Results 1 to 7 of 7

Math Help - First-order ODE.

  1. #1
    Junior Member
    Joined
    Aug 2010
    Posts
    41

    First-order ODE.

    I am trying to solve the following differential equation but cannot get a method that will work.

    dy/dx=(y^2)/(xy-x^2)

    I have tried making it exact as well as various substitutions. All to no avail. I am thinking some sort of combination of variables might work, but I am stumped.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45

    re: First-order ODE.

    The equation has an integrating factor that depends on x .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2010
    Posts
    41

    re: First-order ODE.

    Would you mind a little explanation on how you were able to deduce that?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,653
    Thanks
    1478

    re: First-order ODE.

    Quote Originally Posted by cheme View Post
    I am trying to solve the following differential equation but cannot get a method that will work.

    dy/dx=(y^2)/(xy-x^2)

    I have tried making it exact as well as various substitutions. All to no avail. I am thinking some sort of combination of variables might work, but I am stumped.
    \displaystyle \begin{align*} \frac{dy}{dx} &= \frac{y^2}{xy - x^2} \\ \frac{dx}{dy} &= \frac{xy - x^2}{y^2} \\ \frac{dx}{dy} &= \frac{xy}{y^2} - \frac{x^2}{y^2} \\ \frac{dx}{dy} &= \frac{x}{y} - \left(\frac{x}{y}\right)^2 \\  \end{align*}

    Now make the substitution \displaystyle v = \frac{x}{y} \implies x = vy \implies \frac{dx}{dy} = v + y\,\frac{dv}{dy} and the DE becomes

    \displaystyle \begin{align*} \frac{dx}{dy} &= \frac{x}{y} - \left(\frac{x}{y}\right)^2 \\ v + y\,\frac{dv}{dy} &= v - v^2 \\ y\,\frac{dv}{dy} &= -v^2 \\ v^{-2}\,\frac{dv}{dy} &= -\frac{1}{y} \\ \int{v^{-2}\,\frac{dv}{dy}\,dy} &= \int{-\frac{1}{y}\,dy} \\ \int{v^{-2}\,dv} &= -\ln{|y|} + C_1 \\ \frac{v^{-1}}{-1} + C_2 &= \ln{|y|} + C_1 \\ -\frac{1}{v} &= \ln{|y|} + C\textrm{ where }C = C_1 - C_2\\ -\frac{y}{x} &= \ln{|y|} + C \\ x &= -\left(\frac{y}{\ln{|y|} + C}\right)\end{align*}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45

    Re: First-order ODE.

    Quote Originally Posted by cheme View Post
    Would you mind a little explanation on how you were able to deduce that?
    If Pdx+Qdy=0 has an integrating factor of the form \mu=\mu(x) then (\mu P)_y=(\mu P)_x . Solving you'll obtain \frac{\mu'}{\mu}=\ldots =\dfrac{1}{Q}(P_y-Q_x)=F(x) . In our case , verify that \dfrac{1}{Q}(P_y-Q_x)=\dfrac{-2}{x} .
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2

    Re: First-order ODE.

    Quote Originally Posted by FernandoRevilla View Post
    If Pdx+Qdy=0 has an integrating factor of the form \mu=\mu(x) then (\mu P)_y=(\mu P)_x .
    I'm sure FernandoRevilla meant (\mu P)_{y}=(\mu Q)_{x}.

    Solving you'll obtain \frac{\mu'}{\mu}=\ldots =\dfrac{1}{Q}(P_y-Q_x)=F(x) . In our case , verify that \dfrac{1}{Q}(P_y-Q_x)=\dfrac{-2}{x} .
    One way to get an integrating factor is to posit an integrating factor of the form h(x^{n}y^{m}). Multiply through, take your partial derivatives, turn the crank, etc. The nice thing about doing it this way is that it takes care of a lot of situations in one fell swoop. It won't solve every first-order DE, not even all first-order DE's solvable by means of an integrating factor. But it will solve quite a few.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45

    Re: First-order ODE.

    Quote Originally Posted by Ackbeet View Post
    I'm sure FernandoRevilla meant (\mu P)_{y}=(\mu Q)_{x}.
    I'm sure too.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Re-writing higher order spatial derivatives as lower order system
    Posted in the Differential Equations Forum
    Replies: 11
    Last Post: July 27th 2010, 08:56 AM
  2. Replies: 1
    Last Post: October 27th 2009, 04:03 AM
  3. Replies: 2
    Last Post: February 23rd 2009, 05:54 AM
  4. Replies: 2
    Last Post: November 25th 2008, 09:29 PM
  5. Replies: 4
    Last Post: August 12th 2008, 04:46 AM

Search Tags


/mathhelpforum @mathhelpforum