Expert: Scotto - 7/23/2008

Question
QUESTION: 1/P(dP/dt)= aP + b

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.