Finding an inverse Laplace Transform for a function - solving IVPs with Laplace

In summary, the Laplace transforms can be used to solve initial value problems involving second-order differential equations. For Part A, the equation can be simplified and solved using the Laplace transform and inverse transform. For Part B, the equation can be solved by finding the Laplace transform and then using partial fraction expansion to find the inverse transform.
  • #1
VinnyCee
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Homework Statement



Use Laplace Transforms to solve the following initial value problems

a. [tex]t\,y''\, - \,t\,y'\, + \,y\; = \;2\;\;\;y(0)\;=\;2\;\;\;y'(0)\;=\;-1[/tex]

b. [tex]y''\,+\,2\,y'\,-3\,y\;=\;\delta(t\,-\,1)\,-\,\delta(t\,-\,2)\;\;\;y(0)\;=\;2\;\;\;y'(0)\;=\;-2[/tex]

Homework Equations



Laplace Transforms

The Attempt at a Solution



PART A

[tex] - \frac{d}{{ds}}\left[ {s^2 \,Y\left( s \right)\, - \,s\,y\,\left( 0 \right)\, - \,y'\left( 0 \right)} \right]\, + \,\frac{d}{{ds}}\left[ {s\,Y\left( s \right)\, - \,y\,\left( 0 \right)} \right]\, + \,Y\left( s \right)\; = \;\frac{2}{s}[/tex]

[tex] - \frac{d}{{ds}}\left[ {s^2 \,Y\left( s \right)\, - \,2\,s\, + \,1} \right]\, + \,\frac{d}{{ds}}\left[ {s\,Y\left( s \right)\, - \,2} \right]\, + \,Y\left( s \right)\; = \;\frac{2}{s}[/tex]

[tex]\begin{array}{l}
- \left[ {2\;s\;Y\left( s \right)\; + \;s^2 \;Y'\left( s \right)\; - 2} \right]\; + \;\left[ {Y\left( s \right)\; + \;s\;Y'\left( s \right)} \right]\; + \;Y\left( s \right)\; = \;\frac{2}{s} \\
Y'\left( s \right)\;\left( { - s^2 \; + \;s} \right)\; + \;Y\left( s \right)\;\left( { - 2\;s\; + \;2} \right)\; = \;\frac{2}{s}\;\;\;\; \to \;\;\;\;Y'\; + \;Y\;\left( {\frac{{2\; - \;2\;s}}{{ - s^2 \; + \;s}}} \right)\; = \; - \frac{2}{{s^3 \; - \;s^2 }} \\
\end{array}[/tex]

[tex]\begin{array}{l}
\mu \left( s \right)\; = \;e^{\int {\frac{2}{s}\;ds} } \; = \;s^2 \;\;\;\; \to \;\;\;\;\frac{d}{{ds}}\left\{ {\mu \left( s \right)\;Y\left( s \right)} \right\}\; = \;\mu \left( s \right)\;Q\left( s \right) \\
\mu \left( s \right)\;Y\left( s \right)\; = \;\int { - \frac{2}{{s\; - \;1}}} \;ds\;\;\;\; \to \;\;\;\;Y\left( s \right)\; = \;\frac{{ - 2\;\ln \left( {s\; - \;1} \right)}}{{s^2 }}\; + \;\frac{C}{{s^2 }} \\
\end{array}[/tex]

How do I do an inverse transform for

[tex]\frac{{ - 2\;\ln \left( {s\; - \;1} \right)}}{{s^2 }}\; + \;\frac{C}{{s^2 }}[/tex]PART B

[tex]\begin{array}{l}
\left[ {s^2 \,Y\left( s \right)\, - \,s\,y\,\left( 0 \right)\, - \,y'\left( 0 \right)} \right]\, + \,2\left[ {s\,Y\left( s \right)\, - \,y\,\left( 0 \right)} \right]\, - \,3\,Y\left( s \right)\; = \;e^{ - s} \; - \;e^{ - 2\,s} \\
Y\,\left( s \right)\left( {s^2 \; + \;2\,s\; - \;3} \right)\; - \;2\,s\; - \;2\; = \;e^{ - s} \; - \;e^{ - 2\,s} \;\;\;\; \to \;\;\;\;Y\,\left( s \right)\; = \;\frac{{e^{ - s} \; - \;e^{ - 2\,s} \; + \;2\,s\; + \;2}}{{s^2 \; + \;2\,s\; - \;3}}\end{array}[/tex]

[tex]Y\,\left( s \right)\; = \;\frac{{e^{ - s} \; - \;e^{ - 2\,s} \; + \;2\,s\; + \;2}}{{\left( {s\; - \;1} \right)\,\left( {s\; + \;3} \right)}}[/tex]

How would I go about the partial fraction expansion of the last expression?
 
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  • #2
[tex]f(t) = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}F(s)e^{st} dt[/tex]
 
  • #3
VinnyCee said:

Homework Statement



Use Laplace Transforms to solve the following initial value problems

a. [tex]t\,y''\, - \,t\,y'\, + \,y\; = \;2\;\;\;y(0)\;=\;2\;\;\;y'(0)\;=\;-1[/tex]

b. [tex]y''\,+\,2\,y'\,-3\,y\;=\;\delta(t\,-\,1)\,-\,\delta(t\,-\,2)\;\;\;y(0)\;=\;2\;\;\;y'(0)\;=\;-2[/tex]



Homework Equations



Laplace Transforms



The Attempt at a Solution



PART A

[tex] - \frac{d}{{ds}}\left[ {s^2 \,Y\left( s \right)\, - \,s\,y\,\left( 0 \right)\, - \,y'\left( 0 \right)} \right]\, + \,\frac{d}{{ds}}\left[ {s\,Y\left( s \right)\, - \,y\,\left( 0 \right)} \right]\, + \,Y\left( s \right)\; = \;\frac{2}{s}[/tex]

[tex] - \frac{d}{{ds}}\left[ {s^2 \,Y\left( s \right)\, - \,2\,s\, + \,1} \right]\, + \,\frac{d}{{ds}}\left[ {s\,Y\left( s \right)\, - \,2} \right]\, + \,Y\left( s \right)\; = \;\frac{2}{s}[/tex]
These are meaningless. The Laplace transform of a derivative does not involve a derivative. You seem to be writing "d/dx" of the Laplace transform of the derivative. If that is true you do not want the "d/dx" in the expression.

[tex]\begin{array}{l}
- \left[ {2\;s\;Y\left( s \right)\; + \;s^2 \;Y'\left( s \right)\; - 2} \right]\; + \;\left[ {Y\left( s \right)\; + \;s\;Y'\left( s \right)} \right]\; + \;Y\left( s \right)\; = \;\frac{2}{s} \\
Y'\left( s \right)\;\left( { - s^2 \; + \;s} \right)\; + \;Y\left( s \right)\;\left( { - 2\;s\; + \;2} \right)\; = \;\frac{2}{s}\;\;\;\; \to \;\;\;\;Y'\; + \;Y\;\left( {\frac{{2\; - \;2\;s}}{{ - s^2 \; + \;s}}} \right)\; = \; - \frac{2}{{s^3 \; - \;s^2 }} \\
\end{array}[/tex]

[tex]\begin{array}{l}
\mu \left( s \right)\; = \;e^{\int {\frac{2}{s}\;ds} } \; = \;s^2 \;\;\;\; \to \;\;\;\;\frac{d}{{ds}}\left\{ {\mu \left( s \right)\;Y\left( s \right)} \right\}\; = \;\mu \left( s \right)\;Q\left( s \right) \\
\mu \left( s \right)\;Y\left( s \right)\; = \;\int { - \frac{2}{{s\; - \;1}}} \;ds\;\;\;\; \to \;\;\;\;Y\left( s \right)\; = \;\frac{{ - 2\;\ln \left( {s\; - \;1} \right)}}{{s^2 }}\; + \;\frac{C}{{s^2 }} \\
\end{array}[/tex]

How do I do an inverse transform for

[tex]\frac{{ - 2\;\ln \left( {s\; - \;1} \right)}}{{s^2 }}\; + \;\frac{C}{{s^2 }}[/tex]


PART B

[tex]\begin{array}{l}
\left[ {s^2 \,Y\left( s \right)\, - \,s\,y\,\left( 0 \right)\, - \,y'\left( 0 \right)} \right]\, + \,2\left[ {s\,Y\left( s \right)\, - \,y\,\left( 0 \right)} \right]\, - \,3\,Y\left( s \right)\; = \;e^{ - s} \; - \;e^{ - 2\,s} \\
Y\,\left( s \right)\left( {s^2 \; + \;2\,s\; - \;3} \right)\; - \;2\,s\; - \;2\; = \;e^{ - s} \; - \;e^{ - 2\,s} \;\;\;\; \to \;\;\;\;Y\,\left( s \right)\; = \;\frac{{e^{ - s} \; - \;e^{ - 2\,s} \; + \;2\,s\; + \;2}}{{s^2 \; + \;2\,s\; - \;3}}\end{array}[/tex]

[tex]Y\,\left( s \right)\; = \;\frac{{e^{ - s} \; - \;e^{ - 2\,s} \; + \;2\,s\; + \;2}}{{\left( {s\; - \;1} \right)\,\left( {s\; + \;3} \right)}}[/tex]

How would I go about the partial fraction expansion of the last expression?
 

FAQ: Finding an inverse Laplace Transform for a function - solving IVPs with Laplace

1. How do I find the inverse Laplace Transform for a given function?

The inverse Laplace Transform is calculated by using the formula:

F(t) = (1/2πi) ∫ R e^(st)F(s) ds

where F(t) is the original function and F(s) is its Laplace Transform. The integration is carried out over the line Re(s) = c, where c is a constant chosen such that the integral converges. The result of the integration is the inverse Laplace Transform of the function.

2. Can I use the Laplace Transform to solve initial value problems (IVPs)?

Yes, the Laplace Transform can be used to solve IVPs. By taking the Laplace Transform of both sides of a differential equation, you can convert it into an algebraic equation which is easier to solve. After finding the solution, you can then take the inverse Laplace Transform to get the solution to the original IVP.

3. How do I handle discontinuities or piecewise functions when using the Laplace Transform?

When dealing with discontinuities or piecewise functions, you can use the Laplace Transform on each individual piece and then combine the results using the linearity property of the Laplace Transform. This means that you can find the Laplace Transform for each piece separately and then add them together to get the Laplace Transform for the entire function.

4. Can the Laplace Transform be used for all types of functions?

The Laplace Transform can be used for a wide variety of functions, including piecewise functions, discontinuous functions, and functions with exponential or trigonometric terms. However, it may not be suitable for all types of functions, such as those with rapidly increasing or oscillating behavior.

5. Are there any tools or software that can help with finding inverse Laplace Transforms?

Yes, there are many tools and software available that can help with finding inverse Laplace Transforms, such as Wolfram Alpha, MATLAB, and online Laplace Transform calculators. These tools use algorithms and numerical methods to calculate the inverse Laplace Transform for a given function. However, it is still important to have a good understanding of the underlying principles and techniques for finding inverse Laplace Transforms.

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