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A tricky differential equation

  1. Mar 9, 2004 #1
    Hello :smile:

    I have a differential equations, which I don't know how to solve. Maybe somebody has a tip or two

    The equation is:
    d^2 m / d x^2 = -3x
  2. jcsd
  3. Mar 9, 2004 #2


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    So m is a function of x and its second derivative with respect to x is -3x? What function gives ax when differentiated? And then what function gives THAT function when differentiated? Do you have any initial conditions?
  4. Mar 9, 2004 #3
    Well, I need to find the complete solution. All I know is that I have a function m(x) and that the interval of x is from 0 to 4.
  5. Mar 9, 2004 #4

    matt grime

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    what selfadjoint was getting at was that this is a very easy separable solution, though you've not met those words yet perhaps.

    suppose y' = 2x, ( ' means diff wrt x)

    y= x^2+c

    suppose z'' = y' in the above,
    then z' = x^2+c,

    so what is z?
  6. Mar 13, 2004 #5
    I haven't taken diff. eq. in a while, so I might be wrong, but isn't that like saying:

    m" = -3x ?

    then just integrate: (I'm using | as the symbol for integral)

    |(m" dm) = |(-3x dx)
    m' = -3/2 x^2 + C

    integrate again, gives:
    |(m' dm) = | (-3/2 x^2 + C dx)
    m = -1/2 x^3 + Cx + D

    That's it!

    I don't know what the interval from 0 to 4 falls-in though....
  7. Mar 24, 2004 #6
    Well, wouldn't the interval be just a definite integral from 0 to 4? I may be wrong, if I misread your question. [b(]
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