Solving io ODEs with Laplace

Laplace transforms provide a convenient method for solving input-output (io) ordinary differential equations (ODEs).

Consider a dynamic system described by the io ODE—with \(t\) time, \(y\) the output, \(u\) the input, constant coefficients \(a_i,b_j\), order \(n\), and \(m \le n\) for \(n\in\mathbb{N}_0\)—as: \[\begin{aligned} {8} \label{eq:lap:ioode} \frac{\diff^n y} {\diff t^n} & {}+{} & a_{n-1} \frac{\diff^{n-1} y}{\diff t^{n-1}} & {}+{} & \cdots & {}+{} & a_1 \frac{\diff y} {\diff t} & {}+{} && a_0 y =\nonumber \\ b_m \frac{\diff^m u} {\diff t^m} & {}+{} & b_{m-1} \frac{\diff^{m-1} u}{\diff t^{m-1}} & {}+{} & \cdots & {}+{} & b_1 \frac{\diff u} {\diff t} & {}+{} && b_0 u. \end{aligned}\] Re-written in summation form, \[\begin{aligned} \label{eq:lap:ioode2} \sum_{i=0}^n a_i y^{(i)}(t) &= \sum_{j=0}^m b_j u^{(j)}(t), \end{aligned}\] where we use Lagrange’s notation for derivatives, and where, by convention, \(a_n = 1\).

The Laplace transform \(\mathcal{L}\) of yields $$\begin{align}\label{eq:lap:ioode3} \mathcal{L}\sum_{i=0}^n a_i y^{(i)}(t) &= \mathcal{L}\sum_{j=0}^m b_j u^{(j)}(t) \quad\Rightarrow \\ \sum_{i=0}^n a_i \mathcal{L} \left(y^{(i)}(t) \right) &= \sum_{j=0}^m b_j \mathcal{L} \left(u^{(j)}(t)\right). \tag{linearity} \end{align}$$ In the next move, we recursively apply the differentiation property to yield the following \[\begin{aligned} \label{eq:lap:ioode4} \sum_{i=0}^n a_i \left( s^i Y(s) + \underbrace{\sum_{k=1}^i s^{i-k} y^{(k-1)}(0)}_{I_i(s)} \right) &= \sum_{j=0}^m b_j s^j U(s), \end{aligned}\] where terms in \(I_i(s)\) arise from the initial conditions. Splitting the left outer sum and solving for \(Y(s)\), $$\begin{align}\label{eq:lap:ioode5} \sum_{i=0}^n a_i s^i Y(s) &+ \sum_{i=0}^n a_i I_i(s) = \sum_{j=0}^m b_j s^j U(s) \quad\Rightarrow \\ \sum_{i=0}^n a_i s^i Y(s) &= \sum_{j=0}^m b_j s^j U(s) - \sum_{i=0}^n a_i I_i(s) \quad\Rightarrow \\ Y(s) \sum_{i=0}^n a_i s^i &= U(s) \sum_{j=0}^m b_j s^j - \sum_{i=0}^n a_i I_i(s) \quad\Rightarrow \\ Y(s) &= \underbrace{\frac{\sum_{j=0}^m b_j s^j}{\sum_{i=0}^n a_i s^i} U(s)}_{Y_\text{fo}(s)} + \underbrace{\frac{-\sum_{i=0}^n a_i I_i(s)}{\sum_{i=0}^n a_i s^i}}_{Y_\text{fr}(s)}. \label{eq:laplace_io_ode_sol1} \end{align}$$

So we have derived the Laplace transform image \(Y(s)\) in terms of the forced and free responses (still in the \(s\)-domain, of course)! For a solution in the time-domain, we must inverse Laplace transform: \[\begin{aligned} \label{eq:laplace_io_ode_sol2} y(t) &= \underbrace{(\mathcal{L}^{-1}Y_\text{fo})(t)}_{y_\text{fo}(t)} + \underbrace{(\mathcal{L}^{-1}Y_\text{fr})(t)}_{y_\text{fr}(t)}. \end{aligned}\] This is an important result!