Question: Consider the differential equation y = y 2 subject to y(0) = 2. (a) What is the value of y from the predictorcorrector scheme after

Consider the differential equation y′ = y² subject to y(0) = 2.

(a) What is the value of y from the predictor–corrector scheme after a single step h = 1/4?

(b) It is easy to show that the exact value at x = 1/4 is y = 4. Based on your result for a step size h = 1/4, what step size do we need to reduce the error (defined as |y_numerical − y_exact|) to a value of 1/8?

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