How would I start an attempt at solving this differential equation?

Question

The differential equation is:

${dP\over dT}=kP\Bigl(1-{P\over M}\Bigr)-\alpha P$

where $k > 0, \space M > 0$ and $\alpha > 0$ are constants.

How would I start an approach to solving this differential equation?

Answer 1

This is a version of $y' = ay^2+by $. This can be written as $\dfrac{dy}{ay^2+by} =dx$.

Since $\dfrac1{ay^2+by} =\dfrac1{by}-\dfrac{a}{b(ay+b)} $, $\int\dfrac{dy}{ay^2+by} = \dfrac{\ln(y) - \ln(a y + b)}{b} $ so, letting $c$ be the constant of integration, $x =\dfrac{\ln(y) - \ln(a y + b)}{b}+c $ or $bx-bc =\ln(y) - \ln(a y + b) =\ln(y/(a y + b)) $ or $e^{-bx+bc} =\dfrac{ay+b}{y} =a+\dfrac{b}{y} $ or $e^{-bx+bc}-a =\dfrac{b}{y} $ or $y =\dfrac{b}{e^{-bx+bc}-a} $.

Looks like a version of the logistic curve.

Answer 2

As already said in comment and answer, rewrite $${dP\over dT}=kP\Bigl(1-{P\over M}\Bigr)-\alpha P$$ as $${dT\over dP}=\frac 1{kP\Bigl(1-{P\over M}\Bigr)-\alpha P}=\frac 1 { P (k-\alpha )-\frac{k }{M}P^2}=\frac 1{P\Bigl(k-\alpha-\frac{k }{M}P\Bigr)}=\frac{\frac M k } {P \Bigl(\frac{M(k-\alpha)}k-P\Bigr)}$$ For more legibility, let $a=\frac M k$ and $b=\frac{M(k-\alpha)}k$ to make $${dT\over dP}=\frac a {P(b-P)}$$ and use partial fraction decomposition to get $${dT\over dP}=\frac{a}{b P}-\frac{a}{b (P-b)}\implies \frac b a\,{dT\over dP}=\frac{1}{P}-\frac{1}{P-b}$$ Now, integrate both sides $$\frac b a T+c_1=\log \left(\frac{P}{P-b}\right)\implies \frac{P}{P-b}=c_2\,e^{\frac{b }{a}T}\implies P=\frac{b \,c_2\, e^{\frac{b }{a}T}}{c_2\, e^{\frac{b }{a}T}-1}$$ what you can further simplify as $$P=\frac{b}{1-c_3\,e^{-\frac{b }{a}T}}$$ Replace $a,b$ by their definition and compute $c_3$ from the initial condition.