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Math Help - Help with a basic homogenous ODE

  1. #1
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    Help with a basic homogenous ODE

    I'm having a bit of trouble with a homogeneous ODE:


    \frac{dy}{dx} = \frac{x^2 + 3y^2}{2xy}


    In order to solve this by substituting v = y/x, I first need to manipulate the right side to get all variables in the form of y/x. However, when I simplify it, I get:


    \frac{dy}{dx} = \frac{x}{2y} + \frac{3y}{2x}


    ...which leaves each fraction opposite of the other. What am I missing?
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  2. #2
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    So

    \frac{dy}{dx}=\frac{2}{v}+\frac{3v}{2}.

    The whole point is to get a separable equation, right? Of course, you need to get the derivative in terms of v and x.

    [EDIT]: See below for a correction.
    Last edited by Ackbeet; June 11th 2011 at 02:06 PM.
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    \frac{2}{v} or \frac{1}{2v} ?

    In any case, if I go with the latter, then I can get things here:

    \frac{1}{x} dx = 2v + \frac{2}{3} \frac{1}{v} - \frac{1}{v} dv

    I integrate both sides, move things around, and I get:

    \frac{y^2}{x^2} - \frac{1}{3} ln \left\vert{} \frac{y}{x} \right\vert{}   - ln \left\vert{} x \right\vert{}   = c


    However, the book answer is:

    x^2 + y^2 - cx^3 = 0


    So, either I made some serious mistakes, or there's some algebra magic going on.
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    Quote Originally Posted by Lancet View Post
    \frac{2}{v} or \frac{1}{2v} ?
    The second - you're correct.

    In any case, if I go with the latter, then I can get things here:

    \frac{1}{x} dx = 2v + \frac{2}{3} \frac{1}{v} - \frac{1}{v} dv
    This doesn't look right to me. I get

    \frac{dx}{x}=\frac{2v\,dv}{1+v^{2}}.

    Can you see how I got this? If so, carry this correction through and see what you get.
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    Quote Originally Posted by Ackbeet View Post

    This doesn't look right to me.

    Okay, I'll show the details, and maybe you can help me figure out where I went wrong...

    x \frac{dv}{dx} = \frac{1}{2v} + \frac{3}{2} v - v

    \frac{1}{x}  dx = 2v + \frac{2}{3} * \frac{1}{v} - \frac{1}{v}  dv

    \int \frac{1}{x}  dx = \int 2v + \frac{2}{3} * \frac{1}{v} - \frac{1}{v}  dv

    ln \left\vert{} x \right\vert{} = v^2 + \frac{2}{3} ln \left\vert{} v \right\vert{} - ln \left\vert{} v \right\vert{} + c

    ln \left\vert{} x \right\vert{} = v^2 - \frac{1}{3} ln \left\vert{} v \right\vert{} + c

    \frac{y^2}{x^2} - \frac{1}{3} ln \left\vert{} \frac{y}{x} \right\vert{}   - ln \left\vert{} x \right\vert{}   = c
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    Hmm. Maybe I didn't make myself clear. For your Post # 5, I think the very first equation is wrong. Could you please show me the steps before that equation?
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    Quote Originally Posted by Ackbeet View Post
    Hmm. Maybe I didn't make myself clear. For your Post # 5, I think the very first equation is wrong. Could you please show me the steps before that equation?
    Ah, definitely not what I thought you were referencing. No problem, here we go:


    \frac{dy}{dx} = \frac{x^2 + 3y^2}{2xy}

    \frac{dy}{dx} = \frac{x^2}{2xy} + \frac{3y^2}{2xy}

    \frac{dy}{dx} = \frac{x}{2y} + \frac{3y}{2x}

    v + x \frac{dv}{dx} = \frac{x}{2xv} + \frac{3}{2} v

    x \frac{dv}{dx} = \frac{1}{2v} + \frac{3}{2} v - v
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    Quote Originally Posted by Lancet View Post
    Ah, definitely not what I thought you were referencing. No problem, here we go:


    \frac{dy}{dx} = \frac{x^2 + 3y^2}{2xy}

    \frac{dy}{dx} = \frac{x^2}{2xy} + \frac{3y^2}{2xy}

    \frac{dy}{dx} = \frac{x}{2y} + \frac{3y}{2x}

    v + x \frac{dv}{dx} = \frac{x}{2xv} + \frac{3}{2} v

    x \frac{dv}{dx} = \frac{1}{2v} + \frac{3}{2} v - v
    Ok, I'm with you now. This last equation looks correct to me, now that I look at it some more.

    From your previous post:

    Quote Originally Posted by Lancet
    x \frac{dv}{dx} = \frac{1}{2v} + \frac{3}{2} v - v

    \frac{1}{x}  dx = 2v + \frac{2}{3} * \frac{1}{v} - \frac{1}{v}  dv
    Ok, here's the mistake: addition does not distribute over multiplication like that. You should do this instead:

    x \frac{dv}{dx} = \frac{1}{2v} + \frac{3}{2} v - v=\frac{1}{2v} + \frac{v}{2}

    2x\,v'=\frac{1}{v}+v=\frac{1+v^{2}}{v}.

    Can you continue?
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    Quote Originally Posted by Ackbeet View Post

    Ok, here's the mistake: addition does not distribute over multiplication like that. You should do this instead:

    Ah! I'm not sure I was ever formally taught the trick of flipping numerators/denominators on both sides - I just picked it up, thus why this aspect of it was unknown to me. Thanks!

    So, taking it from here, I now end up with:

    \frac{y^2}{x^2} - x + 1 = c


    Is that the same as the book answer I referenced earlier? Or did I make another mistake? I can't quite eyeball them as being identical, but that doesn't mean anything.
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    Quote Originally Posted by Lancet View Post
    Ah! I'm not sure I was ever formally taught the trick of flipping numerators/denominators on both sides - I just picked it up, thus why this aspect of it was unknown to me. Thanks!

    So, taking it from here, I now end up with:

    \frac{y^2}{x^2} - x + 1 = c


    Is that the same as the book answer I referenced earlier? Or did I make another mistake? I can't quite eyeball them as being identical, but that doesn't mean anything.
    Your answer is much closer, but it's still not the same as the book's answer (which is also what I got). The main difference is the location of the arbitrary constant. Can you post your work again, please?
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    Quote Originally Posted by Lancet View Post
    Ah! I'm not sure I was ever formally taught the trick of flipping numerators/denominators on both sides - I just picked it up, thus why this aspect of it was unknown to me. Thanks!

    So, taking it from here, I now end up with:

    \frac{y^2}{x^2} - x + 1 = c


    Is that the same as the book answer I referenced earlier? Or did I make another mistake? I can't quite eyeball them as being identical, but that doesn't mean anything.
    Your answer is much closer, but it's still not the same as the book's answer (which is also what I got). The main difference is the location of the arbitrary constant. Can you post your work again, please?
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    Quote Originally Posted by Ackbeet View Post
    Your answer is much closer, but it's still not the same as the book's answer (which is also what I got). The main difference is the location of the arbitrary constant. Can you post your work again, please?
    Sure:


    2x \frac{dv}{dx} = \frac{v^2 + 1}{v}

    \frac{1}{2x} dx = \frac{v}{v^2 + 1} dv

    \int \frac{1}{2x} dx = \int \frac{v}{v^2 + 1} dv

    \frac{1}{2} ln \left\vert{} x \right\vert{} = \frac{1}{2} ln \left\vert{} v^2 + 1 \right\vert{} + c

    ln \left\vert{} x \right\vert{} = ln \left\vert{} v^2 + 1 \right\vert{} + c

    e^(ln \left\vert{} x \right\vert{}) = e^(ln \left\vert{} v^2 + 1 \right\vert{} + c)

    x = v^2 + 1 + c

    x = \frac{y^2}{x^2} + 1 + c

    \frac{y^2}{x^2} - x + 1 = c
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    Quote Originally Posted by Lancet View Post
    Sure:


    2x \frac{dv}{dx} = \frac{v^2 + 1}{v}

    \frac{1}{2x} dx = \frac{v}{v^2 + 1} dv

    \int \frac{1}{2x} dx = \int \frac{v}{v^2 + 1} dv

    \frac{1}{2} ln \left\vert{} x \right\vert{} = \frac{1}{2} ln \left\vert{} v^2 + 1 \right\vert{} + c

    ln \left\vert{} x \right\vert{} = ln \left\vert{} v^2 + 1 \right\vert{} + c

    e^(ln \left\vert{} x \right\vert{}) = e^(ln \left\vert{} v^2 + 1 \right\vert{} + c)
    The mistake occurs right here. Whenever you take the exponential of an expression such as x+c, the arbitrary constant, which was additive, becomes multiplicative:

    e^{x+C}=e^{C}e^{x}.

    You can redefine the arbitrary constant as K=e^{C} to get Ke^{x}.

    So, instead of

    x = v^2 + 1 + c,
    you should have gotten

    x=c(v^{2}+1).

    Incidentally, in order to get some larger expression to show up all in an exponent (or subscript), enclose it with curly braces thus: e^{x+y}.
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    Quote Originally Posted by Ackbeet View Post

    The mistake occurs right here. Whenever you take the exponential of an expression such as x+c, the arbitrary constant, which was additive, becomes multiplicative:

    Aha! Thank you, I understand where I went wrong now. With that correction, I can successfully obtain the correct answer. Whew! That was a small bear for me. Thanks for sticking with me through all of that and helping me understand where I was going wrong.
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    Quote Originally Posted by Lancet View Post
    Aha! Thank you, I understand where I went wrong now. With that correction, I can successfully obtain the correct answer. Whew! That was a small bear for me. Thanks for sticking with me through all of that and helping me understand where I was going wrong.
    You're very welcome. Have a good one!
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