# Laplace transform of derivatives: Given the following ODE: x′′−x′−6x=0. In addition we are given the following definitions: Lf′(t)=F(s)−f(0) and Lf′'=s^2−sf(0)−f′(0)

Given the following ODE: ${x}^{″}-{x}^{\prime }-6x=0$. In addition we are given the following definitions:
$L{f}^{\prime }\left(t\right)=sF\left(s\right)-f\left(0\right)$ and
$L{f}^{″}={s}^{2}-sf\left(0\right)-{f}^{\prime }\left(0\right)$
The part I am confused about is what is the transformation of −6x?
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vidovitogv5
${x}^{″}-{x}^{\prime }-6x=0$
Apply the Laplace Transform:
$\mathcal{L}\left({x}^{″}-{x}^{\prime }-6x\right)=0$
$\mathcal{L}\left({x}^{″}\right)-\mathcal{L}\left({x}^{\prime }\right)-\mathcal{L}\left(6x\right)=0$
Since you have that: $L\left[6x\right]\left(s\right)=6{\int }_{0}^{\mathrm{\infty }}{e}^{-ts}\phantom{\rule{thinmathspace}{0ex}}x\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt=6X\left(s\right)$ Hence:
$\mathcal{L}\left({x}^{″}\right)-\mathcal{L}\left({x}^{\prime }\right)-6X\left(s\right)=0$
You know what to do with the first two terms.
$\mathcal{L}\left({x}^{″}\right)={s}^{2}X\left(s\right)-sx\left(0\right)-{x}^{\prime }\left(0\right)$
$\mathcal{L}\left({x}^{\prime }\right)=sX\left(s\right)-x\left(0\right)$
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shaunistayb1
By definition,
$\begin{array}{rl}L\left[6x\right]\left(s\right)& ={\int }_{0}^{\mathrm{\infty }}{e}^{-ts}\phantom{\rule{thinmathspace}{0ex}}6t\phantom{\rule{thinmathspace}{0ex}}dt\\ & \stackrel{IBP}{=}{\left[-\frac{1}{s}{e}^{-ts}6t\right]}_{t=0}^{\mathrm{\infty }}+\frac{6}{s}{\int }_{0}^{\mathrm{\infty }}{e}^{-ts}\phantom{\rule{thinmathspace}{0ex}}dt\end{array}$