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Cannot figure out how to solve this differential equation

  1. Feb 10, 2008 #1
    dy/dx= 1/(x^2-xy)

    I have tried all the basic steps in an attempt to make it take the form of:
    separable
    exact
    homogenous
    bernoulli
    linear

    I have tried substitution with:
    y=ux
    u=x-y
    u=x^2-y

    I cannot for the life of me figure this out, thank you for any help.
     
  2. jcsd
  3. Feb 10, 2008 #2

    arildno

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    Well, you might try do invert it, and solve instead:
    [tex]\frac{dx}{dy}=x^{2}-xy[/tex]
    Isn't that sort of Bernoulli-like? Don't remember all those tricks any longer..

    EDIT:

    If I have done this correctly, we should get something like:
    [tex]x(y)=\frac{e^{-\frac{y^{2}{2}}}{\int{e}^{-\frac{u^{2}}{2}}du[/tex]
     
    Last edited: Feb 10, 2008
  4. Feb 10, 2008 #3
    Arildno, there seems to be something wrong with the formula, the latex image failed to generate. I think you forgot a \frac.

    Anyway, this is the right approach to solving the equation, don't forget the integration constant :-)

    [Edit]

    Some more explanation and intermediate steps might be useful. The equation will become after setting z=x and t=y (in order to avoid mistakes)

    [tex]\frac{dz}{dt}+tz-z^2=0[/tex]

    This is a Riccati type equation, which is in it's general form:

    [tex]\frac{dz}{dt}+Q(t)z+R(t)z^2=P(t)[/tex]

    and can be transformed into a second order linear differential equation by using the substitution:

    [tex]z(t)=\frac{1}{R(t)u}\frac{du(t)}{dt}[/tex]

    In our case this is:

    [tex]z(t)=\frac{-1}{u}\frac{du(t)}{dt}[/tex]

    Differentiating this gives (leaving out the independent variable notation):

    [tex]\frac{dz}{dt}=\frac{1}{u^2}\left(\frac{du}{dt}\right)^2 -\frac{1}{u}\frac{d^2u}{dt^2}[/tex]

    Substituting this in the orginal equation gives:

    [tex]\frac{d^2u}{dt^2}=-t\frac{du}{dt}[/tex]

    Which has the following solution:

    [tex]\frac{du}{dt}=A\cdot e^{-\frac{t^2}{2}}[/tex]

    And thus:

    [tex]u=A\int e^{-\frac{t^2}{2}}dt+B[/tex]

    Using the transformation:

    [tex]z(t)=-\frac{1}{u}\frac{du}{dt}=-\frac{A\cdot e^{-\frac{t^2}{2}}} {A\int e^{-\frac{t^2}{2}}dt+B}[/tex]

    Which is after dividing with A and setting B/A=C:

    [tex]z(t)=-\frac{e^{-\frac{t^2}{2}}} {\int e^{-\frac{t^2}{2}}dt+C}[/tex]

    [Edit 2]

    It is indeed the solution, substituting back in the original equation shows this.

    There might be easier ways to solve this, but I can't think of any.
     
    Last edited: Feb 10, 2008
  5. Feb 10, 2008 #4
    Another way to deal with equation is the following.
    Change the independent variable [itex]x=e^t[/itex], thus [itex]y'=e^{-t}\,\frac{d\,y}{d\,t}[/itex] in order to have for the inital ODE

    [tex]\frac{d\,y}{d\,t}=\frac{1}{e^t-y}\Rightarrow \frac{d\,t}{d\,y}=e^t-y[/tex]

    Now define [itex]t(y)=u(y)-\frac{1}{2}\,y^2[/itex] in order to get rid of [itex]y[/itex] and make the ODE separable, i.e.

    [tex]\frac{d\,u}{d\,y}=e^{u-\frac{1}{2}\,y^2}\Rightarrow e^{-u}\,d\,u=e^{-\frac{1}{2}\,y^2}\,d\,y\Rightarrow -e^{-u}=\int e^{-\frac{1}{2}\,y^2}\,d\,y+C\Rightarrow -e^{t+\frac{1}{2}\,y^2}=\int e^{-\frac{1}{2}\,y^2}\,d\,y+C\Rightarrow -\frac{1}{x}\,e^{\frac{1}{2}\,y^2}=\int e^{-\frac{1}{2}\,y^2}\,d\,y+C[/tex]

    yielding to

    [tex]x=-\frac{e^{-\frac{1}{2}\,y^2}}{\int e^{-\frac{1}{2}\,y^2}\,d\,y+C}[/tex]

    i.e. coomast's solution. :smile:
     
  6. Apr 13, 2009 #5

    arildno

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    Okay, an old Latex error finally rectified..
     
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