To 4 decimals each pls Apply the improved Euler method to approximate the solution on the interval [0, 0.5] with step size h = 0.1. Construct a table showing values of the approximate solution and the actual solution atthe points x = 0.1, 0.2, 0.3, 0.4, 0.5.y' =y-2x-4, y(0) = 3; y(x) = 6+2x-3exComplete the table below.(Round to four decimal places as needed.)XnActual, y (xn)0.10.20.30.40.5

Solution: Given IVP is y1=y-2x-4 , y0=3 x∘=0, y∘=3, h=0.1 , xn=0.5∴fx,y=y-2x-4Formula:…

Answered: Apply the improved Euler method to…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Question

To 4 decimals each pls

Apply the improved Euler method to approximate the solution on the interval [0, 0.5] with step size h = 0.1. Construct a table showing values of the approximate solution and the actual solution at
the points x = 0.1, 0.2, 0.3, 0.4, 0.5.
y' =y-2x-4, y(0) = 3; y(x) = 6+2x-3ex
Complete the table below.
(Round to four decimal places as needed.)
Xn
Actual, y (xn)
0.1
0.2
0.3
0.4
0.5

Expert Solution

Step 1

$S o l u t i o n : G i v e n I V P i s y^{1} = y - 2 x - 4, y (0) = 3 x_{\circ} = 0, y_{\circ} = 3, h = 0.1, x_{n} = 0.5 ∴ f (x, y) = y - 2 x - 4 F o r m u l a : y_{m + 1} = y_{m} + \frac{1}{2} h [f (x_{m}, y_{m}) + f (x_{m} + h, y_{m} + h f (x_{m}, y_{m}))] 1^{s t} s t e p f (x_{\circ}, y_{\circ}) = f (0, 3) = 1 f (x_{\circ} + h, y_{\circ} + h f (x_{\circ}, y_{\circ})) = f (0.1, 2.9) = - 1.3 y_{1} = y_{\circ} + \frac{1}{2} h [f (x_{\circ}, y_{\circ}) + f (x_{\circ} + h, y_{\circ} + h f (x_{\circ}, y_{\circ}))] \Rightarrow y_{1} = 3 + \frac{0.1}{2} (- 1 - 1.3) \Rightarrow y (0.1) = 2.885 2^{n d} s t e p f (x_{1}, y_{1}) = f (0.1, 2.885) = - 1.315 f (x_{1} + h, y_{1} + h f (x_{1}, y_{1})) = f (0.2, 2.7535) = - 1.6465 y_{2} = y_{1} + \frac{1}{2} h [f (x_{1}, y_{1}) + f (x_{1} + h, y_{1} + h f (x_{1}, y_{1}))] \Rightarrow y_{2} = 2.885 + \frac{0.1}{2} (- 1.315 - 1.6465) \Rightarrow y_{2} = 2.7369 \Rightarrow y (0.2) = 2.7369$