# Homework Help: Differential Equations Problem

1. Mar 17, 2013

### Daweih

1. The problem statement, all variables and given/known data
This is actually a question we went over in class, but I kind of spaced out when my teacher was explaining it. I have since solved it myself while I was reviewing for my calculus test, but when I compared my answer with the answer key my teacher provided to our class, I found out my answer was wrong. Here's the question.

"Write and solve the differential equation that models the statement in the following problem.

The rate of change of P with respect to t is proportional to 10-t."

2. Relevant equations
I should add that this is part of our chapter involving exponential growth and decay.

3. The attempt at a solution
Now, I understand how to do this particular problem. You essentially start off with the following differential equation:

dP/dt = K/(10-t)

You then multiply the dt over to isolate all the t variable expressions on the right side.

dP = [K/(10-t)]dt

Then you would take an indefinite integral of both sides respectively. The problem is, my teacher says that the answer should be:

P = -K*ln|10-t| + C

When I solved this on my own, I produced the same result essentially, just without a negative sign in front of the K constant. Am I missing something here? Could someone please explain to me why there should be a negative sign or did my teacher make a mistake?

2. Mar 17, 2013

### eumyang

Did you integrate the right side using u-substitution? After you use that method, the negative will appear in the answer.

3. Mar 17, 2013

### Daweih

Ah, no. I didn't. I tried integrating using substitution just now, and it worked. >.<
Thank you for pointing that out to me. It makes sense now.

4. Mar 18, 2013

### LCKurtz

When you say $y$ is proportional to $x$ that means there is a constant $k$ such that $y=kx$. So if the rate of change of $P$ is proportional to $10-t$, then $\frac{dP}{dt}= k(10-t)$. So you either are solving the problem wrong or stated the problem you are solving wrong.

Last edited: Mar 18, 2013
5. Mar 18, 2013

### Daweih

That's my mistake. I stated the problem incorrectly. It should say "inversely proportional". But regardless, eumyang pointed out what I was missing. I didn't use the substitution method to integrate the differential equation, so I never produced that negative sign.