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Thread: separable first order ODE using y = vx, but all is not well

  1. #1
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    separable first order ODE using y = vx, but all is not well

    Hi folks,


    I am trying to solve the following ordinary differential equation. I am using the substitution y = vx to make it separable.


    $\dfrac{dy}{dx} = \dfrac{(x^2 + y^2)}{x(x + y)} $ ...............(1)


    so substituting $y = vx$ and $\dfrac{dy}{dx} = v + x \dfrac{dv}{dx} $ into (1) gives:


    $v + x \dfrac{dv}{dx} = \dfrac{x^2 (1 + v^2)}{x^2 (1 + v)}$


    such that


    $\int \dfrac {(1 + v)}{(1 - v)} dv = \int \dfrac{dx}{x} $ ............... (2)


    and


    $- \int \dfrac{-1. dv}{(1 - v)} + \int \dfrac{v}{(1 - v)} dv = \int \dfrac{dx}{x} $


    the left hand side can be split, giving:


    $- \ln (1 - v) + \int \dfrac{v}{(1 - v)} dv = \ln |x| + c $ ..................(3) (c can be ln A)


    Now, use substitution $ w = 1 - v, \dfrac{dw}{dv} = -1$


    $\int \dfrac{v}{(1 - v)} dv = \int \dfrac{(w - 1)}{w} \dfrac{dv}{dw} dw = \int dw - \int \dfrac{1}{w} dw = w - \ln |w| $


    such that


    $\int \dfrac{v}{(1 - v)} dv = (1 - v) - \ln (1 - v)$


    from (3) we get


    $ - \ln (1 - v) + (1 - v) - \ln (1 - v) = \ln |x| + c $ ...................(4)


    and


    $- 2 \ln (1 - v) + 1 - v = \ln |Ax| $


    Now using the substitution $v = \frac{y}{x} $


    $ 1 - \dfrac{y}{x} = \ln |Ax| + 2 \ln (1 - \dfrac{y}{x} ) $


    which gives


    $ 1 - \dfrac{y}{x} = \ln |Ax| (\dfrac{x - y}{x})^2 $


    and


    $ y = x(1 - \ln |Ax| (\dfrac{x - y}{x})^2 ) $

    this is the wrong answer. Correct is


    $y = x \ln ( \dfrac{|Ax|}{(x - y)^2})$


    I thought these might be equivalent forms, but I can't get them to match. Can anyone find the error?
    Last edited by s_ingram; May 17th 2018 at 12:13 AM.
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  2. #2
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    Re: separable first order ODE using y = vx, but all is not well

    \begin{align*}
    - \ln |1 - v| + (1 - v) - \ln |1 - v| &= \ln |x| + c \\
    -2\ln |1-v| -v &= \ln|x| + c_1 &\text{where $c_1 = c-1$, a constant} \\
    v &= - \ln |Ax| - \ln 2|1-v| &\text{where $\ln A = \ln (c-1)$, a constant$^*$} \\
    &= -\ln \left(|Ax|(1-v)^2\right) \\
    &= -\ln \left(|Ax|\left(1-\frac{y}{x}\right)^2\right) \\
    &= -\ln \left(|Ax|\frac{1}{x^2}\left(x-y\right)^2\right) \\
    &= -\ln \left(\left|\frac{A}{x}\right|\left(x-y\right)^2\right) \\
    &= -\ln \frac{\left(x-y\right)^2}{|A_1 x|} &\text{where $A_1 = \frac1A$} \\
    &= \ln \frac{|A_1 x|}{\left(x-y\right)^2} \\
    \end{align*}

    $^*$ There's a possibility that we are losing some solutions here, because there are some values of $c$ that lead to a negative constant $c_1$ of which we can't take the logarithm. I'll leave that for you to unpick.
    Last edited by Archie; May 17th 2018 at 07:53 AM.
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    Re: separable first order ODE using y = vx, but all is not well

    I think if you know what the answer is then by choosing the constants carefully, you can get there. I found the 1 term, that you absorbed into the constant, caused by answer to go off in its own direction. This means there is quite some scope for getting different answers! I see what you did and it certainly works. The problem is I cannot see what I have done wrong! Is there any way I can claim that my answer is right?
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  4. #4
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    Re: separable first order ODE using y = vx, but all is not well

    They are the same answer (for suitably chosen values of the constants).

    $$y = x\left(1 - \ln |Ax| \left(\dfrac{x - y}{x}\right)^2 \right) \iff - \ln |1 - v| + (1 - v) - \ln |1 - v| = \ln |x| + c \iff \ln \frac{|A_1 x|}{\left(x-y\right)^2}$$

    The first $\iff$ is from your working (assuming that it's correct) and the second from mine (again assuming it's correct). I have no reason to doubt either derivation although I haven't checked yours in detail (although it looks OK at a quick scan).

    If you pick the same value for both constants, you will almost certainly get a different graph, because you are looking at two distinct members of the solution family.
    Last edited by Archie; May 17th 2018 at 07:59 AM.
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