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2nd order Differential Equation

  • Thread starter xago
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  • #1
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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.
 

Answers and Replies

  • #2
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you can get the solution using Method of variation of parameters. Using Wronskian and all.
 
  • #3
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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
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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
HallsofIvy
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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
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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.
 
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  • #7
Ray Vickson
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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
 
Last edited:
  • #8
HallsofIvy
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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
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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.
 

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