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How do I solve this ODE with numerical methods?

by Inigma
Tags: methods, numerical, solve
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Inigma
#1
Feb23-12, 10:17 PM
P: 4
I have to solve this ODE with numerical methods:
[itex](y^2 - 1)\frac{dy}{dx}=3y[/itex]

I have no initial conditions to solve it like you would normally do. I am hoping to use a numerical method (Euler... Runge Kutta) to approximate the solution. This is if I solve it using numerical methods right? So, I got to the following solution for x, by integrating both sides:
[itex]x=\frac{1}{6}y^2 + \frac{1}{3}lny + c[/itex] where c is the constant after integrating. With c in the way and no initial conditions, how do I then get to go ahead? am I approaching this the wrong way? your help would be greatly appreciated... thanks!
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JJacquelin
#2
Feb24-12, 03:20 AM
P: 756
Quote Quote by Inigma View Post
I have to solve this ODE with numerical methods:
[itex](y^2 - 1)\frac{dy}{dx}=3y[/itex]

I have no initial conditions to solve it like you would normally do. I am hoping to use a numerical method (Euler... Runge Kutta) to approximate the solution. This is if I solve it using numerical methods right? So, I got to the following solution for x, by integrating both sides:
[itex]x=\frac{1}{6}y + 3lny + c[/itex] where c is the constant after integrating. With c in the way and no initial conditions, how do I then get to go ahead? am I approaching this the wrong way? your help would be greatly appreciated... thanks!
Hi !

There is a small mistake in your equation : The coefficient of lny should be 1/3 instead of 3.
The analytical solving is possible, thanks to the Lambert's W function (in attachment).
For numerical solving, you can state an arbitrary initial condition and use any numerical method. This will permit to draw a first curve y(x).
All other curves, corresponding to different values of c, are just translated from the first curve. In fact, giving an initial condition is equivalent to give the translation value. This is because the ODE contains only y and y' but no x.
Attached Thumbnails
Lambert EDO.JPG  
meldraft
#3
Feb24-12, 06:41 AM
P: 280
Actually, unless I am missing something obvious, the correct solution for this equation is:

[tex]x=\frac{y^2}{6}-\frac{lny}{3}+c[/tex]

In order to come to this formulation, however, you have to assume that y cannot be zero.

Also, if you already have the analytical solution, then you no longer need the numerical method, unless of course you explicitly want to solve the equation numerically.

JJacquelin
#4
Feb24-12, 07:11 AM
P: 756
How do I solve this ODE with numerical methods?

Quote Quote by meldraft View Post
Actually, unless I am missing something obvious, the correct solution for this equation is:

[tex]x=\frac{y^2}{6}-\frac{lny}{3}+c[/tex]

In order to come to this formulation, however, you have to assume that y cannot be zero.

Also, if you already have the analytical solution, then you no longer need the numerical method, unless of course you explicitly want to solve the equation numerically.
From the explicit solution given in my preceeding post, the case y=0 doesn't exist.
We can see that if x tends to infinity, exp(-6x) tends to 0, then W(0) tends to 0 and y tends to 0. But tending to 0 doesn't mean equal 0. So, y=0 is a limit case in fact, never reached, in so far as x is finit.
Inigma
#5
Feb24-12, 08:06 AM
P: 4
meldraft: Your answer is correct and I wish to continue beyond the solution you stated. In this case by using a numerical method to plot the appromimate solution.

JJacquelin: Thank you, and I have found a lot of sense from wha you told me. I have also plotted the curves and have a rough idea what to expect in terms of convergence and infinities. I just struggle with the (my preferred method due to higher accuracy) the Runge Kutta method, since there is no x in the differential. I do not know what to do with say [itex]k1=f(x0,y0)=\frac{3y}{y^2 -1}[/itex] would this method work when there is no x in the differential? I guess not...
JJacquelin
#6
Feb24-12, 10:31 AM
P: 756
You can state x0 any value you want. As I already say, you will obtain one first solution i.e. one first curve y(x). All the other solutions are translated from the first one. Alltogether, the result is globally the same : it doesn't mater which is the first solution to start with.


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