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Math Help - Homogeneous differential equations

  1. #1
    Senior Member chella182's Avatar
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    Unhappy Homogeneous differential equations

    I have never understood this. The question I have is...

    Find the general solution of the equation (x-y)\frac{dy}{dx}=x+y.

    So that'd get...

    \frac{dy}{dx}=\frac{x+y}{x-y}

    In my lecture notes I have written...

    How do we solve this type of equation?

    Given y'=F(\frac{y}{x}), set \frac{y(x)}{x}=v(x)

    i.e. y(x)=x v(x), so v+xv'=F(v)

    or xv'=F(v)-v which is a separable equation.

    \int\frac{dv}{F(v)-v}=\int\frac{dx}{x} where x\neq 0; F(v)-v\neq 0


    I understand that that equation is separable and stuff, but the stuff before I just don't get at all. Like where F(v) suddenly comes from and that. Can anyone help?

    Note: y' and v' are derivatives with respect to x.
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  2. #2
    Math Engineering Student
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    \frac{x+y}{x-y}=\frac{1+\dfrac{y}{x}}{1-\dfrac{y}{x}}.

    Put y=xz.
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  3. #3
    Senior Member chella182's Avatar
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    See that's the kinda thing I don't get I think that process is what I need explaining to me.
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  4. #4
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    y'=z+xz' and the ODE becomes z+xz'=\frac{1+z}{1-z} and after some algebra this becomes a separable one!
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  5. #5
    Senior Member chella182's Avatar
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    Nah, still not getting it. Sorry . Where did the y' = whatever come from?
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  6. #6
    Math Engineering Student
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    That's the derivative. We're making a change of variables. We have y'=\frac{dy}{dx} and z' is \frac{dz}{dx}.
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  7. #7
    Senior Member chella182's Avatar
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    Nah, none of this is going in at all, sorry. I'm gonna have to ask tomorrow after his lecture.
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  8. #8
    Math Engineering Student
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    Tell me exactly what you don't get.
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  9. #9
    Senior Member chella182's Avatar
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    I don't know. I don't like the whole \frac{y}{x} thing you've gotta do. I just don't understand any of it. What I've written in my notes just doesn't make sense to me. I also hate the notation y' and what not, but I doubt thats anything to do with it haha.

    Okay, I've gotten this far with it.

    \frac{dy}{dx}=\frac{x+y}{x-y}

    \frac{dy}{dx}=\frac{1+\frac{y}{x}}{1-\frac{y}{x}}

    y=xv so \frac{dy}{dx}=v+x\frac{dv}{dx}

    And now I'm lost...
    Last edited by mr fantastic; February 26th 2009 at 03:03 AM. Reason: Modified some words.
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  10. #10
    Math Engineering Student
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    And that becomes v+x\,\frac{dv}{dx}=\frac{1+v}{1-v}.
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  11. #11
    Senior Member chella182's Avatar
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    And what do I do with that? I just need walking through it all from this point...
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  12. #12
    Math Engineering Student
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    x\,\frac{dv}{dx}=\frac{1+v}{1-v}-v\implies x\,\frac{dv}{dx}=\frac{1+v-v+v^{2}}{1-v}, thus \frac{1-v}{1+v^{2}}\,dv=\frac{dx}x.
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  13. #13
    Senior Member chella182's Avatar
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    Wouldn't it be \int\frac{dx}{x} on the right hand side? I dislike that left-hand side also haha. But I know how to do it (roughly).
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  14. #14
    Math Engineering Student
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    Oh yes, I've edited.
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  15. #15
    Senior Member chella182's Avatar
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    Okay right, on integrating both sides would you get...

    -\frac{1}{2}\ln{(1+v^2)}=\ln{(x)}+A where A = constant of integration?

    The left-hand side is likely to by wrong somehow; I never could do those tricky \int\frac{f'(x)}{f(x)}dx problems.
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