In this note, I’ll describe how to solve a general second order constant coefficient ODE with a general forcing function $f(t)$. That is, we’ll solve \begin{equation}\label{eqn:IVPgen} ay’‘+by’+cy = f(t),\qquad y(0) = y_0, \qquad y’(0) = y_1 . \end{equation}
We’ll require the following few rules for Laplace transforms and their correspinding inverses:
\begin{equation} \label{eqn:convLT} \mathcal{L} (f\ast g) = F(s)G(s), \end{equation}
\begin{equation} \label{eqn:f’} \mathcal{L} (f’) = sF(s) - f(0), \end{equation}
\begin{equation} \label{eqn:f’’} \mathcal{L} (f’’) = s^2 F(s) - sf(0) - f’(0).\end{equation}
Applying rules \eqref{eqn:f’} and \eqref{eqn:f’’} to the IVP in \eqref{eqn:IVPgen} gives
\[\mathcal{L}\left(ay''+by'+cy\right) = \mathcal{L}(f)\] \[a\left(s^2Y(s)-sy(0)-y'(0) \right) + b \left( sY(s) -y(0)\right) + c\left( Y(s) \right) = F(s)\] \[\left(as^2+bs+c\right)Y(s) - \left(asy_0 + ay_1 + by_1\right) = F(s).\]Rearranging gives \begin{equation} \label{eqn:Y} Y(s) = F(s) \left(\frac{1}{as^2+bs+c} \right) + \frac{asy_0+ay_1+by_0}{as^2+bs+c}. \end{equation} We will rewrite this as \begin{equation} \label{eqn:Y2} Y(s) = F(s) H(s) + \Phi(s), \end{equation} where \(H(s) = \frac{1}{as^2+bs+c},\qquad \Phi(s) = \frac{asy_0+ay_1+by_0}{as^2+bs+c}.\)
It is difficult to write out general explicit formulas for the inverse Laplace transforms of $H$ and $\Phi$, so instead we’ll say that we can take the inverse Laplace transform and write it as
\[h(t) = \mathscr{L}^{-1}(H),\qquad \phi(t) = \mathscr{L}^{-1}(\Phi).\]What are the fuctions $\phi$ and $h$ representing? Well, $h$ has Laplace transform
\[\mathcal{L}(h) = \frac{1}{as^2+bs+c},\]which is also the same as the solution to the initial value problem
\[ay''+by'+cy = \delta_0(t),\qquad y(0) = 0,\quad y'(0) = 0.\]Similarly, the function $\phi$ has the same Laplace transform as the solution to the initical value problem
\[ay''+ by'+ cy = 0,\qquad y(0) = y_0, \qquad y'(0) = y_1.\](You can check both of these using rules \eqref{eqn:f’} and \eqref{eqn:f’’}.)
Now we can use the inverse of the rule in \eqref{eqn:convLT} to write
\[\mathcal{L}^{-1} \left( F(s) H(s) \right) = \left( f \ast h \right) (t) = \int_0^t h(\tau) f(t-\tau)\,d\tau.\]Consequently, we can invert the Laplace transform in \eqref{eqn:Y2} and write
\[y(t) = \phi(t) + (f\ast h)(t) = \phi(t) + \int_0^t h(\tau)f(t-\tau)\,d\tau.\]