Suppose $s \neq 0$. Recall the equation we wish to solve $a(t) \frac{d^{2}}{dt^{2}}x(t) +b(t)x(t) + c(t)x(t) = s(t), \quad x(t_{0}) = x_{0}, x'(t_{0})=v_{0} \tag{1}$ > [!Theorem] > Assume that $x_{p}$ is a solution of $(1)$. Then all solutions are of the form $x(t)= x_{p}(t)+x_{c}(t)$where $x_{c}$ is the complementary function of $(4.1)$ (with $s=0$). Hence the solution of the IVP is given by $x(t) = x_{p}(t)+l_{1}x_{1}(t)+l_{2}x_{2}(t), \quad t \in (\alpha,\beta),$where $l_{1},l_{2} \in \mathbb{R}$ are such that $\begin{align} x_{0}=x_{p}(t_{0})+l_{1}x_{1}(t_{0})+x_{2}(t_{0}), \\ v_{0} = \frac{d}{dt} x_{p}(t_{0})+\frac{l_{1}d}{dt} x_{1}(t_{0}) +\frac{l_{2}d}{dt} x_{2}(t_{0}) \end{align}$ > **Proof** > By substitution, show that $x-x_{p}$ satisfies the homogeneous DE. By the above 'general solution theorem', we now then that $x-x_{p}$ is of the form of the complementary function. Therefore indeed $x(t)=x_{p}(t)+x_{c}(t)$. We can also check that the initial conditions are satisfied.