How to solve 2nd order ODE solution eg. te^t+e^t, for t?


by saxm
Tags: ode
saxm
saxm is offline
#1
Mar4-09, 07:06 PM
P: 4
Hi,

I have a second order differential equation with a solution in the form:

[tex]f(t) = Ae^{t}+Bte^{t}[/tex]

I want to solve for t, ie. work out for what value of t does the function f(t) have a particular value. But there seems to be no way (that I know of) to do this. Can anyone give me any pointers to what to do here?

Thanks.
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lurflurf
lurflurf is offline
#2
Mar4-09, 07:52 PM
HW Helper
P: 2,150
You will need a function such as the product-log to do that.
coomast
coomast is offline
#3
Mar5-09, 03:00 PM
P: 279
Quote Quote by saxm View Post
Hi,

I have a second order differential equation with a solution in the form:

[tex]f(t) = Ae^{t}+Bte^{t}[/tex]

I want to solve for t, ie. work out for what value of t does the function f(t) have a particular value. But there seems to be no way (that I know of) to do this. Can anyone give me any pointers to what to do here?

Thanks.
Another way would be to use the Newton-Raphson iterative scheme. Here is a link:

http://nl.wikipedia.org/wiki/Newton-Raphson

Using this for your equation you get:

[tex]f=\alpha=(A+Bt)e^t[/tex]

from which:

[tex]g=\alpha-(A+Bt)e^t=0[/tex]

The function to be solved. The derivative is found to be:

[tex]g'=-(A+B+Bt)e^t[/tex]

The iterative scheme is now:

[tex]t_{n+1}=t_n+\frac{\alpha-(A+Bt)e^t}{(A+B+Bt)e^t}[/tex]

Start with [itex]t_0=0[/itex], giving for the example [itex]A=B=1[/itex], [itex]\alpha=3[/itex]:

[tex]0[/tex]

[tex]1[/tex]

[tex]0.701213[/tex]

[tex]0.622262[/tex]

[tex]0.617657[/tex]

[tex]0.617642[/tex]

Hope this helps,

coomast

element4
element4 is offline
#4
Mar5-09, 04:08 PM
P: 107

How to solve 2nd order ODE solution eg. te^t+e^t, for t?


Maple 12 suggests [tex]t = \text{LambertW}\left( \frac{f\cdot \exp{\frac AB}}B\right) - \frac AB[/tex]

see here.


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