- #1
kostoglotov
- 234
- 6
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
Imgur link to the problem: http://i.imgur.com/c3PgDgU.png
Homework Equations
The Attempt at a Solution
[tex]\int \frac{1}{r-kC} dC = \int dt[/tex]
[tex]\frac{-1}{k} \ln{|r-kC|} = t+K_0[/tex]
[tex]\ln{|r-kC|} = -kt + K_1[/tex]
skipping a couple of steps
[tex]C(t) = \frac{r-Ae^{-kt}}{k}[/tex]
with init. conditions
[tex]C(0) = C_0 = \frac{r-A}{k}[/tex]
[tex]A = r - kC_0[/tex]
So
[tex]C(t) = \frac{r - (r-kC_0)e^{-kt}}{k}[/tex]
[tex]C(t) = C_0 e^{-kt}[/tex]
The answer in the back is
[tex]C(t) = \left(C_0 - \frac{r}{k} \right)e^{-kt} + \frac{r}{k}[/tex]
And the solution given in the solution manual is
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?