Ask Me Help Desk - DP/dt = P(a-bP) using substitution?

dP/dt = P(a-bP) using substitution?

I need to solve this using substitution i.e. treat as a Bernouli equation

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

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

Comment on jcaron2's post

Thank You very much this stuff is still pretty new to me, I appreciate yopur help.

Comment on jcaron2's post

Do you know the proper syntax to solve this using Maple software? I would like to try solving this using Maple as well.

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!

Comment on jcaron2's post

You got it... Thanks Again.

You're welcome!