Answer to Question #227282 in Differential Equations for Ode

Question

Find the relation of the variable w if the growth rate is given by

dw/dt=(kw(α-w))/α, k>0, at t=0,w=α/(1+β)

Answer 1

Let us find the relation of the variable $w$ if the growth rate is given by

$\frac{dw}{dt}=\frac{kw(α-w)}{α},\ \ k>0,$ at $t=0,\ w=\frac{α}{1+β}.$

It follows that

$\frac{\alpha dw}{w(α-w)}=kdt$

$\int\frac{\alpha dw}{w(α-w)}=\int kdt$

$\int(\frac{1}{α-w}+\frac{1}{w})dw=\int kdt$

$\ln|w|-\ln|\alpha -w|=kt+C_1$

$\ln|\frac{w}{\alpha -w}|=kt+C_1$

$\frac{w}{\alpha -w}=C_2e^{kt}$

$w=C_2e^{kt}(\alpha -w)$

$w=C_2e^{kt}\alpha -wC_2e^{kt}$

$w (1+C_2e^{kt})=C_2e^{kt}\alpha$

$w =\frac{C_2\alpha e^{kt}}{1+C_2e^{kt}}$

Since at $t=0,\ w=\frac{α}{1+β},$ we have that

$\frac{α}{1+β}=\frac{C_2\alpha}{1+C_2}$

$\frac{1}{1+β}=\frac{C_2}{1+C_2}$

$1+C_2=C_2+\beta C_2$

$1=\beta C_2$

$C_2=\frac{1}{\beta}$

We conclude that

$w =\frac{\frac{1}{\beta}\alpha e^{kt}}{1+\frac{1}{\beta}e^{kt}}=\frac{\alpha e^{kt}}{\beta+e^{kt}}.$