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May 4th, 2012, 06:11 AM   #1
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High order non-homog odes, general question

Hey,

I have a general question regarding the particular solution to high order odes,

for example of you have an ode y'''' + y''' = x - 2e^(2x)

a particular solution of x^3(Ax + B) + C e^(2x) works,

But I only knew that from pretty much trial and error, and guessing.

Is there a general method of choosing particular solutions for higher order odes?

Thanks,

Linda
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May 4th, 2012, 07:25 AM   #2
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Re: High order non-homog odes, general question

The auxiliary equation associated with the complementary equation has the root r = 0 of multiplicity 3 and the root r = -1 meaning the complementary solution is:



Now, looking at the right side of the ODE, we may assume a particular solution corresponding to the first term of the form:



And for the second term:



Now, for both solutions, the non-negative integer s is chosen to be the smallest integer so that no term in the particular solution is a solution to the corresponding homogeneous equation. Hence, we have:





And so, by superposition, we have:

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May 6th, 2012, 08:46 PM   #3
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Re: High order non-homog odes, general question

Sorry I made a small error in my post,

x^3(Ax + B) + Cxe^(2x) works, I forgot the x next to the exponential term,


but now for

y(x) = x^s(Ce^2x),

so s=1 can work as well? but it's just easier to pick the lowest right so long as it doesn't match the solution to the homogeneous equation?
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May 6th, 2012, 09:21 PM   #4
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Re: High order non-homog odes, general question

Does the exponential term on the right of the equal sign have an x next to it in the ODE? What is the original ODE?
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