ODE question: $y'+A(t) y =B(t)$, with $y(0)=0, B>0$ implies $y\ge 0$; another proof? I am trying to prove that the solution for the different equation
$$y'+A(t) y =B(t)$$ 
with initial condition $y(0)=0$ and the assumption that $B\ge 0$, has non-negative solution for all $t\ge 0$, i.e. $y(t)\ge 0$ for all $t\ge 0$.
The one I know is constructive: Fisrt consider the homogeneous ODE $x'+\frac{1}{2} A(t) x=0$ and let $u'=\frac{B}{x^2}$ with $u(0)=0$. Then $y=\frac{ux^2}{2}$ will satisfy the origional ODE of $y$. Clearly, by the construction of $y$, $y\ge 0$ for $t\ge 0$.
But this proof is not natural in my opnion, there is no reason(at least I didn't see) to construct such $x(t)$ and $u(t)$.
So is there any other way to prove this fact?

Well since it is easy to actually explicitely solve such ODE's, the most natural solution to me would have been to analyse that solution. The solution is $$ y(t) = \frac{ \int^t_0 \exp\left(\int^s_0 A(p) dp \right) B(s) ds }{ \exp \left(\int^t_0 A(p) dp \right)}$$
and since $\exp (x) > 0 $ for all $ x \in \mathbb{R}$ and it is given $B(s) \geq 0 $, the numerator is non-negative while the denominator is strictly positive, and so $y(t) \geq 0.$ 
