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Simple differential equation

  1. Sep 10, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the general solution to the following ODE.

    2. Relevant equations

    [tex]yy'' = 2y'^2[/tex]


    3. The attempt at a solution

    I am not sure how to approach this, but I get the sense that it must, somehow, be separable. I switched variables around in attempt to figure something out, but I got nowhere. I am clearly not seeing something that should be obvious. I should have been able to solve this thing years ago with no problem.

    [tex]\frac{yy''}{y'} = 2y'[/tex]

    But, the [tex]y'[/tex] is still on both sides.

    I then tried doing the [tex]\frac{dy}{dx}[/tex] form as it's easier to see.

    [tex]\frac{yy''}{\frac{dy}{dx}} = 2\frac{dy}{dx}[/tex]

    I still feel like I am getting nowhere.
     
  2. jcsd
  3. Sep 10, 2010 #2
    You know you can directly integrate an expression like:

    [tex]f(y')y''[/tex]

    right? That's just:

    [tex]f(y')dy'[/tex]

    then integrate.

    So what happens if you divide both sides of your DE by [itex]y(y')^2[/itex]? Still not quite right? How about then if you next just multiply both sides by [itex]y'[/itex]. Shouldn't that get differentials on both sides?
     
  4. Sep 10, 2010 #3

    I'm confused. How did you get the equation into the [tex]f(y')[/tex] notation?

    I'll put it in full notation to better highlight my confusion.

    [tex]y\frac{d^2y}{dx^2} = 2(\frac{dy}{dx})^2[/tex]

    In your notation, if you are integrating with the differential being [tex]dy[/tex], where did the [tex]dx[/tex] term go?

    I tried rearranging as you said, and I ended up with:

    [tex]y'' = \frac{2y'^2}{y}[/tex]

    I don't see anywhere to go from here. Maybe the answer lies in the first part of the post. There is definitely a basic thing that I am not understanding that this problem is highlighting.
    Thanks for the help.
     
  5. Sep 11, 2010 #4
    I still have not been able to figure this out. It does not seem separable at all to me.
     
  6. Sep 11, 2010 #5

    hunt_mat

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    Try a solution of the form:
    [tex]
    y=x^{n}
    [/tex]
    then you have an algebraic equation to solve for n. This is as far as I have hot with the problem.
     
  7. Sep 11, 2010 #6

    vela

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    Divide the original equation by yy' to get

    [tex]\frac{y''}{y'} = 2\frac{y'}{y}[/tex]

    You can integrate both sides of that.
     
  8. Sep 11, 2010 #7
    If I put it into that form, I get...

    [tex]\int\frac{1}{y'}y'' = 2\int\frac{1}{y}y'[/tex]

    Integrating out, I (I think) get:

    [tex]\ln(y') + c_1 = 2\ln(y) + c_2[/tex]

    Putting that as an exponential, I get:

    [tex]e^{\ln(y')+c_1} = e^{2\ln(y) + c_2}[/tex]

    This gets me...

    [tex]c_1y' = c_2y[/tex]

    I *think* I'm still running into some algebra problems. I should be coming up with something in terms of x, though maybe it's just late and I'm missing some more algebra. Sorry about this mess.
     
  9. Sep 11, 2010 #8
    That's very good. Combine the constants and obtain:

    [tex]\ln y'=\ln y^2+c[/tex]

    or:

    [tex]\ln\left(\frac{y'}{y^2}\right)=c[/tex]

    Take exponents, still an arbitrary constant on the right. Now you can separate variables right?
     
  10. Sep 11, 2010 #9
    That would be...

    [tex]e^{\ln\frac{y'}{y^2}} = e^{c_1}[/tex]

    [tex]\frac{y'}{y^2} = c_1[/tex]

    [tex]\int\frac{dy}{y^2} = \intc_1dx[/tex]
    [tex]-\frac{1}{y} = c_1x + c_2[/tex]

    I can take out the negative, stick it in the constant.


    [tex]y = \frac{1}{c_1x+c_2}[/tex]

    I *think* that's the answer.
     
  11. Sep 11, 2010 #10

    vela

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    It's easy enough to check. Try plugging it back into the original differential equation and seeing it works.
     
  12. Sep 11, 2010 #11
    Hey, remember when we divided by [itex]y(y')^2[/itex] up there? What are the ramifications of this step?
     
  13. Sep 11, 2010 #12

    hunt_mat

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    That you're not assuming any stationary points or zeros.
     
  14. Sep 11, 2010 #13
    I'm not entirely sure. In fact, that is the step that confuses me the most. I realize what it does algebraically, but I don't see how it was seen that that was what was needed to be done.
     
  15. Sep 11, 2010 #14

    vela

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    It's because you can always integrate something of the form fn(x)f'(x) using the substitution u=f(x):

    [tex]\int f^n(x) f'(x)\,dx = \left\{\begin{array}{lr}\frac{f^{n+1}(x)}{n+1}+c & \textrm{ when }n \ne -1 \\ \log|f(x)|+c & \textrm{ when } n = -1 \end{array}\right.[/tex]

    By dividing by y'y, you turned both sides of the equation into that form.
     
    Last edited: Sep 11, 2010
  16. Sep 11, 2010 #15
    The equation does not contain the argument [itex]x[/itex] explicitly. You can decrease the order by making the parametric substitution:

    [tex]
    p \equiv y'
    [/tex]

    Then:

    [tex]
    y'' = \frac{d y'}{d x} = \frac{d y'}{d y} \frac{d y}{d x} = y' \, \frac{d y'}{d y} = p \frac{d p}{d y}
    [/tex]
     
  17. Sep 12, 2010 #16
    Thanks for the help, everyone. I understand the concepts a little bit better. I think the basic separable differential equations have been the hardest for me, because the algebra involved seems to vary the most.
     
  18. Sep 12, 2010 #17

    hunt_mat

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    This question was harder then the usual ones I have seen. Don't be too disheartened.
     
  19. Sep 12, 2010 #18
    No. You're assuming [itex]y(y')^2 \ne 0[/itex]. Otherwise y=0 or y'=0 or that y=k is a solution which is obvious by inspection since they're all derivatives in the DE. And in this case, y=k happens to be a particular case of the general solution we found. However, there are DEs which have "singular" solutions which are not a particular case of the general solution and this mechanism, that division thing, is one way to find them.
     
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