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Differential equations, euler's method and bisection method

  • Thread starter meddi83
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Homework Statement



Hi guys, I have the following problem and I dont know how to start.

I am given that W = 0.5, X(0) = 0, Iab = 20m and ha = 5m [tex]\frac{dy}{dx}=0[/tex]

and

[tex]
\frac{dy^2}{d^2x}=\frac{W}{T}\sqrt{1 + (\frac{dy}{dx})^2}
[/tex]

I am told to convert the 2nd order ODE to two 1st order ODEs at first. Then I am asked to choose an arbitrary value of T and integrate the system of equations using Euler's method from 0-Iab, where Iab=20, with 3 different step sizes. Then, I have to select the step size that gives an accurate value of y(Iab) (i.e. approximate relative error less than 0.1% if comparing with ha = 5).. Then I am asked to chose two values of T such that one gives y(Iab) smaller than ha and the other one gives y(Iab) bigger than ha and use those two values with the bisection method


Homework Equations



Euler's method: [tex]yi+1 = yi + f(xi,yi)*h[/tex] where [tex]f(x,y) = \frac{dy}{dx}[/tex]



The Attempt at a Solution



At first I broke it to two 1st order ODES:

[tex]\frac{dy}{dx}=g[/tex] and [tex]\frac{dg}{dx}=\frac{W}{T}\sqrt{1 + g^2}[/tex]

Then, for Euler's method: [tex]yi+1 = yi + gi*h[/tex] and [tex]gi+1 = gi + (\frac{W}{T}\sqrt{1+gi^2})*h[/tex]

My problem is, I dont know where to start in order to find y(Iab) (because the problem is asking for an accurate value of y(Iab) )... and what step size (h) to use. .. I also dont know which equation shall be used with the bisection method..

any help would be really appreciated. thank you
 

Answers and Replies

  • #2
HallsofIvy
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Homework Helper
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I would suggest you go back and reread the problem. You say that X(0)= 0 but your differential equation has y as a function of x. What does X(0) mean?
 
  • #3
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Sorry, typo mistake. It says that the slope, at x=0, is dy/dx=0
 
  • #4
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and from the Picture I have, the graph, when x=0 -> y=0.

So I suppose for euler's method, suppose I choose a step size of ie, h=5, then it would be

[tex]y(5) = y(0) + g(0)*h[/tex] and [tex]g(5) = g(0) + (\frac{W}{T}\sqrt{1+g(0)^2})*h[/tex]

Is that correct ? And since I have to integrate from 0-20, I suppose I'll have to do y(10), y(15) and y(20) aswell, right ? (Same for g(10),g(15),g(20) ) ?
 

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