Modelling with higher c.c. LODE’s

class: center, middle, inverse, title-slide

.title[
# Modelling with higher c.c. LODE’s
]
.subtitle[
## Session 05
]
.author[
### Alejandro Ucan
]
.date[
### 2024-09-28
]

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# Goals

* Describe two models of higher order differential equations. <br/><br/>
* Solve the models of higher order differential equations.<br/><br/>

---
# Recall

The second LODE with constant coefficients is given by:

`$$a_2y'' + a_1y' + a_0y + f(x)=0$$`
as engineers we can think that as an equilibrium in a particular problem. For example in case of movement and displacement, we have that the acceleration, velocity and position are in equilibrium.

---
### Control Systems

In Engineering a control system is a device or set of devices to manage, command, direct or regulate the behavior of other devices or systems. For example: <br/><br/>
* Heat and Cooling of machinery in a factory. <br/><br/>
* Current and Power in a factory. <br/><br/>
* Speed and Torque in a motor. <br/><br/>

---
### Main Hypothesis:
`$$a_2y'' + a_1y' + a_0y = f(x)$$`

1. `$y(t)$` is the output of the system (the response). <br/>
2. `$f(t)$` is the input of the system (the excitation or adaptation of the system). <br/>
3. `$a_1$` is the damping ration (resistance or oscilation of the system). <br/>
4. `$a_0$` indicates how fast the system is adapting. <br/>
5. `$a_2$` is the inertia of the system and it indicates the relation between the input and the output.

---
### Example

A heating element in a manufacture process satisfy a second order LODE. We know that the inertia of the system is one, the damping ratio is 4 and the adaptation of the system is 10. Additionaly we have that this heating element is excited with a signal `$f(t)=50$`. If initially the temperature is 20°C and the initial change of rate of the temperature is 0. Find the function that models the temperature, what is the steady temperature of the system?

---
### Solution

We know that the IVP is given by `$$y''+ 4y' + 10y = 50$$` `$$y(0)=20,\, y'(0)=0$$`

The solution of the homogeneous equation is given by `$$y_h(t)=c_1e^{-2t}\cos(\sqrt{6}t)+c_2e^{-2t}\sin(\sqrt{6}t)$$`

The particular solution is given by `$y_p(t)=A,$` solving with undetermined coefficients we get `$A=5$`

The general solution is given by `$$y(t)=c_1e^{-2t}\cos(\sqrt{6}t)+c_2e^{-2t}\sin(\sqrt{6}t)+5$$`

Evaluating the initial conditions we get `$c_1=3$` and `$c_2=\sqrt{6}.$` Therefore the solution is given by `$$y(t)=3e^{-2t}\cos(\sqrt{6}t)+\sqrt{6}e^{-2t}\sin(\sqrt{6}t)+5$$` At infinity, the temperature approaches to 5°C.

---
### Example 2

We want to understand the indoor temperature of a factory, we know that the inertia of the system is 1, the damping ratio is 3 and the adaptation of the system is 2. The factory has a cooling system that excites with a signal `$f(t)=20+\sin(t)$`. If initially the temperature is 30°C and the initial change of rate of the temperature is 1. Find the function that models the temperature, what is the average temperature of the system at infinity?

---
### Solution

We know that the IVP is given by `$$y''+ 3y' + 2y = 20+\sin(t)$$` `$$y(0)=30,\, y'(0)=1$$`

The solution of the homogeneous equation is given by `$$y_h(t)=c_1e^{-t}+c_2e^{-2t}$$`

The particular solution is given by `$y_p(t)=A+B\sin(t)+C\cos(t),$` solving with undetermined coefficients we get `$A=85/10$`, `$B=1/10$` and `$C=-3/10$`.

The general solution is given by `$$y(t)=c_1e^{-t}+c_2e^{-2t}+\frac{85}{10}+\frac{1}{10}\sin(t)-\frac{3}{10}\cos(t)$$`

Evaluating the initial conditions we get `$y(t)=\frac{1}{10} (85 - 177 e^{-2 t} + 345 e^{-t} - 3 \cos(t) + \sin(t))$` At infinity, the temperature oscilates close to 8.5°C.

---
### Example 3

An elevator that weights 500 kg (people included) is in a building. In order to move properly the elevator, we have a resistance 50 `$Ns/m$`, the tension of the cables (adaptation) is given 2000 `$N/m.$` Elevator's motor excites the system with a force of 2 `$m/s^2$`. If the elevator starting movement is from 0 m at a velocity of 0 `$m/s.$` Find the function that models the position of the elevator.

---
### Solution

We know that the IVP is given by `$$500y''+50y'+2000y=2$$` `$$y(0)=0,\, y'(0)=0$$`

The solution of the homogeneous equation is given by `$$y_h(t)=c_1e^{-t/20}\cos(\frac{\sqrt{159}}{20}t)+c_2e^{-t/20}\sin(\frac{\sqrt{159}}{20}t)$$`

The particular solution is given by `$y_p(t)=A,$` solving with undetermined coefficients we get `$A=1/100.$`

The general solution is given by `$$y(t)=c_1e^{-t/20}\cos(\frac{\sqrt{159}}{20}t)+c_2e^{-t/20}\sin(\frac{\sqrt{159}}{20}t)+\frac{1}{100}$$`

Evaluating the initial conditions we get the constants.

---
#### Example 4

An electric motor is used to drive a load, and its speed can be modeled by a second-order linear ordinary differential equation. The damping ratio is 0.2 and the adaptability constant is given by 2. The motor is excited by a signal of `$f(t)=5v.$` If the initial angular speed is 0, and the initial angular speed is 0. Find the function that models the angular speed of the motor.

---
### Solution

We know that the IVP is given by `$$y''+0.2y'+2y=5$$` `$$y(0)=0,\, y'(0)=0$$`

The solution of the homogeneous equation is given by `$$y_h(t)=c_1e^{-t/10}\cos(1.41 t)+c_2e^{-t/10}\sin(1.41 t)$$`

The particular solution is given by `$y_p(t)=A,$` solving with undetermined coefficients we get `$A=5/2.$`

The general solution is given by `$$y(t)=c_1e^{-t/10}\cos(1.41 t)+c_2e^{-t/10}\sin(1.41 t)+\frac{5}{2}$$`

Evaluating the initial conditions we get the constants.