1. Lecture 2 Ordinary Differential Equations fall semester
Instructor: A. S. Brwa / MSc. In Structural Engineering
College of Engineering / Ishik University

2. Ordinary Differential Equations
Ishik University
Ordinary Differential Equations
Ordinary Differential Equation (ODE): is a differential equation containing one or more functions of one independent variable and its derivativeses
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

3. Ishik University
Learning Outcomes
Understand the use of differential equations in modelling engineering systems
Identify the order and type of a differential equation recognize the nature of a general solution
Determine the nature of the appropriate additional conditions which will give a unique solution to the equation
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

4. Ordinary Differential Equations
Ishik University
Ordinary Differential Equations
Rate of Change: A rate of change is a rate that describes how one quantity changes in relation to another quantity. If x is the independent variable and y is the dependent variable, then
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒= 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥 = 𝜕𝑦 𝜕𝑥
Rates of change can be positive or negative. This corresponds to an increase or decrease in the y –value between the two data points. When a quantity does not change over time, it is called zero rate of change
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

5. Ordinary Differential Equations
Ishik University
Ordinary Differential Equations
Many models of engineering systems involve the rate of change of a quantity. There is thus a need to incorporate derivatives into the mathematical model. These mathematical models are examples of differential equations.
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

6. Ordinary Differential Equations
Ishik University
Ordinary Differential Equations
Accompanying the differential equation will be one or more conditions that let us obtain a unique solution to a particular problem. Often we solve the differential equation first to obtain a general solution; then we apply the conditions to obtain the unique solution. It is important to know which conditions must be specified in order to obtain a unique solution.
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

7. Ishik University
Case Study
Newton’s law of cooling as an example When a hot liquid is placed in a cooler environment, experimental observation shows that its temperature decreases to approximately that of its surroundings. A typical graph of the temperature of the liquid plotted against time is shown in Figure 1.
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

8. Ishik University
Faculty of Engineering – Differential Equations – Lecture 2 – 1st Order ODE

9. Ishik University
Case Study
After an initially rapid decrease the temperature changes progressively less rapidly and eventually the curve appears to ‘flatten out’. Newton’s law of cooling states that the rate of cooling of liquid is proportional to the difference between its temperature and the temperature of its environment (the ambient temperature).
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

10. Ishik University
Case Study
To convert this into mathematics, let t be the time elapsed (in seconds, s), y the temperature of the liquid (℃), and 𝒚 𝟎 the temperature of the liquid at the start (t = 0). The temperature of the surroundings is denoted by 𝒚 𝒔 . Now, we can create a mathematical equation which is equivalent to Newton’s law of cooling and state the accompanying condition.
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

11. Ishik University
Case Study
First, find an expression for the rate of cooling, and an expression for the difference between the liquid’s temperature and that of the environment:
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

12. Ishik University
Case Study
The rate of cooling is the rate of change of temperature with time:
𝝏𝒚 𝝏𝒕
The temperature difference is
𝒚−𝒚 𝒔
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

13. Case Study Now formulate Newton’s law of cooling:
Ishik University
Case Study
Now formulate Newton’s law of cooling:
You should obtain equivalently: dθ dt = −k(θ − θs).
𝝏𝒚 𝝏𝒕 ∝ 𝒚−𝒚 𝒔
Which is equivalent to
𝝏𝒚 𝝏𝒕 =−𝜿 𝒚−𝒚 𝒔
𝜿 is a positive constant of proportion
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

14. Ishik University
Case Study
The negative sign ok (−𝜿) is present because (𝒚−𝒚 𝒔 ) is positive, whereas 𝝏𝒚 𝝏𝒕 must be negative, since 𝒚 decreases with time. The units of 𝜿 are 𝑠 −1 . The accompanying condition is 𝒚=𝒚 𝟎 which simply states the temperature of the liquid when the cooling begins.
at t = 0
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

15. 𝒚=𝒚 𝒔 + 𝒚 𝟎 − 𝒚 𝒔 𝒆 −𝜿𝒕 Case Study
Ishik University
Case Study
In the above Task we call t the independent variable and 𝒚 the dependent variable. Since the condition is given at t = 0. so, this is considered as an initial condition. For future reference, the solution of the above differential equation which satisfies the initial condition is.
𝒚=𝒚 𝒔 + 𝒚 𝟎 − 𝒚 𝒔 𝒆 −𝜿𝒕
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

16. General Solution of a ODE
Ishik University
General Solution of a ODE
The general solution of a ODE Consider the equation 𝑦 =𝐴 𝑒 2𝑥 where A is an arbitrary constant. If we differentiate it we obtain 𝑑𝑦 𝑑𝑥 =2𝐴 𝑒 2𝑥 and so, since 𝑦 =𝐴 𝑒 2𝑥 we obtain 𝑑𝑦 𝑑𝑥 =2𝑦
An arbitrary constant is a constant whose value could be assumed to be anything, it doesn't depend on the other variables in an equation or expression. 
A constant that's not arbitrary can usually just take one value, such as 1, 2, or 10
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

17. General Solution of a ODE
Ishik University
General Solution of a ODE
Thus a differential equation satisfied by y is
𝑑𝑦 𝑑𝑥 =2𝑦
Note that we have eliminated the arbitrary constant.
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

18. General Solution of a ODE
Ishik University
General Solution of a ODE
Now consider the equation
𝑦=𝐴𝑐𝑜𝑠3𝑥+𝐵𝑠𝑖𝑛3𝑥
Where, A and B are arbitrary constants.
Differentiate once , we obtain
𝑑𝑦 𝑑𝑥 =−3𝐴𝑠𝑖𝑛3𝑥+3𝐵𝑐𝑜𝑠3𝑥
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

19. General Solution of a ODE
Ishik University
General Solution of a ODE
Differentiating a second time gives 𝑑 2 𝑦 𝑑 𝑥 2 =−9𝐴𝑐𝑜𝑠3𝑥−9𝐵𝑠𝑖𝑛3𝑥 The right hand side is simply (−9) times the expression for y. Hence y satisfies the differential equation 𝑑 2 𝑦 𝑑 𝑥 2 =−9𝑦
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

20. General Solution of a ODE
Ishik University
General Solution of a ODE
Find a differential equation satisfied by 𝑦=𝐴cosh2𝑥 + 𝐵sinh2𝑥 Where, A and B are arbitrary constants. Differentiating once we obtain 𝑑𝑦 𝑑𝑥 =2𝐴𝑠𝑖𝑛h2𝑥 +2𝐵𝑐𝑜𝑛h2𝑥
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

21. General Solution of a ODE
Ishik University
General Solution of a ODE
Differentiating a second time we obtain 𝑑 2 𝑦 𝑑 𝑥 2 =4𝐴cosh2𝑥 +4𝐵sinh2𝑥 𝑑 2 𝑦 𝑑 𝑥 2 =4𝑦
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

22. General Solution of a ODE
Ishik University
General Solution of a ODE
We have seen that an expression including one arbitrary constant required one differentiation to obtain a differential equation which eliminated the arbitrary constant. Where two constants were presented, two differentiations were required.
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

23. General Solution of a ODE
Ishik University
General Solution of a ODE
Integrate twice the differential equation 𝑑 2 𝑦 𝑑 𝑥 2 = 𝑤 2 𝑙𝑥− 𝑥 2 Where, w and L are constants, to find a general solution for y. Integrating once:
where A is an arbitrary constant (of integration)
𝑑𝑦 𝑑𝑥 = 𝑤 2 𝑙 𝑥 2 2 − 𝑥 A
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

24. General Solution of a ODE
Ishik University
General Solution of a ODE
Integrating again: 𝑦= 𝑤 2 𝑙 𝑥 3 6 − 𝑥 Ax+B Where, B is second arbitrary constant (of integration)
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

25. General Solution of a ODE
Ishik University
General Solution of a ODE
Consider the simple differential equation
𝑑𝑦 𝑑𝑥 =2𝑥
On integrating, we obtain the general solution
𝑦= 𝑥 2 +𝐶
Where, C is an arbitrary constant. As C varies we get different solutions, each of which belongs to the family of solutions
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

26. General Solution of a ODE
Ishik University
General Solution of a ODE
Consider the simple differential equation
𝑑𝑦 𝑑𝑥 =2𝑥
On integrating, we obtain the general solution
𝑦= 𝑥 2 +𝐶
Where, C is an arbitrary constant. As C varies we get different solutions, each of which belongs to the family of solutions
Figure-2
Shows different values of C
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations

27. Home Work Find differential equations satisfied by 𝑦=𝐴 cos⁡4𝑥+𝐵 sin⁡4𝑥
Ishik University
Home Work
Find differential equations satisfied by
𝑦=𝐴 cos⁡4𝑥+𝐵 sin⁡4𝑥
𝑦=𝐴 sin⁡𝑥+𝐵 sinh⁡𝑥+𝐶 cos⁡𝑥+𝐷 cosh⁡𝑥
Solve the differential equation
𝑑 2 𝑦 𝑑 𝑥 2 =6𝑥
Faculty of Engineering – Differential Equations – Lecture 2 – Ordinary Differential Equations