Question

Laplace transform

f(x) = { e^2x, 0<=x<1
{ 3x, x>=1

must express in unit step and translations and find the laplace transform

Answer 1

i never did get the knack of the shortcut for this

Answer 2

\[u(x)e^{2x}-u(x-1)e^{2x}+u(x-1)3x\]

Answer 3

how did you arrive at that, or what is u?

Answer 4

i used \(u\) for unit step function

Answer 5

\[\stackrel{e^{2x}\text{ turns on at }x=0}{\overbrace {u(x)e^{2x}}}\quad\stackrel{e^{2x}\text{ turns off at }x=1}{\overbrace {-u(x-1)e^{2x}}}+\stackrel{3x\text{ turns on at }x=1}{\overbrace {u(x-1)3x}}\]

Answer 6

so that is expressing the piecewise as a unit step function and translations, now you go back to the original piecewise and find the laplace, or what do you use to find the laplace transform?

Answer 7

\[\mathcal L\Big\{u(x)e^{2x}-u(x-1)e^{2x}+u(x-1)3x\Big\}\]
\[=\mathcal L\Big\{u(x)e^{2x}\Big\}-\mathcal L\Big\{u(x-1)e^{2x}\Big\}+\mathcal L\Big\{u(x-1)3x\Big\}\]

Answer 8

the first term if you integrate e^-sx u(x) e^2x, to find laplace, what do you do with the u(x)

Answer 9

\[\boxed{\mathcal L\Big\{f(x-a)u(x-a)\Big\}=e^{-as}F(s)}\]

Answer 10

@experimentX do you have time to help me through this problem?