use the improved eulers method with tolerance to approximate the solution to yprime1 yy3 quad y0

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Use the improved Euler's method with tolerance to approximate the solution to $$ y^{\prime}=1-y+y^{3}, \quad y(0)=0 $$ at $ x=1 $. For a tolerance of $ \varepsilon=0.003 $, use a stopping procedure based on the absolute error.