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Tough Q.

  1. Dec 30, 2004 #1
    P.T y=f(x) given by
    [tex]
    \frac {d^2y} {dx^2} + ye^x=0
    [/tex]
    is bounded.ie there exists a c such that f(x)<c for all x.
    I am clueless.How do I begin? :confused: :cry:
    Thanks in advance.
     
  2. jcsd
  3. Dec 30, 2004 #2
    Would it have been better if I had posted it under differential eqns?
    Anyway we are not taught to solve these eqns.
     
    Last edited: Dec 30, 2004
  4. Dec 30, 2004 #3

    Integral

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    You have what is called a seperable equaition. This means you can write it as 2 equations 1 in x the other in y.

    [tex] \frac {d^2y} y + e^x {dx^2}=0 [/tex]

    Now just integrate the pieces.
     
  5. Dec 30, 2004 #4
    Is
    [tex]\int\frac{d^2y}{y} = \frac{dy}{y}+ \int \frac{dy}{y^2}[/tex]
    Thanks for the help
     
    Last edited: Dec 30, 2004
  6. Dec 30, 2004 #5

    Hurkyl

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    This makes me very uneasy -- second derivatives are not nearly as robust as first derivatives with respect to abusing notation... you sure this leads to something reasonable?
     
  7. Dec 30, 2004 #6

    arildno

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    Here's a hint:
    1)Change your independent variable as follows: [tex]u=e^{x}[/tex]

    2) Verify that your differential equation, rewritten in u, satisfies:
    [tex]\frac{d^{2}y}{du^{2}}+\frac{1}{u}\frac{dy}{du}+\frac{1}{u}y=0[/tex]

    3) What sort of dynamical system does this diff. eq. remind you of?
     
  8. Dec 30, 2004 #7
    I am in 12th grade . I am not supposed to know how to solve bessel functions or series solution of D.Es . It will be highly ironic if someone would have to know what he will be taught in the second year in a university to get admission to the university.
    This Q., came in a mock IIT joint entrance exam.
    Is there a way of proving this without solving the eqn?
    Thanks anyway.
    P.S:Ignore my formatting:)
     
  9. Dec 30, 2004 #8

    arildno

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    All right, then:

    But do you see that your ORIGINAL equation looks a lot like the one governing the motion of a spring, with [tex]e^{x}[/tex] instead of a constant?
     
  10. Dec 30, 2004 #9
    True.
    How does it help me prove that the f(x) is bounded?
     
  11. Dec 30, 2004 #10

    dextercioby

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    Well,Hurkyl,let's see if there's something wrong with my & Integral's logics:
    [tex] \frac{d^{2}y}{dx^{2}}=-ye^{x} [/tex]

    [tex] \int (\int \frac{dy}{y}) dy=-\int(\int e^{x}dx) dx [/tex]

    [tex] \int \ln y dy=-e^{x} +Cx+D [/tex]

    [tex] y(\ln y-1)=-e^{x}+Cx+D' [/tex](1)

    It's impossible to put the solution as y=y(x).Since we're interested in the asymptotic behavior of y,le's study (1) when
    a)[tex] x\rightarrow +\infty [/tex]
    Then the RHS of (1) goes to [itex] -\infty [/itex],which means that the 'y' in the LHS goes to 0.
    b) [tex] x\rightarrow -\infty [/tex]
    Then for C>0,RHS goes to [itex] -\infty [/itex],which means 'y' in the LHS goes to 0.,and for C<0,RHS goes to [itex] +\infty [/itex],which means 'y' in the LHS goes to [itex] +\infty [/itex].

    So imposing [itex] C\geq0 [/itex],u have that,in asymptotic limits,'y' is bounded.

    I don't know whether Arildno's approach would yield different conclusions.Afer all,that's a messy equation.If the spring's 'constant' is not constant,but varies exponentially with the departure from the initial (equilibrium) solution,it means that string would stretch to infinity,because the constant would be infinite.However,there's damping,which again varies exponentially with 'x',so the string could go to infinity only under certain circumstances (in my analysis,C<0) and could have a bounded highly damped oscillation for C>=0.

    Daniel.
     
    Last edited: Dec 30, 2004
  12. Dec 30, 2004 #11

    Hurkyl

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    Try it on a simpler problem, like

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

    With your method, I get the equation:

    y ln y - y = (1/2) x^2 + Cx + D

    Neither y = e^x nor y = e^-x are solutions to this equation.
     
    Last edited: Dec 30, 2004
  13. Dec 30, 2004 #12

    Gokul43201

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    I believe poolwin's syllabus does not include differential equations; but does have basic differentiation and integration, including applications to rates, maxima and minima. So, I think his only exposure to the second derivative is its definition as the derivative of the first derivative, that it describes the concavity/convexity of the function, and that its zeros are points of inflexion(inflection).

    Poolwin, will you please clarify if this is correct ? Also, do you only have to show that it is bounded above, and not below too ?
     
    Last edited: Dec 30, 2004
  14. Dec 30, 2004 #13
    My syllabi includes homogenous,linear eqn of 1st orderbut not as I mentioned earlier series soln of D.E or P.D.Es etc in addn to what 43201 said.
    I have given the Qs as is it was given.
    f(x)<c indicates that may be boundedness is only for the upper limiting value

    For the record it came in a mock iit exam held by iit alumni the 'TENSORS'.
     
    Last edited: Dec 31, 2004
  15. Dec 30, 2004 #14
    I forgot to add sadistic.
     
  16. Jan 15, 2005 #15
    I guess the reason why no one is helping is maybe becoz the no: posts is 13.:)
     
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