# Clarification regarding the proper use of constant C.

1. Sep 24, 2011

### smithnya

Please forgive the rather simple nature of my question. I am rather confused about it. I have just started a class on differential equations and I am confused on the proper use of the constant C after integrating. There are occasions in which I end up with something like
y = et + c. If this is the same as etec, then is ec essentially the same as c? Can anyone enlighten me as to the proper use of c in similar situations such as e6t + 6c, etc.?

2. Sep 24, 2011

### Swalker

c is usually a variable you find so that your solution satisfies given initial conditions. Typically c is in front of the exponent. Ie c1 exp^2t.

3. Sep 25, 2011

### AlephZero

Remember that c is an arbitrary constant. There are often different ways to write the "same" constant value.

As a simple example, if you integrate $\int 2x\, dx$, it doesn't really matter if you write the answer is $x^2 + c$ or $x^2 + c + 1$ or $x^2 - c$, or whatever. The expressions give different values for a particular value of c (for example c = 0) but the complete set of expressions for every possible value of c is the same in each case.

But in that example, there isn't any obvious reason to write anything apart from $x^2 + c$, so that is what you will find in a textbook as "the answer".

In your example, the same idea applies, but the alternative ways to write the answer look "more different" than the previous example.

If $y = e^{t+c}$ then you are right that $y = e^t e^c$. You would normally let $e^c = a$ where $a$ is another constant, and write $y = ae^t$ rather than $y = e^t e^c$.

If this was part of a bigger problem, sometimes $y = e^{t+c}$ is easier to work with, and sometimes $y = ae^t$. You have to get used to the fact that in math, there are often different ways of writing "the same thing".

4. Sep 25, 2011

### smithnya

Thank you so much. Great explanation. It answered my question.

5. Sep 25, 2011

### lurflurf

Sometimes a particular form of the constant will cause one to gain or lose solutions.
if
y1=C exp(t)
y2=exp(t-a)
-exp(t) may of may or not be a solution y1 tends to make us think of it as one while y2 may make us forget about it (if working wiht real numbers).
if
y1=(x+B)/(-x+B)
we might forget to consider y=1 as a possible solution while some other form may remind us of that possiblity.