1st Traslation Theorem

class: center, middle, inverse, title-slide

.title[
# 1st Traslation Theorem
]
.subtitle[
## Session 11
]
.author[
### Alejandro Ucan
]
.date[
### 2023-11-12
]

---

<div>
<style type="text/css">.xaringan-extra-logo {
width: 110px;
height: 128px;
z-index: 0;
background-image: url(https://github.com/alxcn/TecLogoEIC/blob/9562a53875418e749a296c85808a19c85fc4be74/IngenieriaCiencias_Horizontal_RGB.png);
background-size: contain;
background-repeat: no-repeat;
position: absolute;
top:2em;right:2em;
}
</style>
<script>(function () {
  let tries = 0
  function addLogo () {
    if (typeof slideshow === 'undefined') {
      tries += 1
      if (tries < 10) {
        setTimeout(addLogo, 100)
      }
    } else {
      document.querySelectorAll('.remark-slide-content:not(.title-slide):not(.inverse):not(.hide_logo)')
        .forEach(function (slide) {
          const logo = document.createElement('div')
          logo.classList = 'xaringan-extra-logo'
          logo.href = null
          slide.appendChild(logo)
        })
    }
  }
  document.addEventListener('DOMContentLoaded', addLogo)
})()</script>
</div>

# Goals

* State 1st Traslation Theorem. <br/><br/>
* Provide some examples on how to apply the theorem. <br/><br/>
* Apply the theorem to solve ODEs.

---
# Motivation:
<br/><br/>

> What is the Laplace transform of the function `$f(t)=e^{5t}t.$`

--
<br/><br/>
Following the definition of the Laplace transform, we have that: `$$\mathcal{L}\{e^{5t}t\}=\int_0^\infty e^{-st}e^{5t}t\,dt=\int_0^\infty te^{-(s-5)t}\,dt$$` `$$\lim_{a\to\infty} -\frac{te^{-(s-5)t}}{s-5}-\frac{e^{-(s-5)t}}{(s-5)^2}\bigg|_0^a=\frac{1}{(s-5)^2}.$$`

---
# 1st Traslation Theorem

> __Theorem:__ Let `$f(t)$` be a function such that `$\mathcal{L}\{f(t)\}=F(s)$` exists. Then `$$\mathcal{L}\{e^{at}f(t)\}=F(s-a).$$`
<br/><br/>
> __Some Implications:__ From now, every time we see a known function "shifted" by an exponential factor, this translates as a shift in `$s$` in the transform.

---
## Applying the Theorem:

### Example (Direct Calculation):

> Compute the Laplace Transform of `$f(t)=e^{5t}t^3,$` `$g(t)=e^{-2t}\cos(4t)$` y `$h(t)=e^{-2t}$` if `$t>1$` and `$h(t)=0$` otherwise.

--
<br/><br/>
##### Solution:

1. The shifted function of `$f$` is `$t^3$` and its transform is `$\frac{3!}{s^4}.$` Therefore the transform of `$f$` is `$\frac{3!}{(s-5)^4}.$` <br/>
2. The shifted function of `$g$` is `$\cos(4t)$` and its transform is `$\frac{s}{s^2+4^2}.$` Therefore the transform of `$g$` is `$\frac{s+2}{(s+2)^2+4^2}.$` <br/>
3. The shifted function of `$h$` is the unit step function and its transform is `$\frac{e^s}{s}.$` Therefore the transform of `$h$` is `$e^{s+2}\frac{1}{s+2}.$` <br/>

---
### Example (Inverse Calculation):

> Find the inverse Laplace transform of: 
`$\mathcal{L}^{-1}\left\{\frac{2s+5}{(s-3)^2}\right\}$`
--
<br/>
##### Solution:
Let's find out the partial fractions: `$\frac{2s+5}{(s-3)^2}=\frac{A}{(s-3)}+\frac{B}{(s-3)^2}=\frac{2}{s-3}+\frac{-1}{(s-3)^2}.$`

Now we notice that those are shifted functions, therefore we look for the corresponding transform and we multiply it by an exponential factor: `$$\mathcal{L}^{-1}\left\{\frac{2s+5}{(s-3)^2}\right\}=\mathcal{L}^{-1}\left\{\frac{2}{s-3}\right\}+\mathcal{L}^{-1}\left\{\frac{-1}{(s-3)^2}\right\}=2e^{3t}-te^{3t}.$$`

---
### Example (Inverse Calculation):

> Find the inverse Laplace transform of:
`$\mathcal{L}^{-1}\left\{\frac{s+5}{s^2+4s+6}\right\}$`
--
<br/>
##### Solution:
First we check if the denominator has real roots: `$s^2+4s+6=0\implies s=\frac{-4\pm\sqrt{16-4(1)(6)}}{2}=-2\pm i.$` Therefore the roots are complex and we need to complete the square: `$$s^2+4s+6=(s+2)^2+4.$$` Therefore the fraction is `$$\frac{s+5}{(s+2)^2+4}=\frac{(s+2)}{(s+2)^2+4}+\frac{3}{(s+2)^2+4}$$` We use the formulaes to find the inverse: `$$\mathcal{L}^{-1}\left\{\frac{(s+2)}{(s+2)^2+4}\right\}=e^{-2t}\cos(2t)\quad \mathcal{L}^{-1}\left\{\frac{3}{(s+2)^2+4}\right\}=\frac{3}{2}e^{-2t}\sin(2t).$$`

---
### Example (Solving IVP)

> Find the solution of the IVP: `$y''-6y'+9y=t^2e^{3t},\quad y(0)=2,\quad y'(0)=17.$`

--
<br/><br/>
First we apply Laplace transform in both sides of the ODE. `$$s^2\mathcal{L}\{y\}-sy(0)-y'(0)-6s\mathcal{L}\{y\}+6y(0)+9\mathcal{L}\{y\}=\mathcal{L}\{t^2e^{3t}\}$$` `$$\implies (s^2-6s+9)\mathcal{L}\{y\}=2s+5+\frac{2}{(s-3)^3}$$` 
We simplify and check if the quadratic polynomial is factorizable:
`$$\mathcal{L}\{y\}=\frac{2s+17}{(s-3)^3}+\frac{2}{(s-3)^3(s-3)^2}$$` We find the inverse Laplace transform.

---
### Example (Solving IVP)

> Find the solution of the IVP: `$$y''+4y'+6y=e^{-t},\quad y(0)=0,\quad y'(0)=0.$$`

--
<br/><br/>
First we apply Laplace transform in both sides fo the ODE `$$s^2\mathcal{L}\{y\}-sy(0)-y'(0)+4s\mathcal{L}\{y\}-4y(0)+6\mathcal{L}\{y\}=\mathcal{L}\{e^{-t}\}$$` `$$\implies (s^2+4s+6)\mathcal{L}\{y\}=\frac{1}{s+1}$$` We use partial fractions: `$$\mathcal{L}\{y\}=\frac{1}{(s+1)(s^2+4s+6)}=\frac{1/3}{s-1}+\frac{s-1/3}{s^2+4s+6}$$` We construct the function using the inverse Laplace transform.