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Differential equation

  1. Dec 22, 2008 #1
    This whole question is to do with a 1D heat equation.

    This is where I get stuck

    He has derived this far

    Q'(t) = -Q(t) * (cb)^2


    P''(x) = -P(x) * b^2

    Now, his differntial equation for Q(t) is

    Q(t) = Ae^-(cb)^2 t


    P(x) = Bcos(bx) + Csin(bx)

    How did he get these?

    Only way I can think of is for the second one where i would rearrange into a homogeneous equation

    P'' + b^2 P = 0

    Where P = 0 or -b^2 (from auxillery eqn)

    so the general solution would be P(x) = B + Ce^-b^2 x

    But this isn't what he found..
  2. jcsd
  3. Dec 22, 2008 #2


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    Homework Helper

    [tex]Q'(t) = -Q(t)(cb)^2[/tex]


    [tex]\int \frac{Q'(t)}{Q(t)}dt= \int -(cb)^2 dt[/tex]

    and now d/dt{lnQ(t)}=Q'(t)/Q(t)

    For P''(x) = -P(x) * b^2

    P''(x)+P(x) * b^2=0

    the auxiliary equation would be r2+b2=0
    What kind of roots would this give?
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