Lecture 25 - Differential Equations PDF

Lecture-25 Mathematics 1 (15B11MA111) CO [C105.5] Module: Ordinary Differential Equations Topic: Homogeneous constant coefficient ODEs Reference for the lecture R.K Jain and S.R.K. Iyenger, “Advanced Engineering Mathematics” fifth edit...

Lecture-25 Mathematics 1 (15B11MA111) CO [C105.5] Module: Ordinary Differential Equations Topic: Homogeneous constant coefficient ODEs Reference for the lecture R.K Jain and S.R.K. Iyenger, “Advanced Engineering Mathematics” fifth edition, Narosa publishing house, 2016. 1 Differential Equations According to Wikipedia In mathematics, a differential equation is an equation that relates one or more functions and their derivatives. The study of differential equations is a wide field in pure and applied mathematics, physics, and engineering. All of these disciplines are concerned with the properties of differential equations of various types. Differential equations play an important role in modeling virtually every physical, technical, or biological process, from celestial motion, to bridge design, to interactions between neurons. 2 3 4 Solution of a Differential Equation An equation containing dependent variable (y) and independent variable (x) and free from derivative, which satisfies the differential equation, is called the solution (primitive) of the differential equation. The solution which contains arbitrary constants is called the general solution of the differential equation. The solution free from arbitrary constants i.e., the solution obtained from the general solution by giving particular values to the arbitrary constants is called a particular solution of the differential equation. You have already studied methods for first order differential equations such as variable separable, Linear ODE etc. Now, we will discuss higher order ODEs (Second and higher order). 5 Higher Order Linear Differential Equations with Constant Coefficients A linear ordinary differential equation of order n, is written as 𝑑𝑛 𝑦 𝑑 𝑛−1 𝑦 𝑑𝑦 𝑎0 𝑥 + 𝑎1 𝑥 + ⋯ + 𝑎𝑛−1 𝑥 + 𝑎𝑛 𝑥 𝑦 = 𝑟(𝑥) 𝑑𝑥 𝑛 𝑑𝑥 𝑛−1 𝑑𝑥 or 𝑎0 𝑥 𝑦 𝑛 𝑥 + 𝑎1 𝑥 𝑦 𝑛−1 𝑥 + ⋯ + 𝑎𝑛−1 𝑥 𝑦 ′ 𝑥 + 𝑎𝑛 𝑥 𝑦 𝑥 =𝑟 𝑥 (1) Where 𝑦 is the dependent variable and 𝑥 is the independent variable and 𝑎0 𝑥 ≠ 0. If 𝑟 𝑥 = 0, then it is called a homogeneous equation, otherwise it is called non-homogeneous equation. 6 For example, a second order homogeneous equation is of the form 𝑎0 𝑥 𝑦 ′′ + 𝑎1 𝑥 𝑦 ′ + 𝑎2 𝑥 𝑦 = 0, 𝑎0 𝑥 ≠ 0 (2) and a non-homogeneous second order equation is of the form 𝑎0 𝑥 𝑦 ′′ + 𝑎1 𝑥 𝑦 ′ + 𝑎2 𝑥 𝑦 = 𝑟(𝑥), 𝑎0 𝑥 ≠ 0 (3) If 𝑎𝑖 𝑥 , 𝑖 = 0,1,2 are constants then the equations are linear second order constant coefficient equations. 7 Differential Operator D Sometimes, it is convenient to write the given linear differential equation in a simple form using the differential operator 𝐷 = 𝑑/𝑑𝑥. We write 𝑑𝑓 𝐷𝑓 𝑥 = 𝐷𝑓 = 𝐷𝑓 𝑥 = = 𝑓 ′. 3 𝑑𝑥 𝑑 We have, for example 𝐷 𝑥𝑛 = 𝑥 𝑛 = 𝑛𝑥 𝑛−1 , 𝑛 constant; 𝑑𝑥 𝐷 𝑐𝑜𝑠𝑥 = −𝑠𝑖𝑛𝑥, etc. Also, 𝐷3 𝑓 = 𝐷 𝐷2 𝑓 = 𝐷 𝑓 ′′ = 𝑓 ′′′ , … , 𝐷𝑘 𝑓 = 𝑓 𝑘. We define 𝐷0 ≡ 1, so that if 1 is the operator defined by 1 𝑓 = 𝑓, we have 𝐷0 𝑓 = 1 𝑓 = 𝑓. 8 We now define the operator L by 𝑑𝑛 𝑑𝑛−1 𝑑 𝐿 = 𝑎0 𝑥 𝑛 + 𝑎1` 𝑥 𝑛−1 + ⋯ + 𝑎𝑛−1 𝑥 + 𝑎𝑛 (𝑥) 𝑑𝑥 𝑑𝑥 𝑑𝑥 = 𝑎0 𝑥 𝐷𝑛 + 𝑎1 𝑥 𝐷𝑛−1 + ⋯ + 𝑎𝑛−1 𝑥 𝐷 + 𝑎𝑛 𝑥 = 𝑃 𝐷 (4) which is a polynomial in D, so that 𝑑𝑛 𝑦 𝑑𝑛−1 𝑦 𝑑𝑦 𝐿𝑦 = 𝑎0 𝑥 𝑛 + 𝑎1` 𝑥 𝑛−1 + ⋯ + 𝑎𝑛−1 𝑥 + 𝑎𝑛 𝑥 𝑦 𝑑𝑥 𝑑𝑥 𝑑𝑥 = 𝑎0 𝑥 𝐷𝑛 𝑦 + 𝑎1 𝑥 𝐷𝑛−1 𝑦 + ⋯ + 𝑎𝑛−1 𝑥 𝐷𝑦 + 𝑎𝑛 𝑥 𝑦 = [𝑎0 𝑥 𝐷𝑛 + 𝑎1 𝑥 𝐷𝑛−1 + ⋯ + 𝑎𝑛−1 𝑥 𝐷 + 𝑎𝑛 𝑥 ]𝑦 = 𝑃(𝐷)𝑦 (5)9 For example, the differential equation 𝑑2 𝑦 𝑑𝑦 2 +5 + 6𝑦 = 0 𝑑𝑥 𝑑𝑥 can be written as 𝐿𝑦 = 𝐷2 + 5𝐷 + 6 𝑦 = 0 (6) where the operator L is given by 𝐿 = 𝑃 𝐷 = 𝐷2 + 5𝐷 + 6. 10 Solution of Higher Order Homogeneous Linear Equations with Constant Coefficients Consider the nth order homogeneous linear equation with constant coefficients 𝑎0 𝑦 (𝑛) + 𝑎1 𝑦 (𝑛−1) + ⋯ + 𝑎𝑛−1 𝑦 ′ + 𝑎𝑛 𝑦 = 0 (7) We attempt to find a solution of the form 𝑦 = 𝑒 𝑚𝑥. Substituting 𝑦 = 𝑒 𝑚𝑥 , 𝑦 (𝑘) = 𝑚𝑘 𝑒 𝑚𝑥 , 𝑘 = 1,2, … , 𝑛 in Eq. (7) and cancelling 𝑒 𝑚𝑥 , we obtain the characteristic equation as 𝑎0 𝑚𝑛 + 𝑎1 𝑚𝑛−1 + ⋯ + 𝑎𝑛−1 𝑚 + 𝑎𝑛 = 0 (8) 11 The degree of this algebraic equation is same as the order of the differential equation. This equation has n roots may be real and distinct, all or some of the roots may be equal, all or some of the roots may be complex. Real and distinct roots Let the polynomial (8) have all real and distinct roots 𝑚1 , 𝑚2 , … , 𝑚𝑛. Then the general solution is given by 𝑦 𝑥 = 𝑐1 𝑒 𝑚1 𝑥 + 𝑐2 𝑒 𝑚2 𝑥 + ⋯ + 𝑐𝑛 𝑒 𝑚𝑛 𝑥 (9) 12 Example 1. Find the general solution of the differential equation 𝑦 ′′′ − 2𝑦 ′′ − 5𝑦 ′ + 6𝑦 = 0. Solution. Substituting 𝑦 = 𝑒 𝑚𝑥 , we obtain the characteristic equation as 𝑚3 − 2𝑚2 − 5𝑚 + 6 = 0. The roots of this equation are 𝑚 = 1, −2,3. since the roots are real and distinct, the general solution of the equation is given by 𝑦 𝑥 = 𝐴𝑒 𝑥 + 𝐵𝑒 −2𝑥 + 𝐶𝑒 3𝑥. 13 Real multiple roots The characteristic equation (8) may have some multiple roots. Let r be the multiplicity of the root 𝑚1. That is root m = 𝑚1 is repeated r times. Let the remaining n-r roots be real and distinct. Substituting m = 𝑚1 we obtain 𝑦1 𝑥 = 𝑒 𝑚1𝑥 as one of the solutions. The remaining r-1 linearly independent solutions corresponding to the multiple root m = 𝑚1 are given by 𝑥𝑦1 , 𝑥 2 𝑦1 , … , 𝑥 𝑟−1 𝑦1. That is, the linearly independent solutions in this case are 𝑒 𝑚1 𝑥 , 𝑥𝑒 𝑚1 𝑥 , 𝑥 2 𝑒 𝑚1𝑥 , … , 𝑥 𝑟−1 𝑒 𝑚1 𝑥 (10) 14 Example 2. Solve the differential equation 𝑦 ′′′ − 3𝑦 ′ − 2𝑦 = 0. Solution. Substituting 𝑦 = 𝑒 𝑚𝑥 , we obtain the characteristic equation as 𝑚3 − 3𝑚 − 2 = 0, or 𝑚 + 1 𝑚2 − 𝑚 − 2 = 0 2 or 𝑚+1 𝑚 − 2 = 0, or 𝑚 = −1, −1, 2. Corresponding to the double root 𝑚 = −1 , the linearly independent solutions 𝑒 −𝑥 and 𝑥𝑒 −𝑥. Hence, the general solution is 𝑦 𝑥 = 𝐴𝑒 2𝑥 + 𝐵𝑥 + 𝐶 𝑒 −𝑥. 15 Complex roots The complex roots occur in conjugate pairs and are of the form 𝑝 ± 𝑖𝑞. Then, the solution of the equation can be written as 𝑝+𝑖𝑞 𝑥 𝑝−𝑖𝑞 𝑥 𝑦 𝑥 = 𝐴𝑒 + 𝐵𝑒 = 𝐴𝑒 𝑝𝑥 𝑒 𝑖𝑞𝑥 + 𝐵𝑒 𝑝𝑥 𝑒 −𝑖𝑞𝑥 = (𝐴𝑒 𝑖𝑞𝑥 + 𝐵𝑒 −𝑖𝑞𝑥 )𝑒 𝑝𝑥 = [𝐴 cos 𝑞𝑥 + 𝑖𝑠𝑖𝑛 𝑞𝑥 + 𝐵 cos 𝑞𝑥 − 𝑖𝑠𝑖𝑛 𝑞𝑥 ]𝑒 𝑝𝑥 Simplifying, we obtain 𝑦 𝑥 = 𝑒 𝑝𝑥 [𝑐1 cos 𝑞𝑥 + 𝑐2 𝑠𝑖𝑛 𝑞𝑥] where 𝑐1 = 𝐴 + 𝐵 and 𝑐2 = 𝑖 𝐴 − 𝐵. 16 Example 3. Find the solution of the initial value problem 𝑦 ′′ + 4𝑦 ′ + 13𝑦 = 0, 𝑦 0 = 0, 𝑦 ′ 0 = 1. Solution. The characteristic equation is given by 𝑚2 + 4𝑚 + 13 = 0, −4 ± 16 − 52 or 𝑚 = = −2 ± 3𝑖 = 𝑝 + 𝑖𝑞. 2 The general solution is given by 𝑦 𝑥 = 𝐴 cos 𝑞𝑥 + 𝐵𝑠𝑖𝑛 𝑞𝑥 𝑒 𝑝𝑥 = 𝑒 −2𝑥 [𝐴 cos 3𝑥 + 𝐵𝑠𝑖𝑛 3𝑥] 17 Substituting in the initial conditions, we obtain 𝑦 0 = 0 = 𝐴, 𝑦 ′ 𝑥 = 𝐵𝑒 −2𝑥 (3 cos 2𝑥 − 2 sin 3𝑥), 𝑦 ′ 0 = 1 = 3𝐵, or 𝐵 = 1/3. The solution of the initial value problem is 𝑦 𝑥 = (𝑒 −2𝑥 sin 3𝑥)/3. 18 Multiple complex roots This case is a combination of the two earlier cases of real multiple roots and simple complex roots. Now, if 𝑝 + 𝑖𝑞 is a multiple root of order 𝑚, then 𝑝 − 𝑖𝑞 is also a multiple root of order 𝑚. For example, if 𝑝1 + 𝑖𝑞1 is a double root, then 𝑝1 − 𝑖𝑞1 is also a double root. The corresponding linearly independent solutions are 𝑒 𝑝1𝑥 cos 𝑞1 𝑥 , 𝑒 𝑝1 𝑥 sin 𝑞1 𝑥 , 𝑥𝑒 𝑝1𝑥 cos 𝑞1 𝑥 , 𝑥𝑒 𝑝1𝑥 sin 𝑞1 𝑥. 19 Example 4. Solve the differential equation 𝑦 𝑖𝑣 + 32𝑦 ′′ + 256𝑦 = 0. Solution. The characteristic equation is 𝑚4 + 32𝑚2 + 256 = 0, or (𝑚2 + 16)2 = 0. The roots of this equation are the double roots 𝑚 = ±4𝑖. Therefore, the general solution is 𝑦 𝑥 = 𝐴𝑥 + 𝐵 cos 4𝑥 + 𝐶𝑥 + 𝐷 sin 4𝑥. Note: The solution of homogeneous part is called as Complementary Function (C. F.) of the differential equation. 20 21