How Do You Find the Inverse Laplace Transform for a Given Differential Equation?

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Homework Help Overview

The discussion revolves around finding the inverse Laplace transform for a differential equation involving second derivatives. The equation presented is a linear combination of derivatives of a function y(t) and an input function x(t), with parameters a and b influencing the behavior of the system.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the Laplace transforms of the function y(t) and its derivatives, questioning the assumptions made about these transforms. There is also an exploration of the Bromwich integral as a method for finding the inverse Laplace transform, along with discussions about the poles of the function and the region of convergence.

Discussion Status

The discussion includes various approaches to understanding the inverse Laplace transform, with some participants suggesting the use of the Bromwich integral. There are questions regarding the correctness of assumptions about the Laplace transforms of derivatives, and some participants express uncertainty about the implications of their findings.

Contextual Notes

Participants are navigating through the definitions and properties of Laplace transforms, particularly in the context of a specific differential equation. There is an emphasis on clarifying the relationships between the transforms of y(t) and its derivatives, as well as the implications of the parameters a and b in the equation.

redundant6939
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Homework Statement


Find H(s) = \frac{Y(s)}{X(s)}
\frac {d^2y(t)}{dt^2} + a\frac {dy(t)}{dt} = x(t) + by(t)

Homework Equations

The Attempt at a Solution


[s^2 + as - b] Y(s) = X(s)
H(s) = \frac{1}{s^2+as-b}

I assume the inverse is a sign or a cosine but unsure which one.
 
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redundant6939 said:

Homework Statement


Find H(s) = \frac{Y(s)}{X(s)}
\frac {d^2y(t)}{dt^2} + a\frac {dy(t)}{dt} = x(t) + by(t)

Homework Equations

The Attempt at a Solution


[s^2 + as - b] Y(s) = X(s)
H(s) = \frac{1}{s^2+as-b}

I assume the inverse is a sign or a cosine but unsure which one.

If ##Y(s)## is the LT of ##y(t)##, what are the LTs of ##y'(t)## and ##y''(t)##?

Hint: not what you wrote.
 
You can use the Bromwich integral to find the inverse Laplace transform.
The poles of ##s## are ##s = \frac{-a\pm\sqrt{a^2 + 4b}}{2}##
\begin{align}
\mathcal{L}^{-1}\Bigl\{\frac{1}{s^2 + as - b}\Bigr\} &= \frac{1}{2\pi i}\int_{\gamma -i\infty}^{\gamma +i\infty}\frac{e^{st}}{s^2 + as - b}ds\\
&= \sum\text{Res}\\
&= \lim_{s\to s_+}(s - s_+)\frac{e^{st}}{s^2 + as - b} + \lim_{s\to s_-}(s - s_-)\frac{e^{st}}{s^2 + as - b}
\end{align}
where ##s_+=\frac{-a + \sqrt{a^2 + 4b}}{2}## and ##s_-=\frac{-a - \sqrt{a^2 + 4b}}{2}##
 
Ray Vickson said:
If ##Y(s)## is the LT of ##y(t)##, what are the LTs of ##y'(t)## and ##y''(t)##?

Hint: not what you wrote.

The LT of ##y'(t)## and ##y''(t)## are sY[s} and ##s^2Y## respectively. I am not sure what you mean exactly.
 
Dustinsfl said:
You can use the Bromwich integral to find the inverse Laplace transform.
The poles of ##s## are ##s = \frac{-a\pm\sqrt{a^2 + 4b}}{2}##
\begin{align}
\mathcal{L}^{-1}\Bigl\{\frac{1}{s^2 + as - b}\Bigr\} &= \frac{1}{2\pi i}\int_{\gamma -i\infty}^{\gamma +i\infty}\frac{e^{st}}{s^2 + as - b}ds\\
&= \sum\text{Res}\\
&= \lim_{s\to s_+}(s - s_+)\frac{e^{st}}{s^2 + as - b} + \lim_{s\to s_-}(s - s_-)\frac{e^{st}}{s^2 + as - b}
\end{align}
where ##s_+=\frac{-a + \sqrt{a^2 + 4b}}{2}## and ##s_-=\frac{-a - \sqrt{a^2 + 4b}}{2}##

The region of convergence then, would it be btw the two values of s?
 
redundant6939 said:
The region of convergence then, would it be btw the two values of s?

You need to find y(t) for the ROC.
 
redundant6939 said:
The LT of ##y'(t)## and ##y''(t)## are sY[s} and ##s^2Y## respectively. I am not sure what you mean exactly.

What I mean is that your statements above are false, in general. Consult a table of Laplace transforms to see what you should have written.
 

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