Where are the particular and homogeneous solution of the ODE when using Laplace?

Question

When solving an ODE with Laplace, it seems as if there is no distinction between the homogeneous and particular solution. As if you calculated both at once.

Is this correct? How does it come? Where are the particular and the homogeneous solution?

Answer 1

I think you can argue that they are still there, just hidden, because you apply the initial conditions in the middle rather than the end.

Consider a non-homogenous initial value problem (with constant coefficients)

$$\sum_i\alpha_i\frac{d^iy}{dt^i}=\theta(t).$$

Take the Laplace Transform of both sides. Note that the Laplace transform of a derivative, after the initial conditions have been substituted, is given by:

$$\mathcal{L}\left\{\frac{d^iy}{dt^i}\right\}=s^iY(s)-p_i(s),$$

where $p_i(s)$ is a polynomial in $s$. Putting this together we have

$$ \begin{align} \sum_i\alpha_i\left(s^iY(s)-p_i(s)\right)&=\Theta(s) \\ \Rightarrow \sum_i \alpha_i s^iY(s)&=\Theta(s)+\sum_i\alpha_ip_i(s) \end{align}$$ Write this as $$\begin{align} q(s)Y(s)&=\Theta(s)+r(s) \\\Rightarrow Y(s)&=\frac{\Theta(s)+r(s)}{q(s)} \\ \Rightarrow Y(s)&=\frac{\Theta(s)}{q(s)}+\frac{r(s)}{q(s)} \\ \Rightarrow Y(s)&=Y_p(s)+Y_H(s).\end{align}$$

I trust you can see the implication of this...