Another Diff Eq's one: am I wrong?

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Homework Help Overview

The discussion revolves around a differential equation problem where the original poster is comparing their solution to that provided in a solution manual. They express uncertainty about the validity of both answers and explore the potential equivalence of their solutions through algebraic manipulation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to reconcile their solution with the one in the manual, questioning whether both can be valid. They explore the implications of combining constants and seek to understand the advantages of the manual's form.

Discussion Status

Participants are engaged in examining the original poster's reasoning and the correctness of their steps. Some suggest verifying the solutions by substituting them back into the original differential equation, while others point out potential mistakes in the original poster's algebra.

Contextual Notes

There is mention of boundary conditions and the impact of initial conditions on the solutions being discussed. The original poster acknowledges making errors during their study, which may affect their reasoning.

kostoglotov
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Homework Statement



Sorry to be bringing these in quick succession, but this one really has me perplexed. Is it possible that both the solution manual and I have two different but valid answers?

I don't want to just go assuming that I'm right...but I think by subsuming certain parts of the solutions manuals constants in it's solution into constants in my solution, that the two can become equivalent...?!

The problem

c3PgDgU.png


Imgur link to the problem: http://i.imgur.com/c3PgDgU.png

Homework Equations

The Attempt at a Solution



\int \frac{1}{r-kC} dC = \int dt

\frac{-1}{k} \ln{|r-kC|} = t+K_0

\ln{|r-kC|} = -kt + K_1

skipping a couple of steps

C(t) = \frac{r-Ae^{-kt}}{k}

with init. conditions

C(0) = C_0 = \frac{r-A}{k}

A = r - kC_0

So

C(t) = \frac{r - (r-kC_0)e^{-kt}}{k}

C(t) = C_0 e^{-kt}

The answer in the back is

C(t) = \left(C_0 - \frac{r}{k} \right)e^{-kt} + \frac{r}{k}

And the solution given in the solution manual is

RVrLe1R.png


Link to solution: http://i.imgur.com/RVrLe1R.png

I can see that they've don'e some minor things different algebraically and with subsuming certain constants into other constants...I feel like my solution should be equivalent though. Is it? If it is, are there any particular advantages to the form the solution manual has given it in? Be they practical advantages or purely mathematical advantages?
 
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You can always check by plugging it back in.
 
Dr. Courtney said:
You can always check by plugging it back in.

assume

C_0e^{-kt} = \left(C_0 - \frac{r}{k}\right)e^{-kt}+\frac{r}{k}

0 = \frac{r}{k} - \frac{r}{k}e^{-kt}

0 = \frac{r}{k}\left(1 - e^{-kt}\right)

So, either r = 0 or e^{-kt} = 1

That doesn't seem right.
 
I meant double check each in the original diff eq.

Then double check boundary conditions.
 
kostoglotov said:
C(t) = \frac{r - (r-kC_0)e^{-kt}}{k}
C(t) = C_0 e^{-kt}
That step is clearly wrong.
 
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vela said:
That step is clearly wrong.

Thank you. Whenever I do study after a night shift I make stupid mistakes. Cheers :)
 

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