Another Diff Eq's one: am I wrong?

In summary, The problem is that two different people have found two different solutions to a differential equation. The attempt at a solution given in the solution manual is wrong because the assumptions made are not valid. However, if you plug in the solution given in the problem, it matches the solution given in the solution manual.
  • #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

c3PgDgU.png


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

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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  • #3
Dr. Courtney said:
You can always check by plugging it back in.

assume

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

[tex]0 = \frac{r}{k} - \frac{r}{k}e^{-kt}[/tex]

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

So, either r = 0 or [itex]e^{-kt} = 1[/itex]

That doesn't seem right.
 
  • #4
I meant double check each in the original diff eq.

Then double check boundary conditions.
 
  • #5
kostoglotov said:
[tex]C(t) = \frac{r - (r-kC_0)e^{-kt}}{k}[/tex]
[tex]C(t) = C_0 e^{-kt}[/tex]
That step is clearly wrong.
 
  • Like
Likes kostoglotov and Dr. Courtney
  • #6
vela said:
That step is clearly wrong.

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

1. Is there only one solution to a differential equation?

No, there can be multiple solutions to a differential equation. The number of solutions depends on the type of equation and the initial conditions.

2. How do you solve a differential equation?

There are various methods for solving a differential equation, including separation of variables, integrating factors, and using power series. The method used depends on the type of equation and its complexity.

3. Can every differential equation be solved analytically?

No, there are some differential equations that cannot be solved analytically. These are known as non-analytical or non-elementary equations. In such cases, numerical methods can be used to approximate the solutions.

4. What is the difference between ordinary and partial differential equations?

The main difference between ordinary and partial differential equations is the number of independent variables. Ordinary differential equations involve only one independent variable, while partial differential equations involve multiple independent variables.

5. Are differential equations only used in math and science?

No, differential equations have applications in various fields such as physics, engineering, economics, and biology. They are used to model and understand real-world phenomena and make predictions about their behavior.

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