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Please help me!

  1. Mar 20, 2005 #1
    Could anyone teach how to solve the following differential equation?

    [tex]\frac{d^2 y}{dx^2}+2\left(\frac{dy}{dx}\right)^2+y=0[/tex]
    Last edited: Mar 20, 2005
  2. jcsd
  3. Mar 20, 2005 #2


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    Well this is a 2nd order nonlinear homogeneous differential equation. Its tougher because the y' is of 2nd power.
    Last edited: Mar 20, 2005
  4. Mar 20, 2005 #3
    Make the substitution [tex] v = \frac{dy}{dx}[/tex] so that [tex]v \frac{dv}{dy} = \frac{d^2y}{dx^2}[/tex]. This results in a first-order ODE (where y is now the indep. variable). Solve this DE for [itex]v(y)[/itex], then sub back in [itex]\frac{dy}{dx}[/itex] for [itex]v[/itex] to get a first-order DE for [itex]y(x)[/itex] and solve (I don't have time to do it right now, so it may not be very pretty).

    The reason that this method works in this case is that [itex]x[/itex] does not appear explicitly in the DE.
  5. Mar 20, 2005 #4


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    How are you going to solve it if its v^2 ?
  6. Mar 20, 2005 #5
    [tex] \frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^2 + y = 0 \Longrightarrow
    v\frac{dv}{dy} + 2v^2 + y = 0[/tex]

    which is an inexact DE, ie. it is in the form [tex] P(y, \ v) + Q( y, \ v)v^\prime = 0[/tex] with [tex] \frac{\partial P}{\partial v} \neq \frac{\partial Q}{\partial y}[/tex]. We look for an integrating factor [itex]I(y)[/itex] such that

    [tex] \ln{I(y)} = \int \frac{P_v -Q_y}{Q} dy[/tex]

    In this case, [tex]Q(y, \ v) = v[/tex] and [tex]P(y, \ v) = 2v^2 + y[/tex] so

    [tex] \ln{I(y)} = \int \frac{ 4v - 0 }{v} dy = \int 4 dy \Longrightarrow I(y) = e^{4y}[/tex]

    note that not including a constant of integration makes no difference for our purposes. From the above, we know that

    [tex]e^{4y}(2v^2 + y) + e^{4y}v\frac{dv}{dy} = 0[/tex]

    is an exact equation.

    Then you need to know how to solve exact equations, but that's not hard~

    Note that I skipped a bunch of steps on the theory of inexact first-order ODEs (as in, why did I look for an integrating factor that way?), but you can look those up or just ask and I can post them~
    Last edited: Mar 20, 2005
  7. Mar 21, 2005 #6


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    Come on guys. I'd like to see this to completion. Maphysique, can you solve the exact equation in v, and get an expression in terms of F(y,v)=c? I end up with an expression for y' that I cannot integrate (a radical). Don't want to interfere with Data helping you though.
  8. Mar 21, 2005 #7


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    Getting an expression for y' that you cannot integrate is fairly typical of problems like this. In general, solutions to non-linear equations cannot be written in terms of elementary functions. Do you have reason to think that this one can be?
  9. Mar 21, 2005 #8
    Yeah, the resulting integral for y is not nice at all. But at least he has it down to an integral, which is perfectly sufficient for approximating a solution :).
  10. Mar 21, 2005 #9


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    Thanks Data. When I first saw the problem, I didn't think to solve it that way. Just in case you guys don't feel like typing it all in, I'll do the next part. Perhaps Maphysique will follow it through:

    I'll write the exact equation as follows:

    [tex][e^{4y}(2v^2 + y]dy + [e^{4y}v]dv = 0[/tex]

    "Exact" means its a total differential of some function F(y,v), that is:

    [tex]dF=[e^{4y}(2v^2 + y]dy + [e^{4y}v]dv[/tex]

    Then surely:

    [tex]\frac{\partial F}{\partial y}=[e^{4y}(2v^2 + y][/tex]


    [tex]\frac{\partial F}{\partial v}=[e^{4y}v][/tex]

    To find F(y,v), let's integrate the partial with respect to y:

    [tex]\int{\frac{\partial F}{\partial y}=\int{[e^{4y}(2v^2 + y]}dy[/tex]

    And thus:


    Where integrating the partial with respect to y necessarilly yields some arbitrary function of v, T(v).

    Differentiating F(y,v) with respect to v now yields:

    [tex]\frac{\partial F}{\partial v}=ve^{4y}+\frac{dT}{dv}[/tex]

    But according to the exact differential, this is also equal to [itex]e^{4y}v[/itex]





    Integrating yields:


    So that:


    Where k is a constant which we'll absorb into the constant c:


    Finally, solving for v (which is y') yields:



    [tex]\frac{dy}{dx}=\pm \sqrt{ce^{-4y}-\frac{1}{2}y+\frac{1}{8}}[/tex]

    But that's ok. You just solve two differential equations now and use whatever part meets the initial conditions. Anyway, I'll spend some time studying this numerically in Mathematica and report some useful results with select initial conditions.
  11. Mar 21, 2005 #10
    Thank you for your replies,and sorry for my delaying reply,everyone.

    I have just solved for myself without using the solution by exact differential equation.

    Here is my solution.

    As pointed by Data, Set


    we can rewrite the differential equation in question


    as follows.


    For [tex]v\neq 0[/tex],

    we have

    [tex]\frac{dv}{dy}+2v=-\frac{1}{v}y\ \cdots [E][/tex]

    Considering the solution of the differential equation


    we set

    [tex]v=ze^{-2y}[/tex],differentiating with respect to [tex]y[/tex]


    plugging this into [tex][E][/tex]

    we have



    integrating with respect to [tex]y[/tex]

    [tex]\frac{1}{2}z^2=-\int ye^{4y}dy[/tex]

    using integral by parts




    where [tex] C=2C' [/tex]

    therefore we obtain



    Consequently,as deduced by saltydog,

    [tex]\frac{dy}{dx}=\pm \sqrt{Ce^{-4y}-\left(\frac{1}{2}y-\frac{1}{8}\right)}[/tex]

    Thank you.

    Last edited: Mar 21, 2005
  12. Mar 21, 2005 #11


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    Hello Maphysique. Glad you're still with us and can explain how you approached it. This is my study of it:

    It's interesting to note that if one would have approached this equation by converting it to two ODEs:



    The constraint on y described below would not have been immediately clear unless the particular application it was describing would have indicated such.

    As stated earlier, the equation is reduced to:


    Solving (for c=1):


    we obtain [itex]x\approx 0.51003[/itex] and becomes negative when x is greater than this value. Thus, the ODE is valid only when y is less than the root of this equation.

    Looking at the ODE again:

    [tex]\frac{d^2 y}{dx^2}+2\left(\frac{dy}{dx}\right)^2+y=0[/tex]

    Since y' is squared, then y'' must necessarilly be negative for the LHS to have any chance of being zero.

    If y'(0)>0 then the graph slopes upward and concaving downward until y reaches the zero of the radical.

    If y'(0)<0 then the graphs slopes downward and concaving downward and becomes more so to balance the square of the derivative.

    Attached are two particular solutions with y'(0)>0 and y'(0)<0 respectively.

    Attached Files:

    Last edited: Mar 21, 2005
  13. Mar 21, 2005 #12
    actually, salty, I believe the solution to the exact DE is

    [tex]\frac{dy}{dx} = \pm \sqrt{ ce^{-4y} - \frac{y}{2} + \frac{1}{8}}[/tex]

    (note the [itex]ce^{-4y}[/itex] as opposed to [itex]ce^{4y}[/itex])

    Edit: I just noticed you had it right in your first post. Undoubtedly just a typo~
    Last edited: Mar 21, 2005
  14. Mar 21, 2005 #13
    And maphysique, that is a nice method. Unfortunately it doesn't lead to any additional simplifications that I can see (you still get the messy square root), but maybe you've figured out how to get around that too! :)
  15. Mar 21, 2005 #14
    I have just edited.

    Thank you, everyone!

  16. Mar 22, 2005 #15
    Un petit coup de Laplace devrait marcher....
  17. Mar 22, 2005 #16
    If your suggestion is to use the Laplace transform, then tell me what the Laplace transform of [itex]f^\prime(x)^2[/itex] is~
  18. Mar 23, 2005 #17


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    U can't use Laplace transformation for nonlinear ODE-s...

  19. Mar 24, 2005 #18
    "U can't use Laplace transformation for nonlinear ODE-s...


  20. Mar 24, 2005 #19
    Like I asked before, what's the Laplace transform of [itex]f^\prime(x)^2[/itex]?
  21. Mar 24, 2005 #20


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    I assume it's very easy to ask questions without doing some documenting first.I think there are plenty of books on the application of integral transformations (Laplace & Fourier) to PDE-s/ODE-s...You can find one,i'm sure.

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