Assumptions about particular solutions

  • Thread starter leehufford
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Hello,

Working through some reduction of order problems. I'm not sure about why the structure of a particular solution is assumed. Here's what I mean:

Given y"-4y = 2 and a known solution is e^-2x, use reduction of order to find a second solution and a particular solution.

Using a formula, the second solution is found quite easily to be y= ce^2x. But the particular solution is assumed to be of the form y = Ax + B. Taking a few derivatives and substituting yields a particular solution of y = -1/2.

But why do we assume the particular solution is of the form y = Ax + B? Thanks in advance,

Lee
 
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Simon Bridge
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Because - historically, that's been found to be a good guess.

To work out what solution to guess, you use the form of the inhomogeniety as a guide.
People have been working on these sorts of problems for long enough for some patterns to emerge so you don't have to use trial and error every single time.

You seem to have a typo in your example.
I'm guessing you meant to write y''+4y=2

The inhomogeniety is a polynomial of degree N=0.
yp=Ax+B would be a polynomial of degree 1=N+1

http://www.math.uah.edu/howell/DEtext/Part3/Guess_Method.pdf [Broken]
if y is any polynomial of degree N , then y , y′ and y′′ are also polynomials of degree N or less.

Since we want this to match the right side of the above differential equation, which is a
polynomial of degree N , it seems reasonable to [guess] a polynomial of degree N...

... seems to disagree with the choice, which is guessing a polynomial of degree N+1.
 
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  • #3
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Thank you for the reply.

Yes that was a typo, and yes your assumption of what I meant to type was correct.

So are you saying when the nonhomogenous part is of degree N I should guess a particular solution with degree N? I pretty much only chose N+1 because thats what my teacher did.

But when equating coefficients the Ax term drops out leaving only B (-1/2). So the conclusion I drew from your response is to basically pick an N equal to the N of the nonhomogenous part and any higher N (like what I did) is unnessesary work?

Thanks again for the swift, accurate, quality response.

-Lee
 
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  • #4
Simon Bridge
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So are you saying when the nonhomogenous part is of degree N I should guess a particular solution with degree N? I pretty much only chose N+1 because that's what my teacher did.
Does it matter ... what efect does it have on the result if you choose a polynomial of degree N or bigger?

Oh I see this already occurred to you and you checked it out - good initiative!
But when equating coefficients the Ax term drops out leaving only B (-1/2). So the conclusion I drew from your response is to basically pick an N equal to the N of the nonhomogenous part and any higher N (like what I did) is unnessesary work?
... that's the one :)
Well done.
 
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