Consider the initial value problem below to answer to following. a) Find the approximations to \( y(0.1) \) and \( y(0.2) \) using Euler's method with time steps of \( \Delta t=0.1,0.05,0.025 \), and \( 0.0125 \). b) Using the exact solution given, compute the errors in the Euler approximations at \( t=0.1 \) and \( t=0.2 \). c) Which time step results in the more accurate approximation? Explain your observations. d) In general, how does halving the time step affect the error at \( t=0.1 \) and \( t=0.2 \) ? \[ y^{\prime}(t)=-6 y, y(0)=1, y(t)=e^{-6 t} \] a) Complete the table below. \begin{tabular}{|c|c|c|} \hline \( \boldsymbol{\Delta t} \) & Approximation to \( \mathbf{y} \mathbf{( 0 . 1 )} \) & Approximation to \( \mathbf{y}(\mathbf{0 . 2} \mathbf{2} \) \\ \hline \( 0.1 \) & \( \square \) & \( \square \) \\ \hline \( 0.05 \) & \( \square \) & \( \square \) \\ \hline \( 0.025 \) & \( \square \) & \( \square \) \\ \hline \( 0.0125 \) & \( \square \) & \( \square \) \\ \hline \end{tabular} (Round to five decimal places as needed.)
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We have the initial value problem: \(y'(t) = -6y\), \(y(0) = 1\), and the exact solution is \(y(t) = e^{-6t}\). Show more…
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