Solving $\frac{b}{a-b}e^{at}=\frac{x(t)}{a-x(t)}$ for $x(t)$ I`ve been trying to solve the differential equation $x(t)'=x(t)(a-x(t)), x(0)=b, t\in [0, \infty]$. Using the technique of seperation of variables, I get $\frac{b}{a-b}e^{at}=\frac{x(t)}{a-x(t)}$. Now I want to solve this for x(t). Any hints?
Why not to use simple method to find $x(t)$ by solving linear equation as follows $$\frac{a}{a-b}e^{at}=\frac{x(t)}{a-x(t)}$$ $$\implies a^2e^{at}-ae^{at}x(t)=(a-b)x(t)$$ $$\implies (a-b)x(t)+ae^{at}x(t)=a^2e^{at}$$ $$\implies \color{blue}{x(t)=\frac{a^2e^{at}}{ae^{at}+a-b}}$$ Now, for the given condition $x(0)=b$, substituting $t=0$, we have $$x(0)=\frac{a^2e^{a(0)}}{ae^{a(0)}+a-b}$$ $$=\frac{a^2}{a+a-b}=\frac{a^2}{2a-b}$$ $$\implies x(0)=b=\frac{a^2}{2a-b}$$ $$ab-b^2=a^2$$ $$a^2+b^2-2ab=0\implies (a-b)^2=0\implies \color{blue}{a=b}$$
since $\dfrac{1}{x(a-x)} = \dfrac{1}{a} (\dfrac{1}{x}+\dfrac{1}{a-x}) $ you have $$ \int (\frac{1}{x}+\frac{1}{a-x} )\mathrm{d}x = a \int \mathrm{d}t $$
so $$ Ln(x)-Ln(a-x) = a t + C $$
or
$$ Ln(\frac{x}{a-x})=at+C $$
so 
$$ x(t) = \frac{a e^C e^{at}}{1+ e^C e^{at} } $$

now $x(0)=b$ so $b=\dfrac{a e ^c}{1+e^c} $ and $ C$ has been determined.