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May 4th, 2012, 06:11 AM  #1 
Newbie Joined: Mar 2012 Posts: 12 Thanks: 0  High order nonhomog 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 
May 4th, 2012, 07:25 AM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: High order nonhomog 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 nonnegative 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: 
May 6th, 2012, 08:46 PM  #3 
Newbie Joined: Mar 2012 Posts: 12 Thanks: 0  Re: High order nonhomog 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? 
May 6th, 2012, 09:21 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: High order nonhomog 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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