Solving Differential Equation: ln |9/64| = k

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kyu
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i got k = ln |9/64|

then how can the next step using ln 0 doesn't make sense. what should i do?
 

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Mark44 said:
Please show us your work.

should be wrong but here goes

dh/dt = -k h^(1/2)
1/h^(1/2) = -k dt
ln 9 - ln 64 = k
ln |9/64| = k

ln 0 - ln 64 = ln |9/64| (t-0)
 
kyu said:
should be wrong but here goes

dh/dt = -k h^(1/2)
1/h^(1/2) = -k dt
Where did the dh go?
kyu said:
ln 9 - ln 64 = k
How did you get this (above)?
kyu said:
ln |9/64| = k

ln 0 - ln 64 = ln |9/64| (t-0)
 
kyu said:
should be wrong but here goes

dh/dt = -k h^(1/2)
1/h^(1/2) = -k dt
ln 9 - ln 64 = k
ln |9/64| = k

ln 0 - ln 64 = ln |9/64| (t-0)

The integral of [itex]h^{-1/2}[/itex] is given by the usual rule for powers, [itex]\int x^\alpha\,dx = \frac{x^{\alpha + 1}}{\alpha + 1} + C[/itex], not the exception [itex]\int x^{-1}\,dx = \ln |x| + C[/itex].
 
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