given-the-differential-equation-yt2yyprimet-2y-and-y04-y0-4-with-exact-solution-

Question

Given the differential equation y′(t)=−$$2yy^{\prime}(t)=-2y$$, and y(0)=4 y(0) = 4 , with exact solution y(t)=4e−2t y(t) = $$4e^{-2t}$$ , what happens to the global error at t=0.5 t = 0.5  if the Euler method's time step Δt=0.1 \Delta t=0.1  is halved?

Accepted Answer

The error is approximately halved

Answer

The error is doubled

Answer

The error is approximately quartered

Answer

The error remains unchanged

Given the differential equation y′(t)=−2yy^{\(\prime\)}(t)=-2y, and y(0)=4 y(0) = 4 , with exact solution y(t)=4e−2t y(t) = 4e^{-2t} , what happens to the global error at t=0.5 t = 0.5 if the Euler method's time step Δt=0.1 \(\Delta\) t=0.1 is halved?

Given the differential equation $y^{'} (t) = - 2 y$ , and $y of 0 equals 4$ , with exact solution $y (t) = 4 e^{- 2 t}$ , what happens to the global error at $t equals 0.5$ if the Euler method's time step $triangle t equals 0.1$ is halved?