Suppose that we use the Improved Euler's method to approximate the solution to the differential equation dy = x - 2.5y; da y(0.4) = 8. Let f(x, y) = x - 2.5y. We let zo = 0.4 and yo = 8 and pick a step size h = 0.25. The improved Euler method is the the following algorithm. From (x, yn), our approximation to the solution of the differential equation at the n-th stage, we find the next stage by computing the x-step n+1 = n + h, and then k₁, the slope at (x, yn). The predicted new value of the solution is Zn+1 = Yn + h·k₁. Then we find the slope at the predicted new point k₂ = f(xn+1, Zn+1) and get the corrected point by averaging slopes Complete the following table: k₁ -19.6 n In Yn 0 0.4 8 1 0.65 4.6625 20.9 2.8021 3 1.15 1.784 4 1.4 1.2463 Zn+1 3.1 -11.80625 1.9109 k₂ -7.1 -3.8773 -6.1051 1.2758 -2.0394 -3.31 0.96 -0.99 6.25 The exact solution can also be found for the linear equation. Write the answer as a function of a. y(x) = 21.7463 2.5x + Yn+1 = Yn + Thus the actual value of the function at the point x = 1.4 is y(1.4) 1.0567 h 2 (k₁ + k₂).
Suppose that we use the Improved Euler's method to approximate the solution to the differential equation dy = x - 2.5y; da y(0.4) = 8. Let f(x, y) = x - 2.5y. We let zo = 0.4 and yo = 8 and pick a step size h = 0.25. The improved Euler method is the the following algorithm. From (x, yn), our approximation to the solution of the differential equation at the n-th stage, we find the next stage by computing the x-step n+1 = n + h, and then k₁, the slope at (x, yn). The predicted new value of the solution is Zn+1 = Yn + h·k₁. Then we find the slope at the predicted new point k₂ = f(xn+1, Zn+1) and get the corrected point by averaging slopes Complete the following table: k₁ -19.6 n In Yn 0 0.4 8 1 0.65 4.6625 20.9 2.8021 3 1.15 1.784 4 1.4 1.2463 Zn+1 3.1 -11.80625 1.9109 k₂ -7.1 -3.8773 -6.1051 1.2758 -2.0394 -3.31 0.96 -0.99 6.25 The exact solution can also be found for the linear equation. Write the answer as a function of a. y(x) = 21.7463 2.5x + Yn+1 = Yn + Thus the actual value of the function at the point x = 1.4 is y(1.4) 1.0567 h 2 (k₁ + k₂).
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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Transcribed Image Text:Suppose that we use the Improved Euler's method to approximate the solution to the differential equation
dy
dx
Complete the following table:
n Xn
Yn
0 0.4
8
1 0.65 4.6625
2 0.9 2.8021
3 1.15
1.784
4 1.4 1.2463
Let f(x, y) = x - 2.5y.
We let x = 0.4 and yo = 8 and pick a step size h = 0.25. The improved Euler method is the the following algorithm. From (an, Yn), our approximation to the solution of the
differential equation at the n-th stage, we find the next stage by computing the x-step n+1 = n + h, and then k₁, the slope at (xn, Yn). The predicted new value of the solution is
Zn+1 = yn + h.k₁. Then we find the slope at the predicted new point k₂ = f(n+1, Zn+1) and get the corrected point by averaging slopes
k₁
- 19.6
-11.80625
-6.1051
-3.31
=
Zn+1
k₂
-7.1
3.1
1.9109
-3.8773
1.2758 -2.0394
0.96
-0.99
+
= 0
Thus the actual value of the function at the point x = 1.4 is
:
y(1.4) 1.0567
2.5y;
The exact solution can also be found for the linear equation. Write the answer as a function of a.
1
y(x) =
21.7463
2.5x
2.5 6.25
y(0.4) = 8.
h
Yn+1 = Yn + (k₁ + k₂).
2
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