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Math Help - How to master testing for homogeneous Diff equations?

  1. #1
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    How to master testing for homogeneous Diff equations?

    Ok guys bear with me because this will be a long post.. I'm confused a little bit because every time I look at a different source material that tries to explain how to test if an equation is homogeneous, they are always explaining it in a different way, although I know that they are actually meaning the same thing, and some I can't understand while others' explanations I can, by the way i have finished integral and differential calculus class but not a straight A student so there are rough patches sometimes.. Let me show you:

    First I will say the way my teacher explained it (at least how i understood it)
    1. first transpose all the terms to one side so that the other side will be equal to 0.
    2. Then multiply and divide both sides by the lcd so that there is no denominator
    3. factor out all of the dy and all of the dx
    4. If it takes the form M(x,y)dx + N(x,y)dy such that M(x,y) and N(x,y) both have the same degree, and the terms in it all have the same exponent, then the equation is homogeneous.
    Is my understanding of my teacher's explanation correct? I'm also still not sure how to apply this,especially when I am running into situations where I am yielding dy^2, dx^2..

    Now here is what I understood from the explanation from a book "Advanced Engineering Mathematics" by O'Neil
    1. You have to equate it to y'
    My problem is that sometimes there appears a (y')^2 so that means that in order to equate it to y' i have to use the quadratic formula?
    2. See if it can have the form y'=f(y/x) by using algebraic manipulation
    My problem is that it doesn't give enough tips and examples showing the "manipulations" that should be done..

    Here is the explanation from sosmath.com
    first it says that
    The differential equation

    is homogeneous if the function f(x,y) is homogeneous, that is-

    and then
    Recognize that your equation is an homogeneous equation; that is, you need to check that f(tx,ty)= f(x,y), meaning that f(tx,ty) is independent of the variable t;
    ---> I don't know how to do what he just mentioned, which is how to check that f(tx,ty)= f(x,y)
    ---> He doesn't show the testing of the equations in his examples he just goes straight to substituting y=ux

    Also how did all the different explanations above end up meaning the same thing? Or are they all three actually different ways of testing? Is there really only one way to test or different ways?

    As you can see just understanding the explanations I already have difficulty.. but I have also tried to test equations taken from some books for homogeneity myself so let me show you where I am having difficulty with examples.
    I have to scan what I have written on paper because I don't know how to use LaTex

    In the following I am just trying to verify the homogeneity of the equations
    First is three problems that I have tried to test using my teacher's method
    How to master testing for homogeneous Diff equations?-img004.jpg
    Did I correctly place them in the form M(x,y)dy + N(x,y)dx = 0?
    If so then I am correct in saying they are all not homogeneous?

    Here are the same problems but this time I am trying to follow the method in the book which is to place them in the form y'=f(y/x)
    How to master testing for homogeneous Diff equations?-img002.jpgHow to master testing for homogeneous Diff equations?-img005.jpg
    Just equating it to y' I already don't get it since it's giving me (y')^2.. also I am not sure in the third problem if it already counts as the form y'=f(y/x)

    Please help me fellow citizens of mathhelpforum
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  2. #2
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    Re: How to master testing for homogeneous Diff equations?

    Frankly, I am puzzled by what you write. You say that "the differential equation \frac{dy}{dx}= f(x,y) is "homogeneous" if f(tx, ty)= f(x,y) but
    (1) you don't write your equations in the form \frac{dy}{dx}= f(x,y) (the one that is given in that form you change to \frac{dy}{dx}- f(x,y)= 0).
    (2) you never replace x and y with tx and ty.

    If you are given a definition use that definition!

    The first equation you give is \frac{y}{x+ \sqrt{xy}}- \frac{dy}{dx}. That can be written in " \frac{dy}{dx}= f(x,y)" form as \frac{dy}{dx}= \frac{y}{x+ \sqrt{xy}} (for some reason you multiply both sides by x+ \sqrt{xy}, changing it from the form you say you want). The "f(x,y)" here is, of course, \frac{y}{x+ \sqrt{xy}}. Replace x by tx and y by ty: \frac{ty}{tx+ \sqrt{(tx)(ty)}}= \frac{ty}{tx+ \sqrt{t^2xy}}= \frac{ty}{tx+ t\sqrt{xy}}= \frac{ty}{t(x+ \sqrt{xy}}= \frac{y}{x+ \sqrt{xy}}.

    Yes, it is also true that a differential equation is "homogeneous" if and only if f(x,y) can be written as a function of the single variable u= x/y. In the case above, f(x,y)= \frac{y}{x+ \sqrt{xy}}. Since there is a single "y" in the numerator and we want that divided by "x", an obvious first step is to divide both numerator and denominator by x: \frac{y}{x+ \sqrt{xy}}= \frac{\frac{y}{x}}{\frac{x+ \sqrt{xy}}{x}} = \frac{\frac{y}{x}}{1+ \frac{\sqrt{xy}}{x}}
    = \frac{\frac{y}{x}}{1+ \sqrt{\frac{xy}{x^2}}}= \frac{\frac{y}{x}}{1+ \sqrt{\frac{y}{x}}}= \frac{u}{1+ \sqrt{u}}

    The others can all be done in much the same way.
    Last edited by HallsofIvy; June 21st 2014 at 04:49 AM.
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    Re: How to master testing for homogeneous Diff equations?

    Quote Originally Posted by HallsofIvy View Post
    Frankly, I am puzzled by what you write. You say that "the differential equation \frac{dy}{dx}= f(x,y) is "homogeneous" if f(tx, ty)= f(x,y) but
    (1) you don't write your equations in the form \frac{dy}{dx}= f(x,y) (the one that is given in that form you change to \frac{dy}{dx}- f(x,y)= 0).
    (2) you never replace x and y with tx and ty.

    If you are given a definition use that definition!

    The first equation you give is \frac{y}{x+ \sqrt{xy}}- \frac{dy}{dx}. That can be written in " \frac{dy}{dx}= f(x,y)" form as \frac{dy}{dx}= \frac{y}{x+ \sqrt{xy}} (for some reason you multiply both sides by x+ \sqrt{xy}, changing it from the form you say you want).
    In that part i was trying to make it match the form of the definition M(x,y)dy + N(x,y)dx = 0
    M(x,y)dy + N(x,y)dx = 0 , where M(x,y) and N(x,y) both have the same degree but I'm not sure exactly how to use this method of testing. This was the method that my teacher discussed in class..
    Does this make sense now?

    Quote Originally Posted by HallsofIvy View Post
    The "f(x,y)" here is, of course, \frac{y}{x+ \sqrt{xy}}. Replace x by tx and y by ty: \frac{ty}{tx+ \sqrt{(tx)(ty)}}= \frac{ty}{tx+ \sqrt{t^2xy}}= \frac{ty}{tx+ t\sqrt{xy}}= \frac{ty}{t(x+ \sqrt{xy}}= \frac{y}{x+ \sqrt{xy}}.

    Yes, it is also true that a differential equation is "homogeneous" if and only if f(x,y) can be written as a function of the single variable u= x/y. In the case above, f(x,y)= \frac{y}{x+ \sqrt{xy}}. Since there is a single "y" in the numerator and we want that divided by "x", an obvious first step is to divide both numerator and denominator by x: \frac{y}{x+ \sqrt{xy}}= \frac{\frac{y}{x}}{\frac{x+ \sqrt{xy}}{x}} = \frac{\frac{y}{x}}{1+ \frac{\sqrt{xy}}{x}}
    = \frac{\frac{y}{x}}{1+ \sqrt{\frac{xy}{x^2}}}= \frac{\frac{y}{x}}{1+ \sqrt{\frac{y}{x}}}= \frac{u}{1+ \sqrt{u}}

    The others can all be done in much the same way.
    I don't understand what we have to look at in the last part: \frac{u}{1+ \sqrt{u}}
    What characteristic of this tells us that it is a homogeneous equation? I thought the degree of the numerator and the degree of the numerator have to be the same, no?
    By the way I totally understood what you're telling me about how to show that f(x,y) = f(tx,ty) so thanks for that!
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    Re: How to master testing for homogeneous Diff equations?

    Oh my I only noticed now some portions of the scan were cut..... but I'm glad HallsofIvy that you were able to discern what was written in the cut part.. yes what was cut in the second image is in fact y'=y/(x+sqrt(xy))
    Well, I guess I will scan it again and post it here for good measure:
    How to master testing for homogeneous Diff equations?-img009.jpg
    I have learned from what you have shown on how to transform the diff eq. to the form y'=f(y/x), so here is another equation that I am trying to test:
    How to master testing for homogeneous Diff equations?-img008.jpg
    I have transformed it to y'=f(y/x) but I don't know how to identify if it can be considered homgeneous, also I am not sure if it is correctly fitting the form f(y/x), pls help.
    Last edited by catenary; June 21st 2014 at 06:51 AM.
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    Re: How to master testing for homogeneous Diff equations?

    So now that $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y}{x} + \sqrt{1 + \left( \frac{y}{x} \right) ^2 } \end{align*}$, make the substitution $\displaystyle \begin{align*} v = \frac{y}{x} \implies y = v\,x \implies \frac{\mathrm{d}y}{\mathrm{d}x} = v + x\,\frac{\mathrm{d}v}{\mathrm{d}x} \end{align*}$ and the DE becomes

    $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{y}{x} + \sqrt{ 1 + \left( \frac{y}{x} \right) ^2 } \\ v + x\,\frac{\mathrm{d}v}{\mathrm{d}x} &= v + \sqrt{1 + v^2 } \\ x\,\frac{\mathrm{d}v}{\mathrm{d}x} &= \sqrt{1 + v^2} \\ \frac{1}{\sqrt{1 + v^2}} \,\frac{\mathrm{d}v}{\mathrm{d}x} &= \frac{1}{x} \end{align*}$

    Now that this is separated you should be able to solve the DE.
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    Re: How to master testing for homogeneous Diff equations?

    Quote Originally Posted by Prove It View Post
    So now that $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y}{x} + \sqrt{1 + \left( \frac{y}{x} \right) ^2 } \end{align*}$, make the substitution $\displaystyle \begin{align*} v = \frac{y}{x} \implies y = v\,x \implies \frac{\mathrm{d}y}{\mathrm{d}x} = v + x\,\frac{\mathrm{d}v}{\mathrm{d}x} \end{align*}$ and the DE becomes

    $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{y}{x} + \sqrt{ 1 + \left( \frac{y}{x} \right) ^2 } \\ v + x\,\frac{\mathrm{d}v}{\mathrm{d}x} &= v + \sqrt{1 + v^2 } \\ x\,\frac{\mathrm{d}v}{\mathrm{d}x} &= \sqrt{1 + v^2} \\ \frac{1}{\sqrt{1 + v^2}} \,\frac{\mathrm{d}v}{\mathrm{d}x} &= \frac{1}{x} \end{align*}$

    Now that this is separated you should be able to solve the DE.
    Thanks Proveit but actually my issue here is not how to solve the DE but how to identify if it is homogeneous..
    By looking at y' = (y/x) + sqrt(1-(y/x)^2) what characteristic of it tells me that it qualifies as a homogeneous equation? I am having trouble testing the equations for homogeneity while I can relatively solve most of it (just integral calculus). by the way what i wrote in that final part it's 1-(y/x)^2 and not a plus sign
    Last edited by catenary; June 21st 2014 at 07:23 AM.
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    Re: How to master testing for homogeneous Diff equations?

    Quote Originally Posted by Prove It View Post
    So now that $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y}{x} + \sqrt{1 + \left( \frac{y}{x} \right) ^2 } \end{align*}$, make the substitution $\displaystyle \begin{align*} v = \frac{y}{x} \implies y = v\,x \implies \frac{\mathrm{d}y}{\mathrm{d}x} = v + x\,\frac{\mathrm{d}v}{\mathrm{d}x} \end{align*}$ and the DE becomes

    $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{y}{x} + \sqrt{ 1 + \left( \frac{y}{x} \right) ^2 } \\ v + x\,\frac{\mathrm{d}v}{\mathrm{d}x} &= v + \sqrt{1 + v^2 } \\ x\,\frac{\mathrm{d}v}{\mathrm{d}x} &= \sqrt{1 + v^2} \\ \frac{1}{\sqrt{1 + v^2}} \,\frac{\mathrm{d}v}{\mathrm{d}x} &= \frac{1}{x} \end{align*}$

    Now that this is separated you should be able to solve the DE.
    Thanks Proveit but actually my issue here is not how to solve the DE but how to identify if it is homogeneous..
    By looking at y' = (y/x) + sqrt(1-(y/x)^2) what characteristic of it tells me that it qualifies as a homogeneous equation? I am having trouble testing the equations for homogeneity while I can relatively solve most of it (just integral calculus).

    So sorry forum moderators it was an accidental double post
    Last edited by catenary; June 21st 2014 at 07:21 AM. Reason: Apologizing
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    Re: How to master testing for homogeneous Diff equations?

    So there really are several approaches to testing a DE for homogeneity? and if there are several is one superior over the other, or how do we know which test will be the easiest to execute for the particular problem? This question I have isn't really explained in the books that I have checked, most just go straight to how to solve the DE and don't really thoroughly help the student on how to test (at least from the books that I opened, maybe other books are nicer)
    Last edited by catenary; June 21st 2014 at 08:50 PM. Reason: grammar
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    Re: How to master testing for homogeneous Diff equations?

    Practicing many different types of problems will help you find patterns. Some problems may be easier with one method while other problems may be easier with another. The more problems you do, the easier it will be for you to see which method might yield results. No test (that I know of) is superior for all problems. This is why books teach several methods.
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    Re: How to master testing for homogeneous Diff equations?

    Quote Originally Posted by SlipEternal View Post
    This is why books teach several methods.
    Those must be nice books. The books i've opened so far barely discuss the testing and focus more on the solving (maybe it's just bad luck so far on my part)..
    Anyway thanks slipeternal for the tips.
    By the way the book Advanced Engineering Mathematics defines the homogeneous equation as taking the form
    y' = f(\frac{y}{x})
    and then it cites the ff. examples:
    y' = sin(\frac{y}{x}) - \frac{x}{y}
    y' = \frac{x^{2}}{y^{2}}
    So if it is the x in the numerator and the y in the denominator that still counts as a form y' = f(\frac{y}{x})?
    Why is it that \frac{x^{2}}{y^{2}} counts as a form f(\frac{y}{x})?
    My main problem is in how to interpret the forms/definitions that the books give in describing the types of homogeneous equations. Where are the rules for this and I believe this particular aspect is actually under the subject of advanced algebra or precalculus? Got any tips or rules for interpreting these forms? By the way thanks so far to all who are helping with what is a tricky topic.
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    Re: How to master testing for homogeneous Diff equations?

    Because it can be written as $\displaystyle \begin{align*} \frac{x^2}{y^2} = \frac{1}{\left( \frac{y}{x} \right) ^2 } \end{align*}$
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    Re: How to master testing for homogeneous Diff equations?

    Quote Originally Posted by Prove It View Post
    Because it can be written as $\displaystyle \begin{align*} \frac{x^2}{y^2} = \frac{1}{\left( \frac{y}{x} \right) ^2 } \end{align*}$
    So I guess in the same vein
    y' = sin(\frac{y}{x}) - \frac{x}{y} can be written as y' = sin(\frac{y}{x}) - (\frac{y}{x})^{-1}
    Oh my god this stuff is tricky.. thanks for guiding me
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    Re: How to master testing for homogeneous Diff equations?

    You said, in your first post, that a differential equation was "homogeneous" if and only if it could be written in the form dy/dx= f(x,y) and that f(tx, ty)= f(x, y) for any (non-zero) t. While there are different tests for "homogeneous", you said you knew that definition so I used that definition to determine whether or not the equation was "homogeneous".

    If \frac{dy}{dx}= f(x, y)= \frac{y+ \sqrt{x^2- y^2}}{x} then

    f(tx, ty)= \frac{ty+ \sqrt{t^2x^2- t^2y^2}}{tx}= \frac{ty+ \sqrt{t^2(x^2- y^2)}}{tx} = \frac{ty+ t\sqrt{x^2- y^2}}{tx}= \frac{t(y+ \sqrt{x^2- y^2})}{tx} = \frac{y+ \sqrt{x^2- y^2}}{x}= f(x,y)

    All of that is basic algebra. If you find that too complicated you need to go back and review algebra.
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    Re: How to master testing for homogeneous Diff equations?

    Just to clarify:
    In my first post I said
    I don't know how to do what he just mentioned, which is how to check that f(tx,ty)= f(x,y)
    The thing is I didn't know how to use the definition
    but thanks to you hallsofivy and everyone else here now i know how and it is very clear.
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