Laplace transform of derivatives

Question

In an example in the book we are given the following ODE: $x''-x'-6x=0$. In addition we are given the following definitions:

$L{f'(t)}=sF(s)-f(0)$ and

$L{f''}=s^2-sf(0)-f'(0)$

The part I am confused about is what is the transformation of $-6x$? I don't see one laid out in the text.

Answer 1

$$x''-x'-6x=0$$ Apply the Laplace Transform: $$\mathcal {L}(x''-x'-6x)=0$$ $$\mathcal {L}(x'')-\mathcal {L}(x')-\mathcal {L}(6x)=0$$ Since you have that: $L[6x](s)=6\int_0^\infty e^{-ts}\,x(t)\,dt =6X(s)$ Hence: $$\mathcal {L}(x'')-\mathcal {L}(x')-6X(s)=0$$ You know what to do with the first two terms. $$\mathcal {L}{(x'')}=s^2X(s)-sx(0)-x'(0)$$ $$\mathcal {L}{(x')}=sX(s)-x(0)$$

Answer 2

By definition, \begin{align} L[6x](s)&=\int_0^\infty e^{-ts}\,6t\,dt \\ &\stackrel{IBP}{=}\left[-\frac{1}{s}e^{-ts}6t\right]_{t=0}^\infty + \frac{6}{s}\int_0^\infty e^{-ts}\,dt \end{align}