I'm trying to solve the ODE
$$y' = \frac{t + 2y}{2t + y}$$
as general homogenous equation. Thus I'm using general initial conditions $y(t_0) = y_0$ with the only restriction $2t_0 + y_0 \neq 0$ (otherwise the denominator would be zero in the defining equation). Notice, that this condition doesn't imply $t_0 \neq 0$.
At this point I want to solve ODE using a transformation into a homogenous equation. Thus look at it like this
$$y' = \frac{1 + 2\frac{y}{t}}{2 + \frac{y}{t}}$$
Expanding with $t$ in the nominator and denominator shows that the equations are equivalent. Then substituting $z = \frac{y}{t}$ yields the separable equation
$$z' = \frac{1}{t} (\frac{1 + 2z}{2 + z} - z)$$
Now I'm wondering about what happens if the initial value $t_0$ was zero. In this case the substitution doesn't work. What am I missing here?
$$y' = \frac{t + 2y}{2t + y}$$ There are many ways to integrate this Differential Equation, @Moo's trick is one of them. Here is another way: $$\frac {dy}{(t + 2y)}=\frac {dt}{(2t + y)}$$ $$\frac {d(t+y)}{3(t + y)}=\frac {d(t-y)}{(t - y)}$$
Integrate: $$\ln \left | \dfrac {t+y}{(t-y)^3} \right|=k \implies \left | \dfrac {t+y}{(t-y)^3} \right|=c$$