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

Solved

Posted on 2006-05-26

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.

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.

27 Comments

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

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

Putting A = 0 would give P = 0

So,

dP

-- = 2P - 2Pt

dt

will be satisfied for that case.

>>-- = 2P - 2Pt

>>dt

>>

>>will be satisfied for that case.

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

>>-- = 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

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?

>>dy

>>--- = 0

>>dx

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

--- = 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

>>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?

>> 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

dP

-- = 2P - 2Pt

dt

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

p = 0 when A = 0

dP

--- = 0

dt

>> 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?

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.

d^2 Y

------- = Y

dt^2

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

I think I'm getting it.

you evaluate both sides of the equation, and determne whether they are equal.

By clicking you are agreeing to Experts Exchange's Terms of Use.

Title | # Comments | Views | Activity |
---|---|---|---|

CAGR Calculation For SIP | 13 | 55 | |

Relative Frequency | 5 | 42 | |

Binomial distribution | 2 | 41 | |

Access question - internal training | 9 | 31 |

Join the community of 500,000 technology professionals and ask your questions.

Connect with top rated Experts

**8** Experts available now in Live!