Posted by Erica on Thursday, January 31, 2013 at 3:41pm.
dy/y = dt/t^2
ln y = -1/t + c
y = e^(-1/t+c) = e^c e^(-1/t)
=C e^(-1/t) agree
C can be anything so
what if C = 0 ?
then y(t) = 0 for all t
if C = 0 for t</= 0 then C can be anything at all for t>0
If y(t) depends on what C is, then how this equation doesn't contradict the uniqueness theorem if it has many solutions?
Because the general solution contains an an arbitrary constant C. The value of C depends on your boundary conditions, for example if y =5 at t = 2
then
5 = C e^-(1/2)
5 = C (.606)
C = 8.24
NOW your solution is unique.
scroll down through this:
http://hyperphysics.phy-astr.gsu.edu/hbase/diff.html
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