AboutScotto Expertise Any kind of mathematics (calculus, analysis, game theory, linear approximation, finite differences, linear regression, linear programming, numerical analysis, probability, statistics, etc.).
I also have answered some questions in
Physics (mass, momentum, falling bodies),
Chemistry (charge, reactions, symbols, molecules), and
Biology.
Experience Experience in the area: I have tutored students in all areas of mathematics for over 20 years.
Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors.
Awards and Honors: I have passed Actuarial tests 100, 110, and 135.
Publications Maybe not a publication, but I have respond to well oveer 3000 questions on the PC.
That's around 2,000 in basic math and 1,000 in advanced math.
Education/Credentials I aquired well over 40 hours of upper division courses. This was well over the number that were required.
I graduated with honors in both my BS and MS degree from Oregon State University.
I was allowed to jump into a few junior level courses my sophomore year.
Awards and Honors I have been nominated as the expert of the month several times.
All of my scores right now are at least a 9.8 average (out of 10).
Past/Present Clients My past clients have been students at OSU, students at the college in South Seattle,
referals from a company, friends and aquantenances, people from my church, and people like you.
I know to separate the variables, but I was told to use partial fractions to solve. I don't see why that would be necassary. When I separate, I get
-aP+1/P dP = b dt
Is this not correct?
ANSWER: When the equation is worked with, we multiply both sides by dt.
This gives (1/P)dP = (aP + b)dt. The aP is times dt, so it can't be moved to the other side.
---------- FOLLOW-UP ----------
QUESTION: Okay, but even when I use partial fractions I get
1/bp-a/b(b+ap)dP = 1 dt
This doesb't seem to have gotten me anywhere, because if
1/[P(b+aP)] requires partial fractions, so does the above expression. I've tries u-du sub, and that doesn't work either. I'm seriously stuck.
Answer Wel, I'm back now and I thought about the problem. Here's how I solved it (what I came up with was t(p):
Continuing with your partial fractions, I get
(1/bp)dP - (a/b)(1/(a+bp))dP = 1 dt.
When this is integrated, we get
ln(p) - (a/b)ln(a+bp)/b = t + C.
You can move the (a/b) up as an exponet, giving
ln(p)- ln(a+bp)^(a/b) = t + C,
and then diivide the two values with a ln, giving
ln(p / (a+bp)^(a/b)) = t + C.
Here we have it with t as a function of p:
t(p) = ln(p / (a+bp)^(a/b)) -C
Trying to solve for p, so that you now have
p/(a+bp)^(a/b) = e^(t+C). This puts p on one side and t on the other. That may be as close as you can get to a function p(t).
That still isn't solved for p, but I'm still not sure that can be done. After all, not many differential equations can be solved when problems have just a little complication to them.
