Respuesta :
Answer:
If [tex]K[/tex] is a constant of integration, then
[tex]P = {\displaystyle \frac{1}{b/a + Ke^{-at}}}[/tex]
Step-by-step explanation:
According to the information of the problem we know that
[tex]{\displaystyle \frac{dP}{dt} = P(a-bP) }[/tex]
Remember that in general a Bernoulli equation is an equation of the type
[tex]y' + p(x)y = q(x)y^n[/tex]
And the idea to solve the equation is to substitute
[tex]{ \displaystyle v = y^{1-n}}[/tex]
Now for this case
[tex]{\displaystyle \frac{dP}{dt} - Pa = -bP^2}[/tex]
Then we substitute
[tex]v = P^{1-2} = P^{-1}[/tex]
Therefore
[tex]P = v^{-1}[/tex]
and if you compute the derivative of that you get that
[tex]{\displaystyle \frac{dP}{dt} = -v^{-2} \frac{dv}{dt}}[/tex]
Now you substitute that onto the original equation and get
[tex]{\displaystyle \frac{dP}{dt} - Pa = -bP^2}[/tex]
[tex]{\displaystyle -v^{-2} \frac{dv}{dt} - v^{-1} = -bv^{-2}[/tex]
If you multiply everything by [tex]-v^2[/tex] you get that
[tex]{\displaystyle \frac{dv}{dt} + v = b }[/tex]
That's a linear differential equation and the solution would be
[tex]v = {\displaystyle \frac{b}{a} + Ke^{-at}} = P^{-1}[/tex]
Where [tex]K[/tex] is a constant of integration, then
[tex]P = {\displaystyle \frac{1}{b/a + Ke^{-at}}}[/tex]
