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

In summary, the conversation discusses different methods for solving a second order differential equation and finding the value of t for a given function. The methods mentioned include using the product-log function and the Newton-Raphson iterative scheme, as well as the use of Maple 12 and the LambertW function. These methods are suggested as potential solutions to the problem at hand.
  • #1
saxm
4
0
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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  • #2
You will need a function such as the product-log to do that.
 
  • #3
saxm said:
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" [Broken]

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
 
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  • #4
Maple 12 suggests [tex]t = \text{LambertW}\left( \frac{f\cdot \exp{\frac AB}}B\right) - \frac AB[/tex]

see http://mathworld.wolfram.com/LambertW-Function.html" [Broken].
 
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1. What is a 2nd order ODE?

A 2nd order ODE (ordinary differential equation) is a mathematical equation that relates a function to its derivatives up to the second order. It is often used to model physical phenomena in various fields such as physics, chemistry, and engineering.

2. What does it mean to solve a 2nd order ODE?

Solving a 2nd order ODE means finding a function that satisfies the equation and its initial or boundary conditions. This function is called the general solution, and it represents all possible solutions to the given ODE.

3. How do you solve a 2nd order ODE?

To solve a 2nd order ODE, you can use various methods such as separation of variables, substitution, or integrating factors. The specific method used depends on the type of ODE and its coefficients. It is also important to check if the solution satisfies the initial or boundary conditions.

4. What is the general form of a 2nd order ODE solution?

The general form of a 2nd order ODE solution is a linear combination of two linearly independent solutions. This means that the solution can be expressed as a combination of two different functions that satisfy the ODE. In the case of a homogeneous ODE, the general solution will be in the form of a linear combination of exponential functions.

5. How do I solve a specific 2nd order ODE, such as te^t + e^t?

To solve a specific 2nd order ODE, you can follow these steps:
1. Rewrite the ODE in standard form, if necessary.
2. Identify the type of ODE and its coefficients.
3. Choose an appropriate method to solve the ODE.
4. Solve for the general solution.
5. Check if the solution satisfies the initial or boundary conditions, and adjust if needed.

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