  3. C.1.25✓ Correct
  4. D.1.15

The idea, in plain words

This method just walks along the tangent line, then recomputes the slope and walks again. Two steps means two slope evaluations and two small additions; there is nothing clever to spot, only arithmetic to keep straight.

Solution

y1=1+0.056≈1.1225;y2=1.1225+0.053.15+2.245+1≈1.1225+0.05⋅2.529≈1.249≈1.25y_{1} = 1 + 0.05\sqrt{6} \approx 1.1225; y_{2} = 1.1225 + 0.05\sqrt{3.15 + 2.245 + 1} \approx 1.1225 + 0.05\cdot2.529 \approx 1.249 \approx 1.25.

Why each option is right or wrong

Checked against 2 symbolic computations
  • A.Wrong

    False.

    1.50 overshoots badly.

    The increments here are small — each step adds roughly 0.05 × 2.5 ≈ 0.125 — so two steps can move y by only about a quarter of a unit.

  • False.

    1.65 is roughly what a single step of size 0.1 would give if the slope were mistakenly evaluated as 6≈2.45\sqrt{6} \approx 2.45 and then multiplied by a larger step, or what four steps would give.

    The problem specifies h = 0.05 from x = 1 to x = 1.1

    which is exactly two steps.

  • C.Correct

    True.

    Step 1:f(1,1)=3+2+1=61: f(1, 1) = \sqrt{3 + 2 + 1} = \sqrt{6}

    so y1=1+0.056≈1.1225(y_{1} = 1 + 0.05\sqrt{6} \approx 1.1225 (verified).

    Step 2 at x=1.05:f=3.15+2⋅1.1225+1=6.395≈2.5288x = 1.05: f = \sqrt{3.15 + 2\cdot1.1225 + 1} = \sqrt{6.395} \approx 2.5288

    so y2≈1.1225+0.1264≈1.2489(y_{2} \approx 1.1225 + 0.1264 \approx 1.2489 (verified exactly as 1+6/20+106+615/200)1 + \sqrt{6}/20 + \sqrt{10\sqrt{6} + 615}/200).

    Rounded, 1.25.

    The trigger: count the steps first — (1.1 − 1)/0.05 = 2 — then update x and y together at each one.

  • False.

    1.15 is close to the result of taking only the first step and stopping (1.1225 rounds to 1.12), or of using a step size half the one specified.

    The second step contributes another 0.126

    which is what separates this from the right answer.

2 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.

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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