The equation is of the form:

$$\frac{dP}{dt} = k.P(1-\frac{P}{b})-h $$

I have expanded to the quadratic form, and then completed the square, to get a new form:

$$\frac{dP}{dt} = c(P-d)^2+g $$

I have then proceeded to solve by dividing through by the RHS, and integrating both sides. During this process I used the subsitution:

$$u = (P-d)$$

such that the integral had the form:

$$ \frac{1}{c}\int\frac{1}{u^2-100}.du $$

So I've ended up with an equation for $t$ in terms of $arctanh(P)$. Referring to initial conditions yields an integration constant also in terms of $arctanh$ values. From my limited understanding of this function, I'm concerned that I have gone wrong somewhere in the above, and that there may be a simpler solution.

I would appreciate any pointers here, or confirmation whether or not the above is a useful way to proceed.



$$\frac{dP}{dt} = c(P-d)^2+g $$ With the change of function $\quad P=d+u$ : $$\frac{du}{dt} = c(u)^2+g \quad\to\quad g\,dt=\frac{du}{\frac{c}{g}u^2+1}$$ Let $\frac{c}{g}u^2=v^2\quad\to\quad u=\sqrt{\frac{g}{c}}v$ $$\sqrt{cg}dt=\frac{dv}{v^2+1} \quad\to\quad \sqrt{cg}\,t=\tan^{-1}(v)+C$$ $$v=\tan((\sqrt{cg}-C)t)$$ $$u=\sqrt{\frac{g}{c}}\tan\left((\sqrt{cg}-C)t\right)$$ $$P=d+\sqrt{\frac{g}{c}}\tan\left((\sqrt{cg}-C)t\right)$$ Note : If $\frac{c}{g}<0\quad$ $\tan$ will be changed to $\tanh$ in the calculus. The result is then : $$P=d+\sqrt{-\frac{g}{c}}\tanh\left((\sqrt{-cg}-C)t\right)$$

$C$ has to be determined according to the initial condition $(P_0,t_0)$ :

$C=\sqrt{cg}\,t_0-\tan^{-1}(v_0)\quad$ with $\quad v_0=\sqrt{\frac{c}{g}}(P_0-d)$

Or, if $\frac{c}{g}<0 \quad\to\quad C=\sqrt{-cg}\,t_0-\tanh^{-1}\left(\sqrt{-\frac{c}{g}}(P_0-d)\right)$


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