Module 1 in DE prelim.pdf

AS1601 INTRODUCTION TO DIFFERENTIAL EQUATIONS Definition 1.1 An equation containing a function and its derivatives, or just derivatives is called a differe...

AS1601 INTRODUCTION TO DIFFERENTIAL EQUATIONS Definition 1.1 An equation containing a function and its derivatives, or just derivatives is called a differential equation. Two (2) Types of Differential Equations  Ordinary Differential Equations (ODEs) are equations containing a function in one (1) variable and its derivatives. Example 1 𝑑𝑓 = 2𝑥 (1) 𝑑𝑥 𝑑𝑓 2𝑥 2 = (2) 𝑑𝑥 3𝑓 3 𝑑2𝑓 𝑑𝑓 3 +2 =𝑓 (3) 𝑑𝑥 2 𝑑𝑥 The three (3) equations contain the function 𝑓 in 𝑥 and its derivatives: 𝑑𝑓 𝑑2𝑓 , 𝑑𝑥 𝑑𝑥 2  Partial Differential Equations (PDEs) are equations containing a function in many variables and its partial derivatives. Example 2 𝜕𝑓 𝜕𝑓 𝜕3𝑓 −3 + = 2𝑓 (4) 𝜕𝑥1 𝜕𝑥2 𝜕𝑥3 𝜕𝑥2 𝜕𝑥1 which contains the function 𝑓 in three (3) variables 𝑥1 , 𝑥2 and 𝑥3 and its partial derivatives: 𝜕𝑓 𝜕𝑓 𝜕3𝑓 , , 𝜕𝑥1 𝜕𝑥2 𝜕𝑥3 𝜕𝑥2 𝜕𝑥1 Definition 1.2 The order of a differential equation is the order of the highest derivative occurring in the equation. If the order is 𝑛, we say that the differential equation is an 𝑛𝑡ℎ -order differential equation. In the previous examples, equations (1) and (2) are first-order differential equations, equation (3) is of 2 nd-order, and equation (4) is of 3rd-order. Definition 1.3 A function 𝑓 in 𝑛 variables 𝑥1 , 𝑥2 , …, and 𝑥𝑛 is a solution to a differential equation if it satisfies the differential equation for all possible values of 𝑥1 , 𝑥2 , …, and 𝑥𝑛. One (1) of the easiest differential equations to solve is a first-order ODE of the form 𝑑𝑓 = 𝑓(𝑥) 𝑑𝑥 of which the equation (1) is an example. 01 Handout 1 *Property of STI Page 1 of 2  AS1601 Example 3 Consider equation (1). 𝑑𝑓 = 2𝑥 𝑑𝑥 Integrating both sides, 𝑓(𝑥) = ∫ 2𝑥𝑑𝑥 𝑓(𝑥) = 𝑥 2 + 𝐶 (5) where 𝐶 may take any constant value. The set of solutions given by (5) is known as the general solution. Specifying the value of 𝐶 gives the particular solutions. For example: 1. 𝑓(𝑥) = 𝑥 2 , where 𝐶 = 0 2. 𝑔(𝑥) = 𝑥 2 + 1, where 𝐶 = 1 3. ℎ(𝑥) = 𝑥 2 − 1, where 𝐶 = −1 To check if the functions are indeed solutions, we differentiate them and see if we arrive with the original problem. Example 4 Consider the previous problem and the particular solution ℎ(𝑥) = 𝑥 2 − 1, 𝑑ℎ 𝑑 2 = (𝑥 − 1) 𝑑𝑥 𝑑𝑥 𝑑 2 𝑑 = (𝑥 ) − (1) 𝑑𝑥 𝑑𝑥 = 2𝑥 which is the original derivative 2𝑥 in equation (1). Example 5 Determine if 𝑓(𝑥) = 𝑥 2 + 2𝑥 + 1 is a solution to 𝑑2 𝑓 𝑑𝑓 3 −𝑥 = 𝑥+8− 𝑑𝑥 𝑑𝑥 Solution: We first arrange the differential equation so that all terms with derivatives are on one side and the remaining terms are on the other. 𝑑 2 𝑓 𝑑𝑓 3 + = 2𝑥 + 8 𝑑𝑥 𝑑𝑥 2 Let 𝑓(𝑥) = 𝑥 + 2𝑥 + 1 𝑑 2 𝑓 𝑑𝑓 𝑑2 𝑑 2 3 + = 3 (𝑥 2 + 2𝑥 + 1) + (𝑥 + 2𝑥 + 1) 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑 = 3 (2𝑥 + 2) + 2𝑥 + 2 𝑑𝑥 = 3(2) + 2𝑥 + 2 = 2𝑥 + 8 which is the original problem itself. REFERENCES: Guterman, M. & Nitecki, Z. (1988). Differential equations a first course 2nd edition. Philadelphia: Saunders College Publishing. Leithold, L. (1996). The calculus 7. Boston: Addison Wesley Longman, Inc.. 01 Handout 1 *Property of STI Page 2 of 2