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Differential Equations (Seperation of Variables)

Posted on 2006-05-26
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Last Modified: 2010-08-05
Find and Sketch the solution to satisfying P = 5 when t = 0.

dP
-- = 2P - 2Pt
dt

The first part of this problem makes plenty of sense:

dP
--- = 2 - 2t dt
 P

ln |P| = 2t - t^2 + C

|P| = e^(2t - t^2 + C)
|P| = e^C * e^(2t - t^2)

A = e^C

|P| = Ae^(2t - t^2)



Now this next part in the book doesnt make sense to me:

"so A > 0 (Makes Sense). In addition, A = 0 gives a solution (WHAT?? HOW?) Thus the general solution to the differential equation is P = Be^(2t - t^2) for any B (Again how so?)"

Then they go on to find the value of B by plugging in t = 0 and P = 5. (this seems easy enough).

I dont get where they just threw in the B from or what they are talking about when they say A = 0 is a solution.
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Comment
Question by:BrianGEFF719
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27 Comments
 
LVL 85

Expert Comment

by:ozo
ID: 16776029
Multipying a soultion by any non zero constant won't change
dP
--
 P
multiplying P by 0 would make sense if the equation was
dP = P * (2 - 2t dt)
for which P=0 is an obvious solution, instead of
dP
--- = 2 - 2t dt
 P
but if P=0 then dividing both sides by P would not be valid
0
 
LVL 19

Author Comment

by:BrianGEFF719
ID: 16776297
Ozo, I dont quite understand. Maybe I dont quite understand what you mean by a solution.
0
 
LVL 37

Expert Comment

by:Harisha M G
ID: 16779661
You have this:

|P| = Ae^(2t - t^2)

Putting A = 0 would give P = 0

So,

dP
-- = 2P - 2Pt
dt

will be satisfied for that case.
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LVL 19

Author Comment

by:BrianGEFF719
ID: 16780383
>>dP
>>-- = 2P - 2Pt
>>dt
>>
>>will be satisfied for that case.


How so, I do not understand., P = 0 Gives dP/dt = 0.
0
 
LVL 19

Author Comment

by:BrianGEFF719
ID: 16780387
how is that a "SOLUTION" though?
0
 
LVL 85

Expert Comment

by:ozo
ID: 16781278
>>dP
>>-- = 2P - 2Pt
>>dt
>>
>>will be satisfied for that case.


How so, I do not understand., P = 0 Gives dP/dt = 0.

P = 0 also gives 2P - 2Pt = 0
so 0 = 0 is satisfied
0
 
LVL 19

Author Comment

by:BrianGEFF719
ID: 16782709
I dont get the idea of how the rate at which a function changes is dependant on the function...it's kinda throwing me through a loop.
0
 
LVL 19

Author Comment

by:BrianGEFF719
ID: 16782716
Also,

I'm not getting these zero solutions.

dy
--- = 0, does not imply that y(x) = 0.
dx




0
 
LVL 85

Expert Comment

by:ozo
ID: 16782740
no, but y(x) = 0 does imply that
dy
--- = 0
dx

so y(x) = 0 satisfies
dy
--- = y(x)
dx
0
 
LVL 19

Author Comment

by:BrianGEFF719
ID: 16782796
Oh I think I got it ozo.

dP
-- = 2P - 2Pt
dt

P is a function of t.

So when p(t) = 0, 2p = 0 and - 2p = 0
thus

dP
--- = 2P - 2Pt  -> is true when P = 0.
dt



Is this logic correct?
0
 
LVL 19

Author Comment

by:BrianGEFF719
ID: 16782799
>>no, but y(x) = 0 does imply that
>>dy
>>--- = 0
>>dx


Okay that confused me, y(x) = x = 0, dy/dx = 1 ?
0
 
LVL 85

Expert Comment

by:ozo
ID: 16782821
dP
--- = 2P - 2Pt  -> is true when P = 0.
dt

Yes.

Okay that confused me, y(x) = x = 0, dy/dx = 1 ?
No, not  y(x) = x
y(x) = 0

y(x) = 0 and P(t) = 0 are the same function
0
 
LVL 19

Author Comment

by:BrianGEFF719
ID: 16782862
>>No, not  y(x) = x
>>y(x) = 0

*DING* Got it. We are saying the function is a horizontal line y = 0. When this is the case then dy/dx is also = 0.

So let ask you:

 dY
 --- = Y + 3
 dt

Does not have a solution of zero because of that + 3?

0
 
LVL 85

Expert Comment

by:ozo
ID: 16782876
Y(t) = 0 is not a solution to
 dY
 --- = Y + 3
 dt
0
 
LVL 85

Expert Comment

by:ozo
ID: 16782880
Y(t) = -3 is  a solution to
 dY
 --- = Y + 3
 dt
0
 
LVL 19

Author Comment

by:BrianGEFF719
ID: 16782884
>>Y(t) = 0 is not a solution to
>> dY
>> --- = Y + 3
>> dt

So, a solution to this problem would require us to know what Y(t) is then, correct? We would then have to differentiate and set it equal to Y + 3, correct?


Brian
0
 
LVL 19

Author Comment

by:BrianGEFF719
ID: 16782893
I think I get why in my original question A = 0 is a solution:

dP
-- = 2P - 2Pt
dt

 P = Ae^(2t - t^2)

 p = 0 when A = 0

dP
--- = 0
dt
0
 
LVL 85

Expert Comment

by:ozo
ID: 16782896
Y(t) = e^t - 3
is another solution to
 dY
 --- = Y + 3
 dt
0
 
LVL 19

Author Comment

by:BrianGEFF719
ID: 16782897
>>Y(t) = -3 is  a solution to
>> dY
>> --- = Y + 3
>> dt


So when Y(t) = -3, dY/dt = 0, how does this qualify as a solution exactly? What are we trying to satisfy by showing that the derivative is equal to zero for a given function value?
0
 
LVL 85

Expert Comment

by:ozo
ID: 16782910
So when Y(t) = -3, dY/dt = 0, and Y+3 = 0
so
dY/dt = Y+3
0
 
LVL 19

Author Comment

by:BrianGEFF719
ID: 16782933
Okay, so to determine if something is a solution.

You say y(x) = {stuff}, then you differentiate y(x) -> y`(x) = {smaller stuff}

and if you have a differential equation

 dy
 ---  = AY + y^2 (or something)
 dx


You would plugin the value of Y and the value of dy/dx and if you have a true statment, y(x) = {stuff} is a solution.


Sorry for the nonsense in this post i'm very tried.
0
 
LVL 19

Author Comment

by:BrianGEFF719
ID: 16782957
So the solution to:


d^2 Y
------- = Y
 dt^2

is y(x) = cosh(x) or y(x) = sinh(x).


I think I'm getting it.
0
 
LVL 85

Accepted Solution

by:
ozo earned 1600 total points
ID: 16782961
to determine if something is a solution,
you evaluate both sides of the equation, and determne whether they are equal.
0
 
LVL 37

Assisted Solution

by:Harisha M G
Harisha M G earned 400 total points
ID: 16785253
So the solution to:


d^2 Y
------- = Y
 dt^2

is y(x) = cosh(x) or y(x) = sinh(x).

________

Also, y = e^t is a solution apart from cosh(x) and sinh(x)

Since,  d²y/dt² = d²(e^t)/dt² = e^t = y
0
 
LVL 19

Author Comment

by:BrianGEFF719
ID: 16785332
e^(-t) is also a solution.
0
 
LVL 37

Expert Comment

by:Harisha M G
ID: 16785486
Yes.. of course :)
0
 
LVL 19

Author Comment

by:BrianGEFF719
ID: 16785577
Thanks for your patience ozo and harish...I really do appricate it.
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