
#1
Mar409, 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. 



#2
Mar409, 07:52 PM

HW Helper
P: 2,150

You will need a function such as the productlog to do that.




#3
Mar509, 03:00 PM

P: 279

http://nl.wikipedia.org/wiki/NewtonRaphson 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 



#4
Mar509, 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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