2nd order Differential Equation

In summary: Then you can calculate the Wronskian for y in the usual way.In summary, Xago has difficulty solving the differential equation for thermal diffusion and is looking for a way to solve it with the help of Wronskian. He incorrectly takes the g(x) to be qt/D, which then eliminates the D from the equation. Once he realizes his mistake, he calculates the Wronskian for y and finds that it is zero.
  • #1
xago
60
0

Homework Statement


Ugh I feel really stupid for posting this but for some reason I can't remember how to solve it.
I am trying to solve the diff eqn: D*[itex]\frac{d^{2}\phi}{dx^{2}}[/itex] - [itex]\Sigma_{a}[/itex]*[itex]\phi(x)[/itex] = -q[itex]_{t}[/itex]
(thermal diffusion equation for neutrons slowing down to thermal energy)

The Attempt at a Solution



Anyways I've taken the roots of the LHS as if it was homogeneous:
([itex]\phi(x)[/itex][itex]^{(2)}[/itex]-[itex]\frac{\Sigma_{a}}{D}[/itex]) = 0

which gives the 2 roots ±[itex]\sqrt{\frac{\Sigma_{a}}{D}}[/itex]

and therefore [itex]\phi(x)[/itex] = C1*[itex]e^{\sqrt{\frac{\Sigma_{a}}{D}}x}[/itex] + C2*[itex]e^{-\sqrt{\frac{\Sigma_{a}}{D}}x}[/itex]

so the only thing that's bothering me is the -q[itex]_{t}[/itex] on the RHS which I don't know what to do with. I've used Maple to get the solution and it gives me
[itex]\phi(x)[/itex] = C1*[itex]e^{\sqrt{\frac{\Sigma_{a}}{D}}x}[/itex] + C2*[itex]e^{-\sqrt{\frac{\Sigma_{a}}{D}}x}[/itex] + [itex]\frac{q_{t}}{\Sigma_{a}}[/itex]

I would just like to know where that last term comes from and how to get it.
 
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  • #2
you can get the solution using Method of variation of parameters. Using Wronskian and all.
 
  • #3
After you calculate the solution to the homogeneous differential equations, the two arbitrary constants are replaced by two functions, A(x) and B(x) of 'x', which calculated by finding Wronskian, and when you will put those functions into the solution obtained for the homogeneous differential equation you will get the last term.
 
  • #4
Ok I've done the problem with method of variation of parameters but I get the particular solution of [itex]\frac{D*q_{t}}{\Sigma_{a}}[/itex] and not [itex]\frac{q_{t}}{\Sigma_{a}}[/itex] like the solution that Maple gives me..
 
  • #5
What, exactly, is [itex]q_t[/itex]? If it is not a function of x, even if it is a function of another variable, such as t, in an equation in which the differentiation is with respect to x, it can be treated like a constant. If that is true then look for a constant "particular solution": [itex]\phi= A[/itex] so that [itex]d^2\phi/dt^2= 0[/itex]. The equation reduces to [itex]-\Sigma_\alpha A= -q_t[/itex] which immediately gives
[tex]A= \frac{q_t}{\Sigma}[/tex]

That, I presume, was how Maple interpreted this problem. If [itex]q_t[/itex] is a function of x, then the solution will depend strongly on exactly what that solution is.
 
  • #6
Yes qt is a constant and your solution is much more practical, but now I'm wondering why I couldn't get the same solution with variation of parameters. It should be possible right?

Edit: I think I see now that variation of parameters cannot be used unless the non-homogenous term is a function of x, unlike here where it is just a constant.
 
Last edited:
  • #7
xago said:
Yes qt is a constant and your solution is much more practical, but now I'm wondering why I couldn't get the same solution with variation of parameters. It should be possible right?

Edit: I think I see now that variation of parameters cannot be used unless the non-homogenous term is a function of x, unlike here where it is just a constant.

If [itex] q_t[/itex] and [itex]\Sigma_a[/itex] are constants, let [itex] y = \phi - q_t/ \Sigma_a[/itex], and look at the DE for y. That is one of the standard tricks used in dealing with DEs.

RGV
 
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  • #8
xago said:
Yes qt is a constant and your solution is much more practical, but now I'm wondering why I couldn't get the same solution with variation of parameters. It should be possible right?

Edit: I think I see now that variation of parameters cannot be used unless the non-homogenous term is a function of x, unlike here where it is just a constant.

No, variation of parameters works in that case- although it is much harder! We can't say why you did not get that solution with variation of parameters because you did not show what you did. When I try using variation of parameters, I get precisely [itex]q_t/\Sigma_\alpha[/itex] as the "particular solution.
 
  • #9
Hey Xago, I got the mistake which you are doing,

Look carefully, that when you are calculating Wronskian, you may be taking the g(x) to be qt, which is wrong because you will have to take it qt/D, as you have earlier divided the whole equation with D to get the characteristic solution (solution to the homogeneous part).
now if you take the g(x) to be qt/D, you will see that the D will go away automatically.
 

1. What is a 2nd order differential equation?

A 2nd order differential equation is a mathematical equation that involves the second derivative of an unknown function. It can be written in the form of y'' = f(x, y, y'), where y is the unknown function, x is the independent variable, y' is the first derivative of y, and f(x, y, y') is a function that relates the variables.

2. What is the order of a differential equation?

The order of a differential equation is the highest derivative present in the equation. In a 2nd order differential equation, the highest derivative is the second derivative.

3. How do you solve a 2nd order differential equation?

There are various methods for solving a 2nd order differential equation, including separation of variables, substitution, and the method of undetermined coefficients. The specific method used depends on the form of the equation and the initial conditions given.

4. What is the role of initial conditions in solving a 2nd order differential equation?

Initial conditions are necessary for solving a 2nd order differential equation because they provide specific values for the unknown function and its derivatives at a given point. These initial conditions help to determine the constants of integration in the solution of the equation.

5. What are some real-life applications of 2nd order differential equations?

2nd order differential equations have various applications in science and engineering, such as modeling motion of objects under the influence of forces, analyzing electrical circuits, predicting population growth, and studying the behavior of springs and pendulums. They are also used in fields like economics and biology to model complex systems.

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