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!