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Homework Help: Differential Equation

  1. Jan 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Solve the differential equation...
    X'' = .5*x^-2
    taken with respect to t
    at t=0, x'=0 and x=10

    2. Relevant equations
    N/A


    3. The attempt at a solution
    I tried to split the variables (ie d^2 X * X^2 = .5 dt^2) but didn't get the right answer with this (i plugged it bak in and it didn't work?)
    How should I start this problem?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 14, 2010 #2
    Show your computations.
     
  4. Jan 14, 2010 #3
    d^2 X * X^2 = .5 dt^2
    integrate both sides...
    dx * x^3/3=(.5*t + C) *dt
    at t=0, dx/dt=0 and x=10...
    0*10^3/3 = .5*0 + C
    C=0

    dx * x^3/3 = .5*t*dt
    integrate both sides again...
    x^4 / 12 = .25*t^2 + C
    same refs....
    10^4/12 = .25*0 +C
    C = 833.3

    x^4/12 = .25*t^2 + 833.3

    algebra...

    x= (3*t^2 + 10000)^(1/4)

    but that doesn't work...
     
  5. Jan 14, 2010 #4

    rock.freak667

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    Do you know the variation of parameters method? That might work.
     
  6. Jan 14, 2010 #5
    I didn't think that would work because I thought the exponent on the x term would have to be 1, while in this case its -2...
     
  7. Jan 14, 2010 #6

    Dick

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    Just integrate (1/2)x^(-2) twice. Don't forget to keep the constants of integration around.
     
  8. Jan 14, 2010 #7
    so....
    integrating once...
    x'=.0345*x^-2*t

    then again...
    x=.0345/2*x^-2*t^2+C\
    ???
     
    Last edited: Jan 14, 2010
  9. Jan 14, 2010 #8

    Dick

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    No, you aren't missing anything. I am. I didn't notice t was the independent variable.
     
  10. Jan 15, 2010 #9

    HallsofIvy

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    NO. You cannot separate a second derivative like this. A second derivative cannot be treated like a fraction.

    Your original equation can be written as [itex]x^2d^2x/dt^2= 1/2[/math]. Let y= dx/dt. Then, by the chain rule, [itex]d^2x/dt= dy/dt= (dy/dx)(dx/dt)= y dy/dx. Your equation becomes y dy/dx= (1/2)x-2. Since that is now a first derivative, it can be treated like a fraction and separated: ydy= (1/2)x-2dx. Integrate that to find y, as a function of x, and then integrate dx/dt= y to find x.` Since you know that x(0)= 10 and x'(0)= 0, you know that y= 0 when x= 10 so can find the first constant of integration before the second integral.

     
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