Euler's method

Problem 6.333 · medium

1. Each step follows the tangent line: yₙ₊₁ = yₙ + h·f(tₙ, yₙ), tₙ₊₁ = tₙ + h.
   Reviewed
2. \[ 1 \cdot \frac{1}{4} \cdot 0 + 2 = 2 \]
   y1 = y0 + h·f(t0, y0) with t0 = 0, slope 0.✓ Proved
3. \[ 1 \cdot \frac{1}{4} \cdot 1 \cdot \frac{1}{2} + 2 = \frac{17}{8} \]
   y2 = y1 + h·f(t1, y1) with t1 = 1/4, slope 1/2.✓ Proved
4. \[ 1 \cdot \frac{1}{4} \cdot 17 \cdot \frac{1}{16} + \frac{17}{8} = \frac{153}{64} \]
   y3 = y2 + h·f(t2, y2) with t2 = 1/2, slope 17/16.✓ Proved
5. \[ 1 \cdot \frac{1}{4} \cdot 459 \cdot \frac{1}{256} + \frac{153}{64} = \frac{2907}{1024} \]
   y4 = y3 + h·f(t3, y3) with t3 = 3/4, slope 459/256.✓ Proved