Answered: Consider the following initial value…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Consider the following initial value problem.
a. Find the approximations to y(0.2) and y(0.4) using Euler’s method
with 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 at
t = 0.2 and t = 0.4?

y '(t) = 4 - y, y(0) = 3; y(t) = 4 - e^-t

Expert Solution

Step 1

Given:

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

To Do:

(a) Find the approximations to $y (0.2) and y (0.4)$ 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.

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

Step 2

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

Let $f (t, y) = 4 - y$ .

Euler's formula is: $y_{i + 1} = y_{i} + ∆ t f (t_{i}, y_{i})$ .

Calculating $y (0.2)$ :

Now, for $∆ t = 0.2$ we have,
$\begin{array}{rcl} y_{1} & = & y_{0} + ∆ t f (t_{0}, y_{0}) \\ = & 3 + (0.2) f (0, 3) \\ = & 3 + (0.2) (1) \\ = & 3 + 0.2 \\ y (0.2) & = & 3.2 \end{array}$

For $∆ t = 0.1$ we have,
$\begin{array}{rcl} y_{1} & = & y_{0} + ∆ t f (t_{0}, y_{0}) \\ = & 3 + (0.1) f (0, 3) \\ = & 3 + (0.1) (1) \\ = & 3 + 0.1 \\ y_{1} & = & 3.1 \end{array}$

$\begin{array}{rcl} y_{2} & = & y_{1} + ∆ t f (t_{1}, y_{1}) \\ = & 3.1 + (0.1) f (0.1, 3.1) \\ = & 3.1 + (0.1) (0.9) \\ = & 3.1 + 0.09 \\ y (0.2) & = & 3.19 \end{array}$

For $∆ t = 0.05$ we have,
$\begin{array}{rcl} y_{1} & = & y_{0} + ∆ t f (t_{0}, y_{0}) \\ = & 3 + (0.05) f (0, 3) \\ = & 3 + (0.05) (1) \\ = & 3 + 0.05 \\ y_{1} & = & 3.05 \end{array}$

$\begin{array}{rcl} y_{2} & = & y_{1} + ∆ t f (t_{1}, y_{1}) \\ = & 3.05 + (0.05) f (0.05, 3.05) \\ = & 3.05 + (0.05) (0.95) \\ = & 3.05 + 0.0475 \\ y_{2} & = & 3.0975 \end{array}$

$\begin{array}{rcl} y_{3} & = & y_{2} + ∆ t f (t_{2}, y_{2}) \\ = & 3.0975 + (0.05) f (0.1, 3.0975) \\ = & 3.0975 + 0.05 (0.9025) \\ = & 3.0975 + 0.0451 \\ y_{3} & = & 3.1426 \end{array}$

$\begin{array}{rcl} y_{4} & = & y_{3} + ∆ t f (t_{3}, y_{3}) \\ = & 3.1426 + (0.05) f (0.15, 3.1426) \\ = & 3.1426 + (0.05) (0.8574) \\ = & 3.1426 + 0.0429 \\ y (0.2) & = & 3.1855 \end{array}$

For $∆ t = 0.025$ we have,
$\begin{array}{rcl} y_{1} & = & y_{0} + ∆ t f (t_{0}, y_{0}) \\ = & 3 + (0.025) f (0, 3) \\ = & 3 + (0.025) (1) \\ = & 3 + 0.025 \\ y_{1} & = & 3.025 \end{array}$

$\begin{array}{rcl} y_{2} & = & y_{1} + ∆ t f (t_{1}, y_{1}) \\ = & 3.025 + (0.025) f (0.025, 3.025) \\ = & 3.025 + (0.025) (0.975) \\ = & 3.025 + 0.0244 \\ y_{2} & = & 3.0494 \end{array}$

$\begin{array}{rcl} y_{3} & = & y_{2} + ∆ t f (t_{2}, y_{2}) \\ = & 3.0494 + (0.025) f (0.05, 3.0494) \\ = & 3.0494 + 0.025 (0.9506) \\ = & 3.0494 + 0.0238 \\ y_{3} & = & 3.0731 \end{array}$

$\begin{array}{rcl} y_{4} & = & y_{3} + ∆ t f (t_{3}, y_{3}) \\ = & 3.0731 + (0.025) f (0.075, 3.0731) \\ = & 3.0731 + 0.025 (0.9269) \\ = & 3.0731 + 0.0232 \\ y_{4} & = & 3.0963 \end{array}$

$\begin{array}{rcl} y_{5} & = & y_{4} + ∆ t f (t_{4}, y_{4}) \\ = & 3.0963 + (0.025) f (0.1, 3.0963) \\ = & 3.0963 + 0.025 (0.9037) \\ = & 3.0963 + 0.0226 \\ y_{5} & = & 3.1189 \end{array}$

$\begin{array}{rcl} y_{6} & = & y_{5} + ∆ t f (t_{5}, y_{5}) \\ = & 3.1189 + (0.025) f (0.125, 3.1189) \\ = & 3.1189 + 0.025 (0.8811) \\ = & 3.1189 + 0.022 \\ y_{6} & = & 3.1409 \end{array}$

$\begin{array}{rcl} y_{7} & = & y_{6} + ∆ t f (t_{6}, y_{6}) \\ = & 3.1409 + (0.025) f (0.15, 3.1409) \\ = & 3.1409 + 0.025 (0.8591) \\ = & 3.1409 + 0.0215 \\ y_{7} & = & 3.1624 \end{array}$

$\begin{array}{rcl} y_{8} & = & y_{7} + ∆ t f (t_{7}, y_{7}) \\ = & 3.1624 + (0.025) f (0.175, 3.1624) \\ = & 3.1624 + (0.025) (0.8376) \\ = & 3.1624 + 0.0209 \\ y (0.2) & = & 3.1833 \end{array}$

Step 3

Now, calculating $y (0.4)$ :
For $∆ t = 0.2$ we have,
$\begin{array}{rcl} y_{1} & = & y_{0} + ∆ t f (t_{0}, y_{0}) \\ = & 3 + (0.2) f (0, 3) \\ = & 3 + (0.2) (1) \\ = & 3 + 0.2 \\ y_{1} & = & 3.2 \end{array}$

$\begin{array}{rcl} y_{2} & = & y_{1} + ∆ t f (t_{1}, y_{1}) \\ = & 3.2 + (0.2) f (0.2, 3.2) \\ = & 3.2 + (0.2) (0.8) \\ = & 3.2 + 0.16 \\ y (0.4) & = & 3.36 \end{array}$

For $∆ t = 0.1$ we have,
$\begin{array}{rcl} y_{1} & = & y_{0} + ∆ t f (t_{0}, y_{0}) \\ = & 3 + (0.1) f (0, 3) \\ = & 3 + (0.1) (1) \\ = & 3 + 0.1 \\ y_{1} & = & 3.1 \end{array}$

$\begin{array}{rcl} y_{2} & = & y_{1} + ∆ t f (t_{1}, y_{1}) \\ = & 3.1 + (0.1) f (0.1, 3.1) \\ = & 3.1 + (0.1) (0.9) \\ = & 3.1 + 0.09 \\ y_{2} & = & 3.19 \end{array}$

$\begin{array}{rcl} y_{3} & = & y_{2} + ∆ t f (t_{2}, y_{2}) \\ = & 3.19 + (0.1) f (0.2, 3.19) \\ = & 3.19 + (0.1) (0.81) \\ = & 3.19 + 0.081 \\ y_{3} & = & 3.271 \end{array}$

$\begin{array}{rcl} y_{4} & = & y_{1} + ∆ t f (t_{1}, y_{1}) \\ = & 3.271 + (0.1) f (0.3, 3.271) \\ = & 3.271 + (0.1) (0.729) \\ = & 3.271 + 0.0729 \\ y (0.4) & = & 3.3439 \end{array}$

For $∆ t = 0.05$ we have,
$\begin{array}{rcl} y_{1} & = & y_{0} + ∆ t f (t_{0}, y_{0}) \\ = & 3 + (0.05) f (0, 3) \\ = & 3 + (0.05) (1) \\ = & 3 + 0.05 \\ y_{1} & = & 3.05 \end{array}$

$\begin{array}{rcl} y_{2} & = & y_{1} + ∆ t f (t_{1}, y_{1}) \\ = & 3.05 + (0.05) f (0.05, 3.05) \\ = & 3.05 + (0.05) (0.95) \\ = & 3.05 + 0.0475 \\ y_{2} & = & 3.0975 \end{array}$

$\begin{array}{rcl} y_{3} & = & y_{2} + ∆ t f (t_{2}, y_{2}) \\ = & 3.0975 + (0.05) f (0.1, 3.0975) \\ = & 3.0975 + 0.05 (0.9025) \\ = & 3.0975 + 0.0451 \\ y_{3} & = & 3.1426 \end{array}$

$\begin{array}{rcl} y_{4} & = & y_{3} + ∆ t f (t_{3}, y_{3}) \\ = & 3.1426 + (0.05) f (0.15, 3.1426) \\ = & 3.1426 + 0.05 (0.8574) \\ = & 3.1426 + 0.0429 \\ y_{4} & = & 3.1855 \end{array}$

$\begin{array}{rcl} y_{5} & = & y_{4} + ∆ t f (t_{4}, y_{4}) \\ = & 3.1855 + (0.05) f (0.2, 3.1855) \\ = & 3.1855 + 0.05 (0.8145) \\ = & 3.1855 + 0.0407 \\ y_{5} & = & 3.2262 \end{array}$

$\begin{array}{rcl} y_{6} & = & y_{5} + ∆ t f (t_{5}, y_{5}) \\ = & 3.2262 + (0.05) f (0.25, 3.2262) \\ = & 3.2262 + 0.05 (0.7738) \\ = & 3.2262 + 0.0387 \\ y_{6} & = & 3.2649 \end{array}$

$\begin{array}{rcl} y_{7} & = & y_{6} + ∆ t f (t_{6}, y_{6}) \\ = & 3.2649 + (0.05) f (0.3, 3.2649) \\ = & 3.2649 + 0.05 (0.7351) \\ = & 3.2649 + 0.0368 \\ y_{7} & = & 3.3017 \end{array}$

$\begin{array}{rcl} y_{8} & = & y_{7} + ∆ t f (t_{7}, y_{7}) \\ = & 3.3017 + (0.05) f (0.35, 3.3017) \\ = & 3.3017 + 0.05 (0.6983) \\ = & 3.3017 + 0.0349 \\ y (0.4) & = & 3.3366 \end{array}$

For $∆ t = 0.025$ we have,
$\begin{array}{rcl} y_{1} & = & y_{0} + ∆ t f (t_{0}, y_{0}) \\ = & 3 + (0.025) f (0, 3) \\ = & 3 + (0.025) (1) \\ = & 3 + 0.025 \\ y_{1} & = & 3.025 \end{array}$

$\begin{array}{rcl} y_{2} & = & y_{1} + ∆ t f (t_{1}, y_{1}) \\ = & 3.025 + (0.025) f (0.025, 3.025) \\ = & 3.025 + (0.025) (0.975) \\ = & 3.025 + 0.0244 \\ y_{2} & = & 3.0494 \end{array}$

$\begin{array}{rcl} y_{3} & = & y_{2} + ∆ t f (t_{2}, y_{2}) \\ = & 3.0494 + (0.025) f (0.05, 3.0494) \\ = & 3.0494 + 0.025 (0.9506) \\ = & 3.0494 + 0.0238 \\ y_{3} & = & 3.0731 \end{array}$

$\begin{array}{rcl} y_{4} & = & y_{3} + ∆ t f (t_{3}, y_{3}) \\ = & 3.0731 + (0.025) f (0.075, 3.0731) \\ = & 3.0731 + 0.025 (0.9269) \\ = & 3.0731 + 0.0232 \\ y_{4} & = & 3.0963 \end{array}$

$\begin{array}{rcl} y_{5} & = & y_{4} + ∆ t f (t_{4}, y_{4}) \\ = & 3.0963 + (0.025) f (0.1, 3.0963) \\ = & 3.0963 + 0.025 (0.9037) \\ = & 3.0963 + 0.0226 \\ y_{5} & = & 3.1189 \end{array}$

$\begin{array}{rcl} y_{6} & = & y_{5} + ∆ t f (t_{5}, y_{5}) \\ = & 3.1189 + (0.025) f (0.125, 3.1189) \\ = & 3.1189 + 0.025 (0.8811) \\ = & 3.1189 + 0.022 \\ y_{6} & = & 3.1409 \end{array}$

$\begin{array}{rcl} y_{7} & = & y_{6} + ∆ t f (t_{6}, y_{6}) \\ = & 3.1409 + (0.025) f (0.15, 3.1409) \\ = & 3.1409 + 0.025 (0.8591) \\ = & 3.1409 + 0.0215 \\ y_{7} & = & 3.1624 \end{array}$

$\begin{array}{rcl} y_{8} & = & y_{7} + ∆ t f (t_{7}, y_{7}) \\ = & 3.1624 + (0.025) f (0.175, 3.1624) \\ = & 3.1624 + (0.025) (0.8376) \\ = & 3.1624 + 0.0209 \\ y_{8} & = & 3.1833 \end{array}$

$\begin{array}{rcl} y_{9} & = & y_{8} + ∆ t f (t_{8}, y_{8}) \\ = & 3.1833 + (0.025) f (0.2, 3.1833) \\ = & 3.1833 + (0.025) (0.8167) \\ = & 3.1833 + 0.0204 \\ y_{9} & = & 3.2038 \end{array}$

$\begin{array}{rcl} y_{10} & = & y_{9} + ∆ t f (t_{9}, y_{9}) \\ = & 3.2038 + (0.025) f (0.225, 3.2038) \\ = & 3.2038 + (0.025) (0.7962) \\ = & 3.2038 + 0.0199 \\ y_{10} & = & 3.2237 \end{array}$

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

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

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

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

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

$\begin{array}{rcl} y_{16} & = & y_{14} + ∆ t f (t_{14}, y_{14}) \\ = & 3.316 + (0.025) f (0.375, 3.316) \\ = & 3.316 + (0.025) (0.684) \\ = & 3.316 + 0.0171 \\ y (0.4) & = & 3.3331 \end{array}$

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