Show that the solution of the initial value problem $$L[y] = y'' + p(t)y' + q(t)y = g(t), y(t_0) = y_0, y'(t_0) = y'_0$$ can be written as $y = u(t) + v(t)$, where $u$ and $v$ are solutions of the two initial value problems $$L[u] = 0, u(t_0) = y_0, u'(t_0) = y'_0$$ $$L[v] = g(t), v(t_0) = 0, v'(t_0) = 0$$ respectively. In other words, the nonhomogeneities in the differential equation and in the initial conditions can be dealt with separately.
What i tried
Since $L[u] = 0$, $u(t)$ is the complemntary solution, hence is the solution to the homogeneous eqn $$L[u] = u'' + p(t)u' + q(t)u = 0, u(t_0) = u_0, u'(t_0) = u'_0$$ and $v(t)$ is the particular solution.
So to show that $u(t)$ is the complementary solution, i subusituted $u(t)$ to the above ODE, Im also thinking should i solve the equation$$y'' + p(t)y' + q(t)y = 0, y(t_0) = y_0, y'(t_0) = y'_0$$ but im unsure how to go about doing it. Could anyone explain. Thanks