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

1. Mar 4, 2009

### saxm

Hi,

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

$$f(t) = Ae^{t}+Bte^{t}$$

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. Mar 4, 2009

### lurflurf

You will need a function such as the product-log to do that.

3. Mar 5, 2009

### coomast

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

http://nl.wikipedia.org/wiki/Newton-Raphson" [Broken]

Using this for your equation you get:

$$f=\alpha=(A+Bt)e^t$$

from which:

$$g=\alpha-(A+Bt)e^t=0$$

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

$$g'=-(A+B+Bt)e^t$$

The iterative scheme is now:

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

Start with $t_0=0$, giving for the example $A=B=1$, $\alpha=3$:

$$0$$

$$1$$

$$0.701213$$

$$0.622262$$

$$0.617657$$

$$0.617642$$

Hope this helps,

coomast

Last edited by a moderator: May 4, 2017
4. Mar 5, 2009

### element4

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

see http://mathworld.wolfram.com/LambertW-Function.html" [Broken].

Last edited by a moderator: May 4, 2017