Goals

• Describe two models of higher order differential equations.
  
  
• Solve the models of higher order differential equations.
  
  

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:

• Heat and Cooling of machinery in a factory.
  
  
• Current and Power in a factory.
  
  
• Speed and Torque in a motor.
  
  

Main Hypothesis:

$$a_2y'' + a_1y' + a_0y = f(x)$$

1. $y(t)$ is the output of the system (the response).
   
2. $f(t)$ is the input of the system (the excitation or adaptation of the system).
   
3. $a_1$ is the damping ration (resistance or oscilation of the system).
   
4. $a_0$ indicates how fast the system is adapting.
   
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.