- 2075 Baisakh · 6 marks
R-K2: y' = y/(x²+y²)
Short answer
Using Heun's RK2 method with h=0.5: k1 = h f(x,y), k2 = h f(x+h, y+k1), y_{n+1}=y_n+(k1+k2)/2. Results: y(0.5) ≈ 1.4000, y(1.0) ≈ 1.6671.
Solve y' = y/(x² + y²), y(0) = 1 using R-K2 method in the range 0, 0.5, 1.
Given y' = y/(x² + y²), y(0) = 1, range 0 to 1 in steps of 0.5, so h = 0.5 (two steps).
The second-order Runge–Kutta (Heun / modified Euler) formulas are:
k₁ = h·f(xₙ, yₙ)
k₂ = h·f(xₙ + h, yₙ + k₁)
yₙ₊₁ = yₙ + ½(k₁ + k₂)
First step
k₁ = h·f(x₀, y₀) = 0.5·f(0.0000, 1.0000) = 0.5000
k₂ = h·f(x₀+h, y₀+k₁) = 0.5·f(0.5000, 1.5000) = 0.3000
y₁ = y₀ + (k₁+k₂)/2 = 1.0000 + (0.5000 + 0.3000)/2 = 1.4000
All steps (h = 0.5)
| step | xₙ | yₙ | k₁ = h·f(xₙ,yₙ) | k₂ = h·f(xₙ+h, yₙ+k₁) | yₙ₊₁ = yₙ + (k₁+k₂)/2 |
|---|---|---|---|---|---|
| 0 | 0.0000 | 1.0000 | 0.5000 | 0.3000 | 1.4000 |
| 1 | 0.5000 | 1.4000 | 0.3167 | 0.2175 | 1.6671 |
Answer: y(0.5) ≈ 1.4000, y(1) ≈ 1.6671
✓ Answer checked by Claude Opus 5.5 (high effort)
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