Consider the following first-order ODE: y+x³ from x=0.5 tox-2.1 1with y(0.5) --1 Solve using classical fourth-order Runge-Kutta method. (h= 0.4) 71ex The analytical solution of the ODE is: y = -x³-3x² - 6x-6+05 In each part, calculate the error between the true solution and the numerical solution at the points where the numerical solution is determined. X 0.5 0.9 1.3 1.7 y -1 -1.35 -1.6 -1.36 error 0 0.0309 0.2831 1.042 X 0.5 0.9 1.3 1.7 y -1 -1.33 -1.38 -0.52 error 0 0.0137 0.0665 0.202 X 0.5 0.9 1.3 1.7 y -1 -1.32 -1.32 -0.32 error 0 0 0.0002 0.000 None of the choices

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Consider the following first-order ODE:
dy
dx
y+x³ from x = 0.5 to x = 2.1 1with y(0.5) = -1
Solve using classical fourth-order Runge-Kutta method. (h= 0.4)
71ex
The analytical solution of the ODE is: y = -x³-3x² - 6x-6+05. In each part, calculate the error between the true solution
and the numerical solution at the points where the numerical solution is determined.
X
0.5
0.9
1.3
1.7
y
-1
-1.35
-1.6
-1.36
error
0
0.0309
0.2831
1.042
X
0.5
0.9
1.3
1.7
y
-1
-1.33
-1.38
-0.52
error
0
0.0137
0.0665
0.202
X
0.5
0.9
1.3
1.7
y
-1
-1.32
-1.32
-0.32
error
0
0
0.0002
0.000
None of the choices
Transcribed Image Text:Consider the following first-order ODE: dy dx y+x³ from x = 0.5 to x = 2.1 1with y(0.5) = -1 Solve using classical fourth-order Runge-Kutta method. (h= 0.4) 71ex The analytical solution of the ODE is: y = -x³-3x² - 6x-6+05. In each part, calculate the error between the true solution and the numerical solution at the points where the numerical solution is determined. X 0.5 0.9 1.3 1.7 y -1 -1.35 -1.6 -1.36 error 0 0.0309 0.2831 1.042 X 0.5 0.9 1.3 1.7 y -1 -1.33 -1.38 -0.52 error 0 0.0137 0.0665 0.202 X 0.5 0.9 1.3 1.7 y -1 -1.32 -1.32 -0.32 error 0 0 0.0002 0.000 None of the choices
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