Exercise · ZW-LOCAL-AND-GLOBAL-ERROR-010 · ★★☆ Intermediate

For \(y'=-2y\), \(y(0)=1\) and \(h=0.1\), the first Euler step gives \(y_1=0.8\). If the local error of that step is \(0.0187\), what is the approximate relative local error with respect to \(y(0.1)=e^{-0.2}\approx0.81873\)?

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Explanation

Dividing \(0.0187\) by \(0.81873\) gives about \(0.0228\), which is the relative error; \(0.0187\) is the absolute one.

The local error is the difference between the exact ODE solution over one step and the method's approximation starting from the exact value.

\[e_{n+1}=y(t_{n+1})-y_{n+1}\]