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View Full Version : DP/dt = P(a-bP) using substitution?


savage44
Feb 23, 2011, 06:43 PM
I need to solve this using substitution i.e. treat as a Bernouli equation

jcaron2
Feb 23, 2011, 10:31 PM
Your differential equation is

P'=P(a-bP)

which can be rewritten as

P'= aP-bP^2

P'-aP= -bP^2

\frac{P'}{P^2}-\frac{a}{P}= -b

If we now do a change of variables

w=\frac{1}{P}

we have

w'=\frac{-P'}{P^2},

which we can then substitute back into the original DE:

\frac{P'}{P^2}-\frac{a}{P}= -b \;\;\; \to \;\;\; w'+aw=b

Now we simply solve the resulting first-order differential equation.

If we multiply everything by e^{at}, we get

w' \cdot e^{at} + aw \cdot e^{at} = b \cdot e^{at}

Now notice that the left side of the equation is equal to the derivative of w \cdot e^{at}. Thus, we can integrate both sides to get:

\int (w' \cdot e^{at} + aw \cdot e^{at}) dt = \int b \cdot e^{at} dt

\int (w \cdot e^{at})' dt = \frac ba \cdot \int a \cdot e^{at} dt

w \cdot e^{at} = \frac ba \cdot e^{at} + C

w = \frac ba +Ce^{-at}

Finally, substituting back in for P, we get

\Large P= \frac 1w\;=\;\frac1{\frac ba +Ce^{-at}}

savage44
Feb 24, 2011, 02:57 AM
Thank You very much this stuff is still pretty new to me, I appreciate yopur help.

savage44
Feb 24, 2011, 03:30 AM
Do you know the proper syntax to solve this using Maple software? I would like to try solving this using Maple as well.

jcaron2
Feb 24, 2011, 07:36 AM
I think the Maple syntax would be as follows:

DE1 := diff(P(x), x) = P(x)*(a - b*P(x));

dsolve(DE1, P(x));


I haven't used Maple in years, and I don't have it installed, so unfortunately I can't check if I got it right or not. Let me know if it gives you the right answer!

savage44
Feb 24, 2011, 11:05 AM
You got it... Thanks Again.

jcaron2
Feb 24, 2011, 11:48 AM
You're welcome!