Solve this: $\frac{dx}{dt} = [a-(b+c)]x-ax^2$ by separation of variables So I have this equation

$$\dfrac{dx}{dt} = [a-(b+c)]x-ax^2$$

which I need to solve by separation of variables. I've never seen an example like this where you have a sum on the RHS of the eq. Usually the examples are just a product on the RHS. 

I have the initial condition of $x(0)=x_0$
Separation of variables means putting everything in $x$ on one side and $t$ on the other, then integrating:  

NOTE I will replace $dx$ with $dX$ and $dt$ with $T$ to avoid having $x$ inside and outside the integral sign.

$\frac{dX}{dT} = (a-(b+c))X-aX^2 \Rightarrow \frac{dX}{(a-(b+c))X-aX^2} = dT$
$\Rightarrow \int\limits_{x0}^x\frac{dX}{(a-(b+c))X-aX^2} =\int\limits_0^tdT$
$\Rightarrow t=\frac{\ln(x)-\ln((a-(b+c))-ax)}{a-(b+c)}-\frac{\ln(x_0)-\ln((a-(b+c))-ax_0)}{a-(b+c)}$  

Finally isolate $x$ to have $x=f(t,a,b,c)$
Separation of variables means putting everything in $x$ on one side and $t$ on the other, then integrating:  

NOTE I will replace $dx$ with $dX$ and $dt$ with $T$ to avoid having $x$ inside and outside the integral sign.

$\frac{dX}{dT} = (a-(b+c))X-aX^2 \Rightarrow \frac{dX}{(a-(b+c))X-aX^2} = dT$
$\Rightarrow \int\limits_{x0}^x\frac{dX}{(a-(b+c))X-aX^2} =\int\limits_0^tdT$
$\Rightarrow t=\frac{\ln(x)-\ln((a-(b+c))-ax)}{a-(b+c)}-\frac{\ln(x_0)-\ln((a-(b+c))-ax_0)}{a-(b+c)}$  

Finally isolate $x$ to have $x=f(t,a,b,c)$