# 14.7: Differentiation Theorem

- Page ID
- 5842

\[\textbf{L} \left( \frac{d^n y}{dt^n}\right) = s^n \bar{y} - s^{n-1}y_0 - s^{n-2} \left(\frac{dy}{dt}\right)_0 - s^{n-3} \left(\frac{d^2y}{dt^2}\right)_0 - ... \quad ... - s \left(\frac{d^{n-2}y}{dt^{n-2}} \right)_0 - \left( \frac{d^{n-1}y}{dt^{n-1}}\right)_0 .\]

This looks formidable, and you will be tempted to skip it – but don't, because it is essential! However, to make it more palatable, I'll point out that one rarely, if ever, needs derivatives higher than the second, so I'll re-write this for the first and second derivatives, and they will look much less frightening.

\[\textbf{L} \dot y = s \bar{y} - y_0\]

and

\[\textbf{L}\ddot y = s^2 \bar{y} - sy_0 - \dot y_0.\]

Here, the subscript zero means "evaluated at *t* = 0".

Equation 14.7.2 is easily proved by integration by parts:

\[\bar{y} = \textbf{L}y = \int_0^\infty ye^{-st}dt = -\frac{1}{s}\int_0^\infty yde^{-st} = -\frac{1}{s} \left[ ye^{-st} \right]_0^\infty + \frac{1}{s} \int_{t=0}^\infty e^{-st} dy =\frac{1}{s}y_0 + \frac{1}{s} \int \dot y dt = \frac{1}{s}y_0 + \frac{1}{s} \textbf{L} \dot y.\]

\[\therefore \qquad \textbf{L}\dot y = s\bar{y} - y_0.\]

From this, \[\textbf{L} \ddot y = \bar{\dot{y}} - \dot y _0 = sL \dot y - \dot y_0 = s(s\bar{y} - y_0) - \dot y_0 = s^2 \bar{y} - sy_0 -\dot y_0.\]

Apply this over and over again, and you arrive at equation 14.7.1