Can an Integrated Factor be Found in Terms of Both Variables?

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Homework Statement



I have been dealing with Exact Equations in my DE class, and I came around this problem.

(t^2-y^2)+(t^2-2ty)(dy/dt)=0

This is obviously not an exact eqn. So I tried using integrated factors on it and try to find this "factor" μ.
But no matter if I did it in terms of t or in terms of y, I couldn't separate it in terms of one variable.

dμ/dt=(-2t)/(t^2-2ty)

or

dμ/dy=(2t)/(t^2-y^2)

Homework Equations



Is there any way that you can find an integrated factor which it is in terms of both variables?
instead of t or y alone, both?



The Attempt at a Solution



I tried everything, and this topic is not even covered in class or in the book. I learned this on my own and I have only learn Integrated factors in terms of y or in terms of t, not both.

Help please.
 
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MMars91 said:

Homework Statement



I have been dealing with Exact Equations in my DE class, and I came around this problem.

(t^2-y^2)+(t^2-2ty)(dy/dt)=0

This is obviously not an exact eqn. So I tried using integrated factors on it and try to find this "factor" μ.
But no matter if I did it in terms of t or in terms of y, I couldn't separate it in terms of one variable.

dμ/dt=(-2t)/(t^2-2ty)

or

dμ/dy=(2t)/(t^2-y^2)

Homework Equations



Is there any way that you can find an integrated factor which it is in terms of both variables?
instead of t or y alone, both?



The Attempt at a Solution



I tried everything, and this topic is not even covered in class or in the book. I learned this on my own and I have only learn Integrated factors in terms of y or in terms of t, not both.

Help please.

A function ##M(x,y)## is said to be homogeneous of order n if ##M(\lambda x,\lambda y)= \lambda^nM(x,y)##. If ##M(x,y)## and ##N(x,y)## are both homogeneous of degree n, then the differential equation ##M(x,y)dx + N(x,y)dy = 0## can be converted to a separable DE with the substitution ##y=ux##.

That applies to your question. Look at http://www.cliffsnotes.com/study_guide/First-Order-Homogeneous-Equations.topicArticleId-19736,articleId-19713.html for a discussion of this type of equation.
 
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Thanks, that just lighted a bulb in my head.
 
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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