[4 points] Consider the differential equation = f(y). A young numerical analyst proposesa new numerical scheme giving rise to the following finite difference approximation:%3D y_{n+1}=y_{n}+\frac{1}{4} h f\left(y_{n+1}\right)+\frac{3}{4} h f\left(y_{n}\right) where yn and yn+1 represent solutions at times tn and tr+1, respectively, and h = tn+1 - tn is the step size. Let f(y) = -iy with (i = v-1); identify, with proper mathematical reasoning and analysis, whether or not h = 1 will provide a stable solution. Show all steps. Box your-answer.

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