Integrating Factor: Solving without Trial & Error

barksdalemc
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Is there a way other than trial and error to tell whether an integrating factor h is a function of x only, y only, or of x and y?
 
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What do you mean? An integrating factor is simply an algorithm.

Also, integrating factors are completely worthless because they are impossible to retain. You should know variation of parameters because it works in more general cases, and actually makes sense.
 
I think Mindscrape is misunderstanding the word "integrating factor". In particular, I cannot see what "variation of parameters" could have to do with "integrating factor". He seems to be confusing "integrating factor" with the method of "undetermined coefficients".

An "integrating factor" is a function of the variables which, if you multiply a differential equation by it, makes the equation "exact"- or, in other terms, if you multiply a differential by it, makes the differential and exact differential. To answer your question, bartsdalemc, no, there is no general method of determining whether an integrating factor involves only one or several of the variables. If there were, the problem of solving a general first order differential equation would become trivial- and, believe me, it is not!
 
I know what an integrating factor is...

Let me elaborate. I think integrating factors are terrible because they are for evaluating, specifically, first order constant coefficient ODEs. Variation of parameters works for any order and nonconstant coefficients, and in general makes more sense. In my opinion, it is much better to learn the general case.

What do you mean variation of parameters has nothing to do with integrating factors? Integrating factors are merely a shortcut for variation of parameters. While you gain a specific formula for the integrating factor, the way to obtain that formula follows a certain algorithm (maybe a term that made you think I was confused?): multiply by the integrating factor µ(x), turn the equation into an exact differential, use the fact that the integrating factor is an exponential, and obtain the solution.

Variation of parameters will also give the solution, but may take longer. Method of undetermined coefficients, if you are good at guessing, will solve it too.

I just don't understand what the original poster is talking about. By the context of the differential equation you know variables are involved.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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