Solving a non linear ODE problem Please I will like to solve this non linear ODE problem 
$$
y'(x) = e^{-b(x)} \times \left( p + \left( \frac{q}{n} y(x) \right)\right)\left( n - y(x) \right). 
$$
Can anyone help? Thank you
I made some correction to the equation
 A: EDIT: Answer addresses the problem prior to a recent edit that changed the equation of the question.
From 
$$
y'(x) = \left( p + \frac{q}{n} \right) y(x)(n - y(x))*e^{-b(x)}
$$
we can solve it by putting it in the form 
$$
\int \frac{y'(x)}{y(x)(n - y(x))} \, \mathrm{d} x = \left( p + \frac{q}{n} \right) \int e^{-b(x)} \, \mathrm{d} x = \xi (x)
$$
and as the integral on the left becomes 
$$
\int \frac{y'(x)}{y(x)(n - y(x))} \, \mathrm{d} x = \frac{1}{n} \int y'(x) \cdot \left( \frac{1}{y(x)} + \frac{1}{n - y(x)} \right) \, \mathrm{d} x = \frac{1}{n} \ln \left| \frac{y(x)}{n - y(x)}  \right|,
$$
then we have that $y(x)$ satisfies
$$
\frac{1}{n} \ln \left| \frac{y(x)}{n - y(x)} \right| = \xi(x) + C
$$
and 
$$
y(x) = (n - y(x)) Ae^{n \xi(x)} \implies y(x) = \frac{nA e^{n \xi(x)}}{1 + Ae^{n \xi(x)}}.
$$
[$A$ of course is simply $e^{nC}$.]
A: Using the same approach as izoec, we can arrive to$$\frac{n (n (p+q) \log (n p+q y)-q y)}{q^2}= \int e^{-b(x)} \, \mathrm{d} x = \xi (x)$$ and the solution of $y$ is $$y=- \frac {n} {q} \left(p+(p+q) W\left(-\frac{\left(e^{\frac{\xi(x)  q^2}{n^2
   p}-1}\right)^{\frac{p}{p+q}}}{n (p+q)}\right)\right)$$
I hope and wish that I did not make any mistake.
