Differential equations, euler's method and bisection method

meddi83
Messages
3
Reaction score
0

Homework Statement



Hi guys, I have the following problem and I don't know how to start.

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

and

<br /> \frac{dy^2}{d^2x}=\frac{W}{T}\sqrt{1 + (\frac{dy}{dx})^2}<br />

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: yi+1 = yi + f(xi,yi)*h where f(x,y) = \frac{dy}{dx}



The Attempt at a Solution



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

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

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

My problem is, I don't 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 don't know which equation shall be used with the bisection method..

any help would be really appreciated. thank you
 
Physics news on Phys.org
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?
 
Sorry, typo mistake. It says that the slope, at x=0, is dy/dx=0
 
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

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

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) ) ?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top