Initial value problem for $y' = \alpha y - y^2$ with $y(0)=y_0 \neq 0$ and $\alpha > 0$

Question

I'm solving past exam questions in preparation for an Applied Mathematics course. I came to the following question. If it's any indication of difficulty, the exercise is only Part 1-A of the sheet so it's supposed to be pretty easy

Solve the initial value problem $y' = \alpha y - y^2$ with $y(0)=y_0 \neq 0$ and $\alpha > 0$

The problem is, I can't seem to be able to deal with that $y^2$, as in my methodology I am urged to integrate over $\frac{dy}{dt}$, and the square ruins that.

Any help would be greatly appreciated.

Answer 1

$$y'=\alpha y-y^2$$ $$y'-\frac{\alpha^2}{4}=\alpha y-y^2-\frac{\alpha^2}{4}$$ $$\frac{dy}{dx}-\frac{\alpha^2}{4}=-(y-\frac{\alpha}{2})^2$$ $$\int\frac{dy}{\frac{\alpha^2}{4}-(y-\frac{\alpha}{2})^2}=\int dx$$ which can then be solved.

Answer 2

$$\frac{dy}{dt} = ay -y^2 \Rightarrow \frac{dy}{ay -y^2} = dt$$

$$\int_{y(0)}^{y(t)}\frac{dy}{ay -y^2} = \int_0^t dt \Rightarrow \\ \int_{y(0)}^{y(t)}\left[\frac{1}{ay} - \frac{1}{a(y-a)}\right]dy = t \Rightarrow \\ \frac{1}{a}[\log(y(t)) - \log(y(0))] - \frac{1}{a}[\log(y(t)-a) - \log(y(0)-a)] = t \Rightarrow \\ \log\left(\frac{y(t)(y(0)-a)}{y(0)(y(t)-a)}\right) = at \Rightarrow \\ \frac{y(t)}{y(t)-a} = e^{at}\frac{y(0)}{y(0)-a} \Rightarrow \\ y(t)\left[1-e^{at}\frac{y(0)}{y(0)-a}\right] = -ae^{at}\frac{y(0)}{y(0)-a} \Rightarrow y(t) = \frac{-ae^{at}y(0)}{y(0)-a-e^{at}y(0)}.$$