emmostatwf

2022-09-01

I have an expression such as ${\int}_{0}^{x+l}y(z)g(x-z)dz$ and I want to evaluate its Laplace transform w.r.t x in terms of the Laplace transform of y(x). I know that I can substitute $t=x+l$, and coerce it into the standard from to get the Laplace transform w.r.t t, but I need its transform w.r.t x.

Motivation:

I would like to solve an integral equation: $y(x)=f(x)+{\int}_{0}^{x+l}y(z)g(x-z)dz$

If the integral limit had been to x, we would have had $y(x)=f(x)+{\int}_{0}^{x}y(z)g(x-z)dz$. This leads to $Y(s)=\frac{F(s)}{1+K(s)}$

Motivation:

I would like to solve an integral equation: $y(x)=f(x)+{\int}_{0}^{x+l}y(z)g(x-z)dz$

If the integral limit had been to x, we would have had $y(x)=f(x)+{\int}_{0}^{x}y(z)g(x-z)dz$. This leads to $Y(s)=\frac{F(s)}{1+K(s)}$

lascosasdeali3v

Beginner2022-09-02Added 10 answers

Hint:

$y(x)=f(x)+{\int}_{0}^{x+l}y(z)g(x-z)\text{}dz$

$y(x)=f(x)+{\int}_{x}^{-l}y(x-t)g(t)\text{}d(x-t)$

$y(x)=f(x)-{\int}_{x}^{-l}y(x-t)g(t)\text{}dt$

$y(x)=f(x)+{\int}_{-l}^{x}y(x-t)g(t)\text{}dt$

$y(x)=f(x)+{\int}_{0}^{x+l}y(z)g(x-z)\text{}dz$

$y(x)=f(x)+{\int}_{x}^{-l}y(x-t)g(t)\text{}d(x-t)$

$y(x)=f(x)-{\int}_{x}^{-l}y(x-t)g(t)\text{}dt$

$y(x)=f(x)+{\int}_{-l}^{x}y(x-t)g(t)\text{}dt$

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My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$