Here's the ode i set up,

assuming i set it up properly, how would i deal with the x in the denominator?

Substituting y= c-x gives this:

the y in the denominator has me mixed up

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- March 12th 2010, 03:53 AMphycdudeUnfamiliar ode
Here's the ode i set up,

assuming i set it up properly, how would i deal with the x in the denominator?

Substituting y= c-x gives this:

the y in the denominator has me mixed up - March 12th 2010, 04:19 AMcrymorenoobs
Why not just substitute y= (1/c-x) instead, giving:

(-d2y)/(dt2) + ky = 0 - March 12th 2010, 04:46 AMHallsofIvy
- March 12th 2010, 05:06 AMshawsend
That's a case of independent variable missing. So if and I let , then I get the new equation:

with variables separated. That's ok if we get a . Just take it's square root and solve both first-order odes resulting from the root.

Guess it's kinda' messy though and you'll have to invert the expression and get it in terms of an inverse function. - March 12th 2010, 05:30 AMchisigma
The second order DE has the form...

(1)

... where...

(2)

The approach to solve (1) is the following: because is...

(3)

... the (1) becomes...

(4)

... so that we have a first order DE in and . Integrating (4) we obtain...

(5)

... where is a primitive of . Separating variables again in (5) we obtain...

(6)

... so that is...

(7)

Now what remains to do is compute a primitive of given by (2), insert it in (7), and then integrating again... very easy (Wink) ...

Kind regards

- March 12th 2010, 12:35 PMHallsofIvy
Shawsend and ChiSigma are giving essentially the same answer.

- March 12th 2010, 01:03 PMphycdude
- March 12th 2010, 01:06 PMphycdude
- March 12th 2010, 02:24 PMshawsend
Hi. How about start at the end and work backwards. Suppose I solve a DE for and I end up with a messy implicit expression in terms of like and I can't invert it to get x(t). One thing to do is just create a table of x and t values starting with the initial value x(0)=x_0. Something like:

myInversionTable=Table[{myFunction[x]-c,x},{x,x_0,xend}]

Now, I've just constructed an "inverse" table of (t,x(t)) values. Now, in the range of values that is one-to-one, I can then fit that table to some function like a polynomial of high degree and that polynomial will probably be a good empirical representation of the solution for some range of values of t.

So, we have:

Let's integrate that explicitly based on the initial conditions :

Now integrate explicitly again:

- March 13th 2010, 12:29 AMchisigma
Lets try to effectively perform double integration. From (2) we have...

(8)

... so that (7) becomes...

(9)

Now first we set so that the integral in (9) becomes...

(10)

... and then so that the integral (10) becomes...

(11)

... so that the solution of (1) is...

(12)

Kind regards