It's simple with Operator Calculus .
Quite analogous to Example 1 on page 3 of this PDF we have:
$$ y'' + a y' + b y = x(t)$$
Operator Calculus enables us to "abstract" as much as possible from the solution $y(t)$ :
$$ \left[ \left(\frac{d}{dt}\right)^2 + a \frac{d}{dt} + b \right] y(t) = x(t)$$
What we are going to do next is decompose into
factors. We are (almost) forced to do so, because the operator $(d/dt)$ and
the constants $a$ and $b$ mutually behave as if they were ordinary
numbers: the commutator of a differentiation and a constant is zero. But
the commutative law for ( linear ) operators is the only thing which could
be deviant from algebra with ordinary numbers.
$$ \left(\frac{d}{dt} \right)^2 + a \left(\frac{d}{dt} \right) + b =
\left(\frac{d}{dt} - \lambda_1 \right)
\left(\frac{d}{dt} - \lambda_2 \right) $$
$\lambda_1$ and $\lambda_2$ are roots of the so-called characteristic
equation : $$ \lambda^2 + a \lambda + b = 0 \quad \Longrightarrow \quad \lambda_{12}= (-a \pm \sqrt{a^2 - 4 b})/2 $$
Herewith, the differential equation can be rewritten as:
$$ \left(\frac{d}{dt} - \lambda_1 \right)
\left(\frac{d}{dt} - \lambda_2 \right) y(t) = x(t) $$
We are going to use now the "very useful formula" from Operator Calculus
(please find it on page 2 of the abovementioned document) :
$$
\frac{d}{dt} + f = e^{-\int f\,dt}\; \frac{d}{dt} \; e^{+\int f\,dt}
$$
In our case:
$$ \frac{d}{dt} - \lambda_{1,2} =
e^{ - \int - \lambda_{1,2} \, dt} \ \frac{d}{dt}\
e^{ + \int - \lambda_{1,2} \, dt} =
e^{\, \lambda_{1,2} t } \ \frac{d}{dt}\ e^{\, - \lambda_{1,2} t } $$
Giving, at last, for the O.D.E. :
$$ e^{ \lambda_1 t } \frac{d}{dt}\ e^{ - \lambda_1 t }\;
e^{ \lambda_2 t } \frac{d}{dt}\ e^{ - \lambda_2 t } \; y(t) = x(t) $$
Systematic integration is possible now:
$$ e^{-\lambda_1 t} e^{\lambda_2 t} \frac{d}{dt}\ e^{-\lambda_2 t} y(t) =
\int x(t) e^{ - \lambda_1 t }\,dt $$
$$ y(t) = e^{\lambda_2 t} \int e^{(\lambda_1-\lambda_2) t}
\left[ \int x(t) e^{ - \lambda_1 t }\,dt \right] \, dt $$
As has been said, $\lambda$ can be solved from $\lambda^2 + a\lambda + b = 0$,
a quadratic equation with discriminant: $ D=a^2-4b$.
Real valued as well as complex solutions are possible. A special case for $D=0$
may be distinguished; in this case $ \lambda_1 = \lambda_2 $
and $e^{(\lambda_1-\lambda_2)t} = 1$ .
EDIT. Almost forgot the boundary conditions $y(0)=y'(0)=0$ to take into account:
$$ y(t) = e^{\lambda_2 t} \int_0^t e^{(\lambda_1-\lambda_2) u}
\left[ \int_0^u x(v) e^{ - \lambda_1 v }\,dv \right] \, du $$
Note. If "nothing about the form of x is known" - but why a bounded & non-negative requirement
- then $y''(0) = 0$ is in general not true , unless $x(0) = 0$ :
$$
y''(t) = \lambda_2^2 e^{\lambda_2 t} \int_0^t e^{(\lambda_1-\lambda_2) u}
\left[ \int_0^u x(v) e^{ - \lambda_1 v }\,dv \right] \, du +
(\lambda_1 + \lambda_2) e^{\lambda_1 t} \int_0^t x(v) e^{ - \lambda_1 v }\,dv
\; + \; x(t)
$$
With differential equations of the kind, $\,y''(0) = 0\,$ cannot serve as a boundary condition anyway.