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

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SUMMARY

The discussion focuses on finding the inverse Laplace transform for the differential equation given by \(\frac{d^2y(t)}{dt^2} + a\frac{dy(t)}{dt} = x(t) + by(t)\). The transfer function is expressed as \(H(s) = \frac{Y(s)}{X(s)} = \frac{1}{s^2 + as - b}\). Participants emphasize the use of the Bromwich integral to compute the inverse transform and identify the poles at \(s = \frac{-a \pm \sqrt{a^2 + 4b}}{2}\). The Laplace transforms of the derivatives \(y'(t)\) and \(y''(t)\) are confirmed as \(sY(s)\) and \(s^2Y(s)\), respectively.

PREREQUISITES
  • Understanding of Laplace transforms and their properties
  • Familiarity with differential equations and their solutions
  • Knowledge of the Bromwich integral for inverse transforms
  • Ability to identify and compute poles of complex functions
NEXT STEPS
  • Study the Bromwich integral method for inverse Laplace transforms
  • Learn how to identify and analyze poles in complex functions
  • Consult a comprehensive table of Laplace transforms for reference
  • Practice solving differential equations using Laplace transforms
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Students, engineers, and mathematicians involved in control systems, signal processing, or any field requiring the application of Laplace transforms to solve differential equations.

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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