How to determine particular solutions for cauchy euler

  • Thread starter ericm1234
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If given a cauchy euler equation (non-homogeneous) equation, does the approach in looking for a particular solution (in order to solve the non-homogeneous part), differ from normal?
I am also in general confused about how to assign a particular solution form, in many cases. I have yet to find a good, general explanation for how this is determined, covering all cases.
 

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  • #2
LCKurtz
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If given a cauchy euler equation (non-homogeneous) equation, does the approach in looking for a particular solution (in order to solve the non-homogeneous part), differ from normal?
I am also in general confused about how to assign a particular solution form, in many cases. I have yet to find a good, general explanation for how this is determined, covering all cases.

These equations are closely related to constant coefficient equations. The substitution ##x = e^t## in a Cauchy-Euler equation in x will change it into a constant coefficient DE in ##t##. You can solve that with the usual methods, then convert back with ##t =\ln x##. Given the few times this comes up, it is probably just as well to do it that way instead of memorizing the corresponding rules for the C-E equation. For example, take the equation$$
x^2y''(x) + xy'(x) + 4y(x) = \cos(2\ln x)$$The substitution ##x=e^t## gives this equation in ##t##:$$
y''(t) + 4y(t) = \cos(2t)$$(If you don't know how to do that, I can expand on it). This is one where the NH term contains part of the complementary solution. But the point is, you probably know how to handle it, right? So solve this for ##y(t)## and substitute ##t = \ln x## to get the solution to the original DE.
 
  • #3
jasonRF
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A general technique for finding the particular solution of a linear ode (such as Cauchy-Euler) is the method of "variation of parameters". Any differential equations book should discuss this - google probably will find something as well. Good luck!

jason
 
  • #4
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Ok thanks guys, that substitution idea works pretty well actually.
VOP can sometimes be very difficult with certain non-homogeneous terms in these CE problems.
 

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