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Differential EquationsHardNumerical Methods: Euler to Runge–Kutta

Numerical Methods: Euler to Runge–Kutta · Hard problem

For y′=yy' = y, y(0)=1y(0) = 1, Euler's method gives y(1)≈2.25y(1) \approx 2.25 with h=12h = \dfrac{1}{2} and y(1)≈2.44140625y(1) \approx 2.44140625 with h=14h = \dfrac{1}{4}. Since Euler's error is approximately ChCh, combine the two values by Richardson extrapolation to get an improved estimate of y(1)y(1). Round to four decimal places.