Euler's method

Problem 6.331 · medium

1. Each step follows the tangent line: yₙ₊₁ = yₙ + h·f(tₙ, yₙ), tₙ₊₁ = tₙ + h.
   Reviewed
2. \[ 0 + 1 \cdot \frac{1}{5} \cdot 0 = 0 \]
   y1 = y0 + h·f(t0, y0) with t0 = 0, slope 0.✓ Proved
3. \[ 1 \cdot \frac{1}{5} \left(-1\right) \frac{1}{25} + 0 = - \frac{1}{125} \]
   y2 = y1 + h·f(t1, y1) with t1 = 1/5, slope -1/25.✓ Proved
4. \[ 1 \cdot \frac{1}{5} \left(-21\right) \frac{1}{125} - 1 \cdot \frac{1}{125} = - \frac{26}{625} \]
   y3 = y2 + h·f(t2, y2) with t2 = 2/5, slope -21/125.✓ Proved
5. \[ 1 \cdot \frac{1}{5} \left(-251\right) \frac{1}{625} - \frac{26}{625} = - \frac{381}{3125} \]
   y4 = y3 + h·f(t3, y3) with t3 = 3/5, slope -251/625.✓ Proved