# Homework Help: Constant of integration's sign

1. Mar 20, 2014

### DocZaius

1. The problem statement, all variables and given/known data

Solve for A:

dA/ds = -k s

2. Relevant equations

See problem statement

3. The attempt at a solution

I switched the equation around:
dA= -k s ds

Integrated:

A= -(k/2) (s^2+c)

Apparently, that is wrong and I see that the answer should be:

A= -(k/2) (s^2-b^2)

My question is: Why must the constant of integration subtract from s^2? Why couldn't it add to it?

2. Mar 20, 2014

### LCKurtz

You wouldn't write the constant as $-b^2$ unless you had another condition given, such as an initial condition. Writing it that way implies the constant must be negative (or positive if you multiply the other minus sign through). In general you wouldn't do that. It doesn't matter whether you add or subtract the constant of integration, but it does matter if you square it, making the term positive.

3. Mar 20, 2014

### DocZaius

Couldn't the constant of integration be complex? Are you allowed to have that when s is real?

4. Mar 20, 2014

### LCKurtz

If, as in your example, the functions in the DE are real and the boundary conditions are real, the solution will be real. You would typically write the solution as$$y = -\frac{ks^2}{2} + C$$Equivalently you could write it as$$y = -\frac{ks^2}{2} - C$$In neither case would you use $C^2$ without additional information about the DE. I thought that was what you were concerned about.

5. Mar 20, 2014

### DocZaius

It was what I was concerned about. I was trying to think of some reason why -b^2 was there, and a complex b was the only thing that could make it work, since that squared must be real and could be either positive or negative. Oh well, thank you for your help.

edit: Nevermind even the square of b could be complex if b was complex. I was thinking of the absolute value. I give up!

Last edited: Mar 20, 2014
6. Mar 20, 2014

### LCKurtz

You're welcome. The upshot of this is that you wouldn't normally use that $-b^2$ without more info. So you were right on the money to wonder about it. There is probably more to that problem.