Question: I'm new to solving differential equations.

How would we go about solving a differential equation like

$y'=-4+5y-y^2$ ?

Answer: 
$$\frac{dy}{dt} = -4+5y-y^2$$

$$\frac{dy}{-4+5y-y^2}=dt$$

$$\int \frac{dy}{4+5y-y^2} = \int dt$$

$$\frac{1}{3}\text{ln}(|\frac{y-1}{y-4}|)=t+c\_1$$

$$\text{ln}(|\frac{3}{y-4} + 1|)=3t+c\_2$$

From here just you can isolate for y, or just leave it implicit.
One more way, use $y=u'/u$, then
$$y'=-4+5y-y^2 \implies \frac{u u''-u'^2}{u^2}=-4+5\frac{u'}{u}-\frac{u'^2}{u^2} $$
$$\implies u''-5u'+4=0$$ Let $u=e^{mx}$, then
$m^2-5m+2=0 \implies m=2,3$.
So $u(x)=C\_1e^{2x}+C\_2 e^{3x}$, which gives
$$y(x)=\frac{2C\_1 e^{2x}+ 3 C\_2 e^{3x}}{C\_1 e^{2x}+ C\_2 e^{3x}}
=\frac{2+3Ce^{x}}{1+Ce^{x}}$$