Consider the following initial value problem.a. Find the approximations to y(0.2) and y(0.4) using Euler’s methodwith time steps of Δt = 0.2, 0.1, 0.05, and 0.025.b. Using the exact solution given, compute the errors in the Euler approximations at t = 0.2 and t = 0.4.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 att = 0.2 and t = 0.4? y '(t) = 4 - y, y(0) = 3; y(t) = 4 - e-t

Given:

$y\text{'}\left(t\right)=4-y, y\left(0\right)=3; y\left(t\right)=4-e^{-t}$.

To Do:

(a) Find the approximations to $y\left(0.2\right) \mathrm{and} y\left(0.4\right)$ using the Euler's method with time steps $∆t=0.2, 0.1, 0.05, 0.025$.

(b) Using the exact solution calculate the errors.

(c) Which time step gives more accurate approximation?

(d) In general, how does halving the time step affect the error at $t=0.2 \mathrm{and} t=0.4$?

(a)
We have,
$y\text{'}\left(t\right)=4-y, y\left(0\right)=3; y\left(t\right)=4-e^{-t}$.

Let $f\left(t,y\right)=4-y$.

Euler's formula is: $y_{i+1}=y_i+∆t f\left(t_i, y_i\right)$.

Calculating $y\left(0.2\right)$:

Now, for $\mathbf{∆}\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}$ we have,
$\begin{array}{rcl}{y}_1& =& y_0+∆t f\left(t_0, y_0\right)\\ & =& 3+\left(0.2\right)f\left(0,3\right)\\ & =& 3+\left(0.2\right)\left(1\right)\\ & =& 3+0.2\\ \mathit{y}\left(0.2\right)& \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{2}\end{array}$

For $\mathbf{∆}\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}$ we have,
$\begin{array}{rcl}{y}_1& =& y_0+∆t f\left(t_0, y_0\right)\\ & =& 3+\left(0.1\right)f\left(0,3\right)\\ & =& 3+\left(0.1\right)\left(1\right)\\ & =& 3+0.1\\ y_1& =& 3.1\end{array}$

$\begin{array}{rcl}{y}_2& =& y_1+∆t f\left(t_1, y_1\right)\\ & =& 3.1+\left(0.1\right)f\left(0.1, 3.1\right)\\ & =& 3.1+\left(0.1\right)\left(0.9\right)\\ & =& 3.1+0.09\\ \mathit{y}\left(0.2\right)& \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{19}\end{array}$

For $\mathbf{∆}\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$ we have,
$\begin{array}{rcl}{y}_1& =& y_0+∆t f\left(t_0, y_0\right)\\ & =& 3+\left(0.05\right)f\left(0,3\right)\\ & =& 3+\left(0.05\right)\left(1\right)\\ & =& 3+0.05\\ y_1& =& 3.05\end{array}$

$\begin{array}{rcl}{y}_2& =& y_1+∆t f\left(t_1, y_1\right)\\ & =& 3.05+\left(0.05\right)f\left(0.05,3.05\right)\\ & =& 3.05+\left(0.05\right)\left(0.95\right)\\ & =& 3.05+0.0475\\ y_2& =& 3.0975\end{array}$

$\begin{array}{rcl}{y}_3& =& y_2+∆t f\left(t_2, y_2\right)\\ & =& 3.0975+\left(0.05\right)f\left(0.1,3.0975\right)\\ & =& 3.0975+0.05\left(0.9025\right)\\ & =& 3.0975+0.0451\\ y_3& =& 3.1426\end{array}$

$\begin{array}{rcl}{y}_4& =& y_3+∆t f\left(t_3, y_3\right)\\ & =& 3.1426+\left(0.05\right)f\left(0.15, 3.1426\right)\\ & =& 3.1426+\left(0.05\right)\left(0.8574\right)\\ & =& 3.1426+0.0429\\ \mathit{y}\left(0.2\right)& \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{1855}\end{array}$

For $\mathbf{∆}\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{025}$ we have,
$\begin{array}{rcl}{y}_1& =& y_0+∆t f\left(t_0, y_0\right)\\ & =& 3+\left(0.025\right)f\left(0,3\right)\\ & =& 3+\left(0.025\right)\left(1\right)\\ & =& 3+0.025\\ y_1& =& 3.025\end{array}$

$\begin{array}{rcl}{y}_2& =& y_1+∆t f\left(t_1, y_1\right)\\ & =& 3.025+\left(0.025\right)f\left(0.025,3.025\right)\\ & =& 3.025+\left(0.025\right)\left(0.975\right)\\ & =& 3.025+0.0244\\ y_2& =& 3.0494\end{array}$

$\begin{array}{rcl}{y}_3& =& y_2+∆t f\left(t_2, y_2\right)\\ & =& 3.0494+\left(0.025\right)f\left(0.05,3.0494\right)\\ & =& 3.0494+0.025\left(0.9506\right)\\ & =& 3.0494+0.0238\\ y_3& =& 3.0731\end{array}$

$\begin{array}{rcl}{y}_4& =& y_3+∆t f\left(t_3, y_3\right)\\ & =& 3.0731+\left(0.025\right)f\left(0.075,3.0731\right)\\ & =& 3.0731+0.025\left(0.9269\right)\\ & =& 3.0731+0.0232\\ y_4& =& 3.0963\end{array}$

$\begin{array}{rcl}{y}_5& =& y_4+∆t f\left(t_4, y_4\right)\\ & =& 3.0963+\left(0.025\right)f\left(0.1, 3.0963\right)\\ & =& 3.0963+0.025\left(0.9037\right)\\ & =& 3.0963+0.0226\\ y_5& =& 3.1189\end{array}$

$\begin{array}{rcl}{y}_6& =& y_5+∆t f\left(t_5, y_5\right)\\ & =& 3.1189+\left(0.025\right)f\left(0.125, 3.1189\right)\\ & =& 3.1189+0.025\left(0.8811\right)\\ & =& 3.1189+0.022\\ y_6& =& 3.1409\end{array}$

$\begin{array}{rcl}{y}_7& =& y_6+∆t f\left(t_6, y_6\right)\\ & =& 3.1409+\left(0.025\right)f\left(0.15, 3.1409\right)\\ & =& 3.1409+0.025\left(0.8591\right)\\ & =& 3.1409+0.0215\\ y_7& =& 3.1624\end{array}$

$\begin{array}{rcl}{y}_8& =& y_7+∆t f\left(t_7, y_7\right)\\ & =& 3.1624+\left(0.025\right)f\left(0.175, 3.1624\right)\\ & =& 3.1624+\left(0.025\right)\left(0.8376\right)\\ & =& 3.1624+0.0209\\ \mathit{y}\left(0.2\right)& \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{1833}\end{array}$

Now, calculating $y\left(0.4\right)$:
For $\mathbf{∆}\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}$ we have,
$\begin{array}{rcl}{y}_1& =& y_0+∆t f\left(t_0, y_0\right)\\ & =& 3+\left(0.2\right)f\left(0,3\right)\\ & =& 3+\left(0.2\right)\left(1\right)\\ & =& 3+0.2\\ y_1& =& 3.2\end{array}$

$\begin{array}{rcl}{y}_2& =& y_1+∆t f\left(t_1, y_1\right)\\ & =& 3.2+\left(0.2\right)f\left(0.2,3.2\right)\\ & =& 3.2+\left(0.2\right)\left(0.8\right)\\ & =& 3.2+0.16\\ \mathit{y}\left(0.4\right)& \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{36}\end{array}$

For $\mathbf{∆}\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}$ we have,
$\begin{array}{rcl}{y}_1& =& y_0+∆t f\left(t_0, y_0\right)\\ & =& 3+\left(0.1\right)f\left(0,3\right)\\ & =& 3+\left(0.1\right)\left(1\right)\\ & =& 3+0.1\\ y_1& =& 3.1\end{array}$

$\begin{array}{rcl}{y}_2& =& y_1+∆t f\left(t_1, y_1\right)\\ & =& 3.1+\left(0.1\right)f\left(0.1, 3.1\right)\\ & =& 3.1+\left(0.1\right)\left(0.9\right)\\ & =& 3.1+0.09\\ y_2& =& 3.19\end{array}$

$\begin{array}{rcl}{y}_3& =& y_2+∆t f\left(t_2, y_2\right)\\ & =& 3.19+\left(0.1\right)f\left(0.2, 3.19\right)\\ & =& 3.19+\left(0.1\right)\left(0.81\right)\\ & =& 3.19+0.081\\ y_3& =& 3.271\end{array}$

$\begin{array}{rcl}{y}_4& =& y_1+∆t f\left(t_1, y_1\right)\\ & =& 3.271+\left(0.1\right)f\left(0.3, 3.271\right)\\ & =& 3.271+\left(0.1\right)\left(0.729\right)\\ & =& 3.271+0.0729\\ \mathit{y}\left(0.4\right)& \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{3439}\end{array}$ 

For $\mathbf{∆}\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$ we have,
$\begin{array}{rcl}{y}_1& =& y_0+∆t f\left(t_0, y_0\right)\\ & =& 3+\left(0.05\right)f\left(0,3\right)\\ & =& 3+\left(0.05\right)\left(1\right)\\ & =& 3+0.05\\ y_1& =& 3.05\end{array}$

$\begin{array}{rcl}{y}_2& =& y_1+∆t f\left(t_1, y_1\right)\\ & =& 3.05+\left(0.05\right)f\left(0.05,3.05\right)\\ & =& 3.05+\left(0.05\right)\left(0.95\right)\\ & =& 3.05+0.0475\\ y_2& =& 3.0975\end{array}$

$\begin{array}{rcl}{y}_3& =& y_2+∆t f\left(t_2, y_2\right)\\ & =& 3.0975+\left(0.05\right)f\left(0.1,3.0975\right)\\ & =& 3.0975+0.05\left(0.9025\right)\\ & =& 3.0975+0.0451\\ y_3& =& 3.1426\end{array}$

$\begin{array}{rcl}{y}_4& =& y_3+∆t f\left(t_3, y_3\right)\\ & =& 3.1426+\left(0.05\right)f\left(0.15,3.1426\right)\\ & =& 3.1426+0.05\left(0.8574\right)\\ & =& 3.1426+0.0429\\ y_4& =& 3.1855\end{array}$

$\begin{array}{rcl}{y}_5& =& y_4+∆t f\left(t_4, y_4\right)\\ & =& 3.1855+\left(0.05\right)f\left(0.2, 3.1855\right)\\ & =& 3.1855+0.05\left(0.8145\right)\\ & =& 3.1855+0.0407\\ y_5& =& 3.2262\end{array}$

$\begin{array}{rcl}{y}_6& =& y_5+∆t f\left(t_5, y_5\right)\\ & =& 3.2262+\left(0.05\right)f\left(0.25, 3.2262\right)\\ & =& 3.2262+0.05\left(0.7738\right)\\ & =& 3.2262+0.0387\\ y_6& =& 3.2649\end{array}$

$\begin{array}{rcl}{y}_7& =& y_6+∆t f\left(t_6, y_6\right)\\ & =& 3.2649+\left(0.05\right)f\left(0.3, 3.2649\right)\\ & =& 3.2649+0.05\left(0.7351\right)\\ & =& 3.2649+0.0368\\ y_7& =& 3.3017\end{array}$

$\begin{array}{rcl}{y}_8& =& y_7+∆t f\left(t_7, y_7\right)\\ & =& 3.3017+\left(0.05\right)f\left(0.35, 3.3017\right)\\ & =& 3.3017+0.05\left(0.6983\right)\\ & =& 3.3017+0.0349\\ \mathit{y}\left(0.4\right)& \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{3366}\end{array}$

For $\mathbf{∆}\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{025}$ we have,
$\begin{array}{rcl}{y}_1& =& y_0+∆t f\left(t_0, y_0\right)\\ & =& 3+\left(0.025\right)f\left(0,3\right)\\ & =& 3+\left(0.025\right)\left(1\right)\\ & =& 3+0.025\\ y_1& =& 3.025\end{array}$

$\begin{array}{rcl}{y}_2& =& y_1+∆t f\left(t_1, y_1\right)\\ & =& 3.025+\left(0.025\right)f\left(0.025,3.025\right)\\ & =& 3.025+\left(0.025\right)\left(0.975\right)\\ & =& 3.025+0.0244\\ y_2& =& 3.0494\end{array}$

$\begin{array}{rcl}{y}_3& =& y_2+∆t f\left(t_2, y_2\right)\\ & =& 3.0494+\left(0.025\right)f\left(0.05,3.0494\right)\\ & =& 3.0494+0.025\left(0.9506\right)\\ & =& 3.0494+0.0238\\ y_3& =& 3.0731\end{array}$

$\begin{array}{rcl}{y}_4& =& y_3+∆t f\left(t_3, y_3\right)\\ & =& 3.0731+\left(0.025\right)f\left(0.075,3.0731\right)\\ & =& 3.0731+0.025\left(0.9269\right)\\ & =& 3.0731+0.0232\\ y_4& =& 3.0963\end{array}$

$\begin{array}{rcl}{y}_5& =& y_4+∆t f\left(t_4, y_4\right)\\ & =& 3.0963+\left(0.025\right)f\left(0.1, 3.0963\right)\\ & =& 3.0963+0.025\left(0.9037\right)\\ & =& 3.0963+0.0226\\ y_5& =& 3.1189\end{array}$

$\begin{array}{rcl}{y}_6& =& y_5+∆t f\left(t_5, y_5\right)\\ & =& 3.1189+\left(0.025\right)f\left(0.125, 3.1189\right)\\ & =& 3.1189+0.025\left(0.8811\right)\\ & =& 3.1189+0.022\\ y_6& =& 3.1409\end{array}$

$\begin{array}{rcl}{y}_7& =& y_6+∆t f\left(t_6, y_6\right)\\ & =& 3.1409+\left(0.025\right)f\left(0.15, 3.1409\right)\\ & =& 3.1409+0.025\left(0.8591\right)\\ & =& 3.1409+0.0215\\ y_7& =& 3.1624\end{array}$

$\begin{array}{rcl}{y}_8& =& y_7+∆t f\left(t_7, y_7\right)\\ & =& 3.1624+\left(0.025\right)f\left(0.175, 3.1624\right)\\ & =& 3.1624+\left(0.025\right)\left(0.8376\right)\\ & =& 3.1624+0.0209\\ y_8& =& 3.1833\end{array}$

$\begin{array}{rcl}{y}_9& =& y_8+∆t f\left(t_8, y_8\right)\\ & =& 3.1833+\left(0.025\right)f\left(0.2, 3.1833\right)\\ & =& 3.1833+\left(0.025\right)\left(0.8167\right)\\ & =& 3.1833+0.0204\\ y_9& =& 3.2038\end{array}$

$\begin{array}{rcl}{y}_{10}& =& y_9+∆t f\left(t_9, y_9\right)\\ & =& 3.2038+\left(0.025\right)f\left(0.225, 3.2038\right)\\ & =& 3.2038+\left(0.025\right)\left(0.7962\right)\\ & =& 3.2038+0.0199\\ y_{10}& =& 3.2237\end{array}$

$\begin{array}{rcl}{y}_{11}& =& y_{10}+∆t f\left(t_{10}, y_{10}\right)\\ & =& 3.2237+\left(0.025\right)f\left(0.25, 3.2237\right)\\ & =& 3.2237+\left(0.025\right)\left(0.7763\right)\\ & =& 3.2237+0.0194\\ y_{11}& =& 3.2431\end{array}$

$\begin{array}{rcl}{y}_{12}& =& y_{11}+∆t f\left(t_{11}, y_{11}\right)\\ & =& 3.2431+\left(0.025\right)f\left(0.275, 3.2431\right)\\ & =& 3.2431+\left(0.025\right)\left(0.7569\right)\\ & =& 3.2431+0.0189\\ y_{12}& =& 3.262\end{array}$

$\begin{array}{rcl}{y}_{13}& =& y_{12}+∆t f\left(t_{12}, y_{12}\right)\\ & =& 3.262+\left(0.025\right)f\left(0.3, 3.262\right)\\ & =& 3.262+\left(0.025\right)\left(0.738\right)\\ & =& 3.262+0.0184\\ y_{13}& =& 3.2805\end{array}$

$\begin{array}{rcl}{y}_{14}& =& y_{13}+∆t f\left(t_{13}, y_{13}\right)\\ & =& 3.2805+\left(0.025\right)f\left(0.325, 3.2805\right)\\ & =& 3.2805+\left(0.025\right)\left(0.7195\right)\\ & =& 3.2805+0.018\\ y_{14}& =& 3.2984\end{array}$

$\begin{array}{rcl}{y}_{15}& =& y_{14}+∆t f\left(t_{14}, y_{14}\right)\\ & =& 3.2984+\left(0.025\right)f\left(0.35, 3.2984\right)\\ & =& 3.2984+\left(0.025\right)\left(0.7016\right)\\ & =& 3.2984+0.0175\\ y_{15}& =& 3.316\end{array}$

$\begin{array}{rcl}{y}_{16}& =& y_{14}+∆t f\left(t_{14}, y_{14}\right)\\ & =& 3.316+\left(0.025\right)f\left(0.375, 3.316\right)\\ & =& 3.316+\left(0.025\right)\left(0.684\right)\\ & =& 3.316+0.0171\\ \mathit{y}\left(0.4\right)& \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{3331}\end{array}$