<s>[INST] Solving differential equation by separation of variables In the following question I am trying to solve the following ODE by the method of separation of variables. From there I want to use the initial condition given to estimate $y(8)$
The given equation is,
$$\frac{dy}{dx}=\frac{y(2-y)}{x+3}$$
Separating the variables gives:
$$\int \frac{1}{y(2-y)}dy=\int \frac{1}{x+3}dx$$
Using partial fractions for the left side gives,
$$\frac {1}{y(2-y)}=\frac{A}{y}+\frac{B}{2-y}$$
$$=-\frac{\frac{1}{2}}{y}+ \frac{\frac{1}{2}}{2-y}$$
$$y=-\frac{1}{2y}+ \frac{1}{2(y-2)}$$
So therefore,
$$\int \frac{1}{y(2-y)}dy=\int \frac{1}{x+3}dx=$$
$$-\frac{1}{2}lny-\frac{1}{2}ln(y-2)=ln(x+3)+c$$
$$ln(y^2-2y)=\frac{e^c}{(x+3)^2}$$
$$y^2-2y=\frac{e^c}{(x+3)^2}$$
$$y^2-2y-\frac{e^c}{(x+3)^2}=0$$
Applying the quadratic formula,
$$y=1+\sqrt{1+\frac{e^c}{(x+3)^2}}$$
$$y=1-\sqrt{1+\frac{e^c}{(x+3)^2}}$$
So my thinking is that $y$ has to be $>2$ in order for $ln(y-2)$ to be defined
Using the initial condition $y(0)=6$ gives 
$$6=1+\sqrt{1+\frac{e^c}{(0+3)^2}}$$
$$c=ln216$$
and hench $y(8)=1+\sqrt {\frac{337}{121}}$
So I am just wondering if I did this correctly, and the quadratic formula gave two answers so I'm wondering if I chose the correct one.
 [/INST] $$\frac {1}{y(2-y)}=\frac{A}{y}+\frac{B}{2-y}=\frac {2A+(B-A)y}{y(2-y)}$$
means $2A=1$ and $A=B$. Then
$$\dfrac12\left(\frac{1}{y}+\frac{1}{2-y}\right)dy=\dfrac{1}{x+3}dx$$
after integration
$$\dfrac12\ln\frac{y}{2-y}=\ln(x+3)+\ln C$$
or
$$\frac{y}{2-y}=C(x+3)^2$$
with $y(0)=6$ you find $\dfrac{y}{2-y}=-\dfrac16(x+3)^2$ and finally $x=8$ gives us $\color{blue}{y(8)=\dfrac{242}{115}}$.
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