← 2013 Paper 2
UPSC 2013 Maths Optional Paper 2 Q7b — Step-by-Step Solution
15 marks · Section B
Euler's method (and modified Euler) · Numerical Analysis · asked 2× in 13 yrs · Read the full method →
Question
Use Euler’s method with step size h=0.15 to compute the approximate value of y(0.6), correct up to five decimal places from the initial value problem
y′=x(y+x)−2,y(0)=2.
Technique
Standard forward Euler — explicit, first-order accurate.
Solution
Setup. f(x,y)=xy+x2−2. Step size h=0.15. Number of steps: (0.6−0)/0.15=4.
Euler’s recurrence: yn+1=yn+hf(xn,yn).
Iterations
| Step | xn | yn | f(xn,yn) | hf |
|---|
| 0 | 0.00 | 2.00000 | 0⋅2+0−2=−2.00000 | −0.30000 |
| 1 | 0.15 | 1.70000 | 0.15(1.85)−2=−1.72250 | −0.25838 |
| 2 | 0.30 | 1.44163 | 0.30(1.74163)−2=−1.47751 | −0.22163 |
| 3 | 0.45 | 1.22000 | 0.45(1.67000)−2=−1.24850 | −0.18728 |
| 4 | 0.60 | 1.03272 | — | — |
Sample computations
Step 1 → Step 2 (x1=0.15, y1=1.70000):
- f(0.15,1.70)=0.15(1.70+0.15)−2=0.15(1.85)−2=0.2775−2=−1.72250.
- y2=1.70000+0.15(−1.72250)=1.70000−0.25838=1.44163 (rounded to 5 dp).
Step 2 → Step 3 (x2=0.30, y2=1.44163, using full precision 1.441625):
- f(0.30,1.441625)=0.30(1.741625)−2=0.5224875−2=−1.4775125.
- y3=1.441625−0.221627=1.219998 (5 dp).
Step 3 → Step 4 (x3=0.45, y3=1.219998):
- f(0.45,1.219998)=0.45(1.669998)−2=0.751499−2=−1.248501.
- y4=1.219998−0.187275=1.032723.
Final answer
Answer
y(0.6)≈1.03272.