Finding Approximate Solutions using the Runge-Kutta Method

In summary, the Runge-Kutta method is used to approximate 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. The book's solution uses a more accurate method than RK2 with step size 0.2, while my solution uses RK2 with step size 0.1.
  • #1
Math10
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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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  • #2
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$$y_{n+1} = y_n + hf(x_n + \frac h 2,y_n + \frac h 2f(x_n,y_n))\text{?}$$
 
  • #3
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.
 
  • #4
@LCKurtz , yes, that's the right formula that I'm using.
 
  • #5
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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  • #6
So which answer is right? The book's answer or yours?
 
  • #7
By the way, do I need to set the mode to radians or degrees on my calc?
 
  • #8
What's RK2 and RK4?
 
  • #9
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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