Expert: Paul Klarreich - 2/19/2007

Question
hi i would like some help with this question.            dV/dt= -sqrt(V)
find an expression for V in terms of t.
this is what i tried: dt=dV/-sqrt(V); i then integrated both sides where 'int' represents the 'integral symbol'
int dt = int dV/-sqrt(V)
t+C(constant)=-2V^(1/2)
[(t+C)^2]/[(-2)^2] = (V^1/2)^2
[(t+C)^2]/4=V
this seems correct to me but the problem i'm having is that if i use at the beginning int -dt = int dV/sqrt(V) instead of int dt = int dV/-sqrt(V) my answer is V=[(c-t)^2]/4  why is this the case please help which answer is correct or are they both? thankz in advance!

Answer
Questioner:   A
Category:  Advanced Math
Question:  hi i would like some help with this question.            dV/dt= -sqrt(V)
find an expression for V in terms of t.
this is what i tried: dt=dV/-sqrt(V); i then integrated both sides where 'int' represents the 'integral symbol'
int dt = int dV/-sqrt(V)
t+C(constant)=-2V^(1/2)
[(t+C)^2]/[(-2)^2] = (V^1/2)^2
[(t+C)^2]/4=V
this seems correct to me but the problem i'm having is that if i use at the beginning int -dt = int dV/sqrt(V) instead of int dt = int dV/-sqrt(V) my answer is V=[(c-t)^2]/4  why is this the case please help which answer is correct or are they both? thankz in advance!
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Hi, A,

A first note:  Was there anything wrong with the first answer? I mean the one on related rates.  I haven't heard anything from you; perhaps you didn't get the email notice of the answer.

About this one -- usually there are more than one variations on the 'separation of variables' method, which is what you are using.  Ordinarily, there will be more than one place to put the constant, for example, and there is likely to be some initial condition (like,  V = something when t = something.), which you don't have here.

So my suggestion(s):

A. Note that  (t + C)^2,  (t - C)^2,  (C - t)^2, etc, are all the same expression when it comes down to it.

B. Take each one and do what your fourth-grade teacher said when she was teaching you subtraction:  CHECK.  Take your answer: V(t) and substitute into the differential equation and see if it works.  

[Also -- If you make the subject line refer to the problem, that makes it more useful.  A subject line like "math" or "Please help" is useless.  No big deal, though -- I can change it myself, and frequently do.]