Applied Laplace Transform

class: center, middle, inverse, title-slide

.title[
# Applied Laplace Transform
]
.subtitle[
## Session 10
]
.author[
### Alejandro Ucan
]
.date[
### 2023-11-05
]

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# Goals

* Introduce the inverse of a Laplace transform. <br/><br/>
* Introduce the relationship between the Laplace transform and the derivative of a function. <br/><br/>
* Solve initial value problems using Laplace transforms. <br/><br/>

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# Most commun Laplace transforms:

The next list has the Laplace transforms for the most commun functions

* `$\mathcal{L}\left\{t^n\right\}=\frac{n!}{s^{n+1}}$` 
  * `$\mathcal{L}\left\{e^{at}\right\}=\frac{1}{s-a}$`
  * `$\mathcal{L}\left\{\sin(kt)\right\}=\frac{k}{s^2+k^2}$`
  * `$\mathcal{L}\left\{\cos(kt)\right\}=\frac{s}{s^2+k^2}$`
  * `$\mathcal{L}\left\{\sinh(kt)\right\}=\frac{k}{s^2-k^2}$`
  * `$\mathcal{L}\left\{\cos(kt)\right\}=\frac{s}{s^2-k^2}$`
  
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# Inverse Laplace Transform

> Suppose that we know the Laplace transform of an unknown function, how to determine the function that produces the Laplace transform? <br/><br/>
> __Definition:__ The inverse Laplace transform of `$F(s)$` is the function `$f(t)$` such that `$$\mathcal{L}\{f(t)\}=F(s).$$`
  
--
<br/><br/>  
  * `$F(s)=\frac{1}{s^5}$`
  * `$F(s)=\frac{1}{s^2+7}$`
  * `$F(s)=\frac{s^2+6s+9}{(s-1)(s-2)(s+4)}$`
  
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# Laplace transform of a Derivative:

> __Theorem:__ If `$\mathcal{L}\{f(t)\}=F(s)$` exists, then `$$\mathcal{L}\{f'(t)\}=sF(s)-f(0)$$` `$$\mathcal{L}\{f^(n)(t)\}=s^nF(s)-s^{n-1}f(0)-s^{n-2}f'(0)-\cdots- f^{(n-1)}(0).$$`

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<br/><br/>
### Example:

Find the solution of the initial value problem `$y'+3y=13\sin(2t)\quad y(0)=6.$`

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#### Solution:

1. We apply Laplace transform in both sides of the equation:
`$$\mathcal{L}(y')+3\mathcal{L}(y)=\mathcal{13\sin(2t)}$$`
2. We use derivative theorem to express the transform of the derivative.
`$$s^2 \mathcal{L}(y)-sy(0)=\frac{26}{s^2+4}$$`
3. We solve for `$\mathcal{L}(y)$` `$$(s^2-6s)\mathcal{L}(y)=\frac{26}{s^2+4}\Rightarrow \mathcal{L}(y)=\frac{26}{s(s-6)(s^2+4)}$$`

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#### Solution:

4. We use partial fractions to express `$\mathcal{L}(y)$` as a sum of simpler fractions 
`$$\mathcal{L}(y)=\frac{13 (3 s - 2)}{40 (s^2 + 4)} + \frac{13}{120 (s - 6)} - \frac{13}{12 s}$$`

5. Then we use the table of Laplace transforms to find the inverse transform.

`$$y(t)=-\frac{13}{12}+\frac{13}{120}e^{6t}+\frac{13}{40}\left(3\cos(2t)-\frac{1}{2}\sin(2t)\right)$$`

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### Example

Find the solution of the following initial value problem `$$y''-3y'+2y=e^{-4t},\quad y(0)=1,\quad y'(0)=5.$$`

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### Example

Find the solution of the following initial value problem `$$2y'''+3y''-3y'-2y=e^{-t},\, y(0)=0,\,y'(0)=0,\,y''(0)=1.$$`

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