- #1
Firepanda
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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
and
P''(x) = -P(x) * b^2
Now, his differntial equation for Q(t) is
Q(t) = Ae^-(cb)^2 t
and
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..
This is where I get stuck
He has derived this far
Q'(t) = -Q(t) * (cb)^2
and
P''(x) = -P(x) * b^2
Now, his differntial equation for Q(t) is
Q(t) = Ae^-(cb)^2 t
and
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..