Find the inverse Laplace Tranformation by using convolution theorem for the function frac{1}{s^3(s-5)}

Question
Laplace transform
Find the inverse Laplace Tranformation by using convolution theorem for the function $$\frac{1}{s^3(s-5)}$$

2021-03-06
Step 1
According to the given information, it is required to find the inverse of Laplace transform using convolution theorem.
$$\frac{1}{s^3(s-5)}$$
it is required to find $$L^{-1}\left(\frac{1}{s^3(s-5)}\right)$$
Step 2
Solving further:
$$\text{let } G(s)=\frac{1}{s^3} , H(s)=\frac{1}{(s-5)}$$
the inverse laplace of G(s) and H(s) are
$$L^{-1}(G(s))=L^{-1}\left(\frac{1}{s^3}\right)=\frac{t^2}{2}=g(t)$$
$$L^{-1}(H(s))=L^{-1}\left(\frac{1}{(s-5)}\right)=e^{5t}=f(t)$$
Step 3
The convolution theorem states that if G(s) and H(s) are the Laplace transform of g(t) and h(t) then,
$$L((g*h)(t))=G(s)H(s)$$
$$\Rightarrow (g*t)(t)=L^{-1}(G(s)H(s))$$
$$\Rightarrow g(t)h(t)=L^{-1}(F(s))=\int_0^t f(t-u)g(u)du$$
Step 4
Now solve the integral to find the inverse.
$$\int_0^t e^{5(t-u)} \frac{u^2}{2} du = \frac{1}{2} e^{5t} \int_0^t(e^{-5u} \cdot u^2)du$$
$$= \frac{1}{2}e^{5t}\left[-\frac{1}{125}(25t^2e^{-5t}-2(-5te^{-5t}-e^{-5t})-2)\right]$$
$$= -\frac{1}{250}e^{5t}\left[25t^2e^{-5t}+10te^{-5t}+2e^{-5t}-2\right]$$
$$\int_0^t e^{5(t-u)} \frac{u^2}{2} du = L^{-1}(\frac{1}{s^3(s-5)})=-\frac{1}{10}t^2-\frac{1}{25}t-\frac{1}{125}+\frac{1}{125}e^{5t}$$

Relevant Questions

Find the inverse Laplace transform of the given function by using the convolution theorem. $${F}{\left({s}\right)}=\frac{s}{{{\left({s}+{1}\right)}{\left({s}^{2}+{4}\right)}}}$$
Write down the qualitative form of the inverse Laplace transform of the following function. For each question first write down the poles of the function , X(s)
a) $$X(s)=\frac{s+1}{(s+2)(s^2+2s+2)(s^2+4)}$$
b) $$X(s)=\frac{1}{(2s^2+8s+20)(s^2+2s+2)(s+8)}$$
c) $$X(s)=\frac{1}{s^2(s^2+2s+5)(s+3)}$$
$$\text{Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by }$$
$$F(s)=\int_0^\infty e^{-st} f(t)dt \(\text{where we assume s is a positive real number. For example, to find the Laplace transform of } f(t)=e^{-t} \text{ , the following improper integral is evaluated using integration by parts:} \(F(s)=\int_0^\infty e^{-st}e^{-t}dt=\int_0^\infty e^{-(s+1)t}dt=\frac{1}{s+1}$$
$$\text{ Verify the following Laplace transforms, where u is a real number. }$$
$$f(t)=t \rightarrow F(s)=\frac{1}{s^2}$$
Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by
$$F(s)=\int_0^\infty e^{-st}f(t)dt$$
where we assume s is a positive real number. For example, to find the Laplace transform of f(t) = e^{-t}, the following improper integral is evaluated using integration by parts:
$$F(s)=\int_0^\infty e^{-st}e^{-t}dt=\int_0^\infty e^{-(s+1)t}dt=\frac{1}{(s+1)}$$
Verify the following Laplace transforms, where u is a real number.
$$f(t)=1 \rightarrow F(s)=\frac{1}{s}$$
Given that $$f{{\left({t}\right)}}={4}{e}^{{-{3}{\left({t}-{4}\right)}}}$$
a) Find $${L}{\left[\frac{{{d} f{{\left({t}\right)}}}}{{{\left.{d}{t}\right.}}}\right]}$$ by differentiating f(t) and then using the Laplace transform tables in lecture notes.
b) Find $${L}{\left[\frac{{{d} f{{\left({t}\right)}}}}{{{\left.{d}{t}\right.}}}\right]}$$ using the theorem for differentiation
c) Repeat a) and b) for the case that $$f{{\left({t}\right)}}={4}{e}^{{-{3}{\left({t}-{4}\right)}}}{u}{\left({t}-{4}\right)}$$
In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative $$\frac{dy}{dt}$$ also appears. Consider the following initial value problem, defined for t > 0:
$$\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{4}{\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}={3},{y}{\left({0}\right)}={0}$$
a) Use convolution and Laplace transforms to find the Laplace transform of the solution.
$${Y}{\left({s}\right)}={L}{\left\lbrace{y}{\left({t}\right)}\right)}{\rbrace}-?$$
b) Obtain the solution y(t).
y(t) - ?
$$y"+\omega^{2}y=\sin \gamma t , y(0)=0,y'(0)=0$$
$$1)\ y(t)=L^{-1}\bigg(\frac{\gamma}{(s^{2}+\omega^{2})^{2}}\bigg) \(2)\ y(t)=L^{-1}\bigg(\frac{\gamma}{s^{2}+\omega^{2}}\bigg) \(3)\ y(t)=L^{-1}\bigg(\frac{\gamma}{(s^{2}+\gamma^{2})^{2}}\bigg) \(4)\ y(t)=L^{-1}\bigg(\frac{\gamma}{(s^{2}+\gamma^{2})(s^{2}+\omega^{2})}\bigg)$$
Find inverse Laplace transform $$L^{-1}\left\{\frac{s-5}{s^2+5s-24}\right\}$$ Please provide supporting details for your answer
$$\displaystyle{Y}{\left({s}\right)}=\frac{8}{{s}}+\frac{1}{{{s}-{3}}}\cdot{e}^{{-{2}{s}}}$$
$${F}{\left({s}\right)}=\frac{{{2}{s}-{3}}}{{{s}^{2}-{4}}}$$