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Recursionhard
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The forward Euler method for dydt=ky\frac{dy}{dt} = kydtdy​=ky with step size hhh gives yn+1=yn(1+kh)y_{n+1} = y_n(1 + kh)yn+1​=yn​(1+kh). Applying this to dydt=−3y\frac{dy}{dt} = -3ydtdy​=−3y with h=0.2h = 0.2h=0.2 and y(0)=2y(0) = 2y(0)=2, which recurrence results?