Finding Approximate Solutions using the Runge-Kutta Method

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Homework Help Overview

The discussion revolves around using the Runge-Kutta method to approximate solutions for the initial-value problem y'+(x^2)y=sin xy, with the initial condition y(1)=pi and a step size of h=0.2. Participants are comparing their results with those from a textbook.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants share their calculations using the Runge-Kutta method, specifically RK2, and compare their results with those from the textbook. Questions arise regarding the correctness of the formula used and the potential for different methods affecting the results.

Discussion Status

There is an ongoing exploration of the results obtained by different participants, with some expressing confidence in their calculations while questioning the accuracy of the textbook answers. Multiple interpretations of the formula and method are being discussed, and some participants have confirmed agreement on their results.

Contextual Notes

Participants are considering the implications of step size on accuracy and whether the calculations should be performed in radians or degrees. There is also mention of different Runge-Kutta methods (RK2 and RK4) and their potential impact on the results.

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Homework Statement


Use the Runge-Kutta method to find approximate values of the solution of the initial-value problem y'+(x^2)y=sin xy, y(1)=pi; h=0.2 at the points xi=x0+ih, where x0 is the point where the initial condition is imposed and I=1, 2.

Homework Equations


yn+1=yn+hf(xn+1/2h, yn+(1/2)hf(xn, yn))
f(x, y)=sin xy-(x^2)y
h=0.2, x0=1, y0=pi.

The Attempt at a Solution


y1=pi+0.2f(1+0.1, pi+0.1f(1, pi))
=pi+0.2f(1.1, 2.83291)
=2.4669

y2=2.4669+0.2f(1.3, 2.4669+0.1f(1.2, 2.4669))
=2.4669+0.2f(1.3, 2.11683)
=1.76101

But the answer is:
y1=2.475605264, y2=1.825992433.
I got y1=2.4669 and y2=1.76101.
Which is the correct answer? Mine or the book's answer? If I'm wrong, please correct me.
 
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Math10 said:

Homework Statement


Use the Runge-Kutta method to find approximate values of the solution of the initial-value problem y'+(x^2)y=sin xy, y(1)=pi; h=0.2 at the points xi=x0+ih, where x0 is the point where the initial condition is imposed and I=1, 2.

Homework Equations


yn+1=yn+hf(xn+1/2h, yn+(1/2)hf(xn, yn))
Does that formula mean$$y_{n+1} = y_n + hf(x_n + \frac h 2,y_n + \frac h 2f(x_n,y_n))\text{?}$$
 
Using RK2 and OpenOffice Calc, I get y1=2.4636, y2 =1.8181, which are closer to your answers than to the book's.
However, if I cut the step size to 0.1 and, correspondingly, look at y2, y4 I get nearly the book answers: 2.4734, 1.8247.
This suggests the book is using a more accurate method than RK2 with step size 0.2 Maybe it uses RK4 with step size 0.2 - I haven't tried that.
 
@LCKurtz , yes, that's the right formula that I'm using.
 
LCKurtz said:
Does that formula mean$$y_{n+1} = y_n + hf(x_n + \frac h 2,y_n + \frac h 2f(x_n,y_n))\text{?}$$

Math10 said:
@LCKurtz , yes, that's the right formula that I'm using.

OK. I put that formula in Maple and I get: ##(x_1,y_1)= (1.2, 2.463635926),~(x_2,y_2) = (1.4, 1.818074242)## which agree with Haruspex's results.
 
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So which answer is right? The book's answer or yours?
 
By the way, do I need to set the mode to radians or degrees on my calc?
 
What's RK2 and RK4?
 
All I can tell you for sure, is that using the formula you gave us, both Haruspex and I get identical answers independently. I doubt we are both wrong. And, speaking for myself, I am assuming the formula you gave is the appropriate one. If it isn't, all bets are off.
 
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  • #10
Math10 said:
By the way, do I need to set the mode to radians or degrees on my calc?
Radians, of course!
 
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  • #12
LCKurtz said:
I am assuming the formula you gave is the appropriate one
It is RK2.
 
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  • #13
Thank you guys for the help!
 

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