Identifying homogenous differential equations Again, with respect to the same question I posted some minutes ago (Identifying separable differential equations), I discovered that the textbook provided to me was indeed wrong. Now, I'm beginning to question some of it's answers.
Taking the same function from the previous question,
$$\frac{dy}{dt}+p(t)y(t)=q(t)y(t)+3y(t)$$
refactoring $p(t)y(t)−q(t)y(t) + 3 y(t)$ as $(3 - r(t)) y(t)$ and I came to
$$\frac{dy}{dt} = (3−r(t))y(t),\text{ with }r(t)=q(t)−p(t)$$
Given that I don't know my $y(t)$, I'm not able to verify that is homogeneous, right? Considering that by definition, it's only homogeneous if $f(tx, ty) = t^n  f(x,y)$. In this case, I only know my $dy/dt$, and it isn't a $F(y/x)$ type function. I have a pretty solid guess that is non homogeneous, but I don't know the way to prove it.

Whether or not your equation is homogeneous doesn't actually have anything to do with the form of the solution. If an equation is homogeneous, that basically just means that all of the terms have $y$ in them in some way or another.
In this case, we have an equation of the form $y' + f(t)y = 0$, letting $f(t) = r(t) - 3 = p(t) - q(t) - 3$. This is homogeneous because there is no source term in terms of $t$ on the right-hand side.
If we instead had an equation like $y' + f(t)y = \sin t$ or $y' + f(t)y = 1$, then these would be nonhomogeneous because there is a source term, a function of $t$ on the right-hand side which we can't get into a term with $y$.
