Suppose we have the initial value problem $$\frac{d^2y}{dt^2}=h(t),$$ for some given function $h$, with initial conditions $y(0)=y'(0)=0$.
Using the boundary conditions $G(0;\tau)=G'(0;\tau)=0$, I have derived the Green's function:
$$G(t;\tau)= \begin{cases} 0,&0\le t<\tau,\\ t-\tau,&\tau<t<\infty. \end{cases} $$ Then I used the general equation for the solution of the problem, $$y(t)=\int_0^\infty G(t;\tau)h(\tau)d\tau$$ to get
$$y(t)=t\int_0^th(\tau)d\tau-\int_0^t\tau h(\tau)d\tau.$$
But, how can I verify directly that this function satisfies both the differential equation and the initial value conditions?
To check it, find the derivatives: by the "fundamental theorem of Calculus", the derivative of $\int_a^t f(\tau)d\tau$, with respect to t, is f(t) so the derivative of $t\int_0^t h(\tau)d\tau$, by the product rule is $\int_0^t h(\tau) d\tau+ t h(t)$. Similarly, the derivative of $\int_0^t \tau h(\tau) d\tau$ is $t h(t)$.