Non-homogeneus Linear ODEs

class: center, middle, inverse, title-slide

.title[
# Non-homogeneus Linear ODEs
]
.subtitle[
## Session 02
]
.author[
### Alejandro Ucan
]
.date[
### 2023-09-17
]

---

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# Session's Goals

* Learn the generic form of a linear ODE <br/><br/>
* Learn how to solve linear ODEs <br/><br/>
* Learn the method of variation of parameters <br/><br/>

---
# Linear ODE.

> The generic form of a linear ODE is `$$a_1(x)y'+a_0(x)y+g(x)=0,$$` where `$a_0(x)$` and `$a_1(x)$` are the coefficients. <br/><br/>
If `$g(x)=0,$` then the ODE is said to be __homogeneous__, otherwise it is __non-homogeneous.__

--
##### Example:

Homogeneuos: `$y'-2xy-2=0,\, xy'-4y-x^6e^x=0.$` <br/><br/>
Non-homogeneuos: `$y'-2xy-2=1,\, xy'-4y-x^6e^x=1.$`
---

## How to solve those ODEs?

> Put the ODE in form `$$y'+P(x)y=f(x).$$` Our solutions have to be in an interval `$I$` where `$P(x)$` and `$f(x)$` are continuous functions. <br/><br/>
The solution is given by `$$y(x)=y_h(x)+y_p(x),$$` where `$y_h$` is a solution to the homogeneous version of the ODE and `$y_p$` is a particular solution of the ODE.

---
## Method: Integrating Factor

> We need to find a solution of the form `$y_p=u(x)y_1(x)$` where `$y_1$` is a solution of the homogeneous version of the ODE. <br/><br/>
Método:
  1. Escribir la EDO de la forma: `$$y'+P(x)y=f(x).$$`
  1. The coefficient function `$P(x)$` will help us to compute our __integrating factor__ `$$e^{\int P(x)dx}.$$`
  1. If we multiply our ODE by the integrating factor, we obtain `$$\frac{d}{dx}\left[e^{\int P(x)dx} y\right]= e^{\int P(x)dx}f(x).$$`
  1. Integrate in both sides and solve for `$y.$`
---
### Example 1

> Solve `$4y'-3y=4.$`

1. Putting the ODE in the form `$y'+P(x)y=f(x)$` we have `$y'-\frac{3}{4}y=1.$`

1. `$P(x)=-\frac{3}{4},$` therefore the integrating factor `$$e^{\int \frac{-3}{4}dx}=e^{\frac{-3}{4}x}.$$`

1. We multiply our ODE by the integrating factor and we obtain `$$\frac{d}{dx}\left[e^{\frac{-3}{4}x}y\right]=e^{\frac{-3}{4}x}.$$`

1. We integrate and solve for `$y.$` `$$e^{\frac{-3}{4}x}y=c\Rightarrow y=ce^{\frac{3}{4}x}.$$`
Integramos y despejamos: `$$e^{-3x}y=c\Rightarrow y=c e^{3x}.$$`

---
### Example 2

> Solve `$y'-2y=6.$`

1. It is already in its standard form.

1. `$P(x)=-2,$` therefore the integrating factor is `$$e^{\int -2dx}=e^{-2x}.$$`

1. We multiply our ODE by the integrating factor and we obtain `$$\frac{d}{dx}\left[e^{-2x}y\right]=6e^{-2x}.$$`

1. We integrate both sides and solve for `$y$`: `$$e^{-2x}y=-3e^{-2x}+c\Rightarrow y=-3+ce^{-2x}.$$`

---
### Ejemplo 3

> Resuelva la ecuación `$y'-2xy=2x.$`

---
# Non-free Falling Body

> Consider a falling body but the medium where it falls presents a resistance (let's say the air). If the velocity with which the body falls is given by `$v(t)$` then it satisfies the non-homogeneous linear ODE: `$$mv'+\beta v=mg,$$` where `$m$` is the mass of the object, `$\beta$` is the resistance constant, and `$g$` is the gravity force.
---
## Lets solve it!

Using __integrating factor__ we can solve this ODE, let's see:

1. `$P(x)=\beta/m,$` `$f(x)=g$` y the integrating factor: `$$e^{\frac{\beta t}{m}}.$$`

1. Multiplying the integrating factor and transform the equation `$$\frac{d}{dt}\left[e^{\frac{\beta t}{m}}v\right]=ge^{\frac{\beta t}{m}}.$$`

1. Integrating both sides and solve for `$v$`: `$$v(t)=\frac{mg}{\beta}+ce^{-\frac{\beta t}{m}}.$$`