MATH 308 Differential Equations

Transcription

1 MATH 308 Differential Equations Summer, 2014, SET 6 JoungDong Kim Set 6: Section 3.3, 3.4, 3.5, 3.6 Section 3.3 Complex Roots of the Characteristic Equation Recall that a second order ODE with constant coefficients; ay +by +cy = 0, (1) its characteristic equation is ar 2 +br +c = 0. (2) If the characteristic equation has two different roots r 1 and r 2, then the general solution is y = c 1 e r1t +c 2 e r2t. But in general, the equation (2) may have complex roots or repeated root. 1

2 is Characteristic equation with complex roots. If the roots of equation (2) are complex numbers, they must be Conjugate complex numbers, that Therefore, two solutions of equation (1) are r 1 = λ+iµ, r 2 = λ iµ. y 1 (t) = e (λ+iµ)t = e λt (cos(µt)+isin(µt)), y 2 (t) = e (λ iµ)t = e λt (cos(µt) isin(µt)). Unfortunately, the solutions y 1 and y 2 are complex-vlaued functions, whereas in general we would prefer to have real-valued solutions. Find a fundamental set of real-valued solutions. and y 1 +y 2 = 2e λt cosµt, y 1 y 2 = 2ie λt sinµt. Hence, we have obtained a pair of real-valued solutions u(t) = e λt cos(µt), v(t) = e λt sin(µt). These two real valued functions also solve the original equation (1). So the general solution of equation (1) can be written as, y = c 1 e λt cos(µt)+c 2 e λt sin(µt). Ex1) Find the general solution of the following equation, y +2y +2y = 0. 2

3 Ex2) Solve the initial value problem, 16y 8y +145y = 0, y(0) = 2, y (0) = 1. 3

4 Section 3.4 Repeated Roots; Reduction of Order In this section, we study the case of repeated roots of characteristic equation. Ex3) Solve the initial value problem, y +4y +4y = 0, y(0) = 1, y (0) = 0. 4

5 In general, if the characteristic equation has two repeated roots, r 1,2 = r 0, then the general solution of the second order ODE has the following form: y = c 1 e r0t +c 2 te r0t, c 1 and c 2 can be uniquely determined if two initial conditions are given. Ex4) Find the solution of the initial value problem, y y +0.25y = 0, y(0) = 2, y (0) = 1/3. 5

6 Solve this problem with a different initial condition y (0) = 2, compare two solutions by the graphs and explain why they behave so different as t increases. 6

7 Reduction of Order Let us recall the way that we seek the other special solution y = te 2t in Ex3. We start by guessing that the other solution is in the form as y = v(t)e 2t, plug this function into the equation and derive a new equation for v(t). This is a useful technique to look for solutions for second order linear ODE if one solution is given. Ex5) Given that y 1 (t) = t 1 is a solution of 2t 2 y +3ty y = 0, t > 0. Find another solution which is not linearly dependent with y 1. 7

8 Summary The structure of the general solutions solely depends on the characteristic equation. 1. If r 1 and r 2 are different real numbers. The general solution is y = c 1 e r 1t +c 2 e r 2t. 2. If r 1 and r 2 are two conjugate complex numbers, r 1,2 = λ±iµ. The general solution is y = c 1 e λt cos(µt)+c 2 e λt sin(µt). 3. If the coefficients are not constant, we can determine a different solution if one is already given. These two solutions form a fundamental set of solutions. 8

9 Section 3.5 Nonhomogeneous Equations: Method of Undetermined Coefficients. In section 3.5 and 3.6, we will study two methods for solving nonhomogeneous equations. We begin by looking at the structure of the general solutions for nonhomogeneous equations. Consider the general form of a nonhomogeneous linear second order ODE: Its corresponding homogeneous equation is: y +p(t)y +q(t)y = g(t). (3) y +p(t)y +q(t)y = 0. (4) Theorem 0.1. If Y 1 and Y 2 are two solutions of the nonhomogeneous equation (3), then their difference Y 1 Y 2 is a solution of the corresponding homogeneous equation (4). If we assume that y 1 and y 2 are a fundamental set of solutions of equation (4), then we have Y 1 Y 2 = c 1 y 1 +c 2 y 2. Theorem 0.2 (General Solution). The general solution of the nonhomogeneous equation (3) can be written in the form: y = c 1 y 1 +c 2 y 2 +Y p, where y 1 and y 2 are a fundamental set of solutions of the homogeneous equation (4), c 1 and c 2 are arbitrary constants, and Y p is some particular solution of the nonhomogeneous equation (3). This theorem actually gives us a strategy for finding all solutions of equation (3): 1. We first solve the homogeneous equation (4), find a fundamental set of solutions y 1 and y Find one solution Y p for the nonhomogeneous equation (3). And add together the functions found in the two preceding steps. 3. For given initial conditions, we can use the initial data to determine c 1 and c 2. 9

10 Method of undetermined coefficients. When the functions p(t) and q(t) are constants, we know how to find a fundamental set of solutions from previous sections. So the key to solve equation (3) is to find a special solution Y p. The first and easiest way to find such a solution is Guessing. However, it is more efficient if we have some rules in mind. This is the motivation of the method of undetermined coefficients. Ex6) Find a particular solution of y 3y 4y = 3e 2t. 10

11 Ex7) Find a particular solution of y 3y 4y = 2sint. 11

12 Ex8) Find a particular solution of y 3y 4y = 8e t cos(2t). 12

13 Ex9) Find a particular solution of y 3y 4y = 3e 2t +2sint 8e t cos(2t). 13

14 Ex10) Find a particular solution of y 3y 4y = 2e t. 14

15 The particular solution of ay +by +cy = g(t) is in the table given. Table 1: The particular solution of ay +by +cy = g(t). g(t) Y(t) P n (t) = a 0 t n +a 1 t n 1 + +a n t s (A 0 t n +A 1 t n 1 + +A n ) P n (t)e αt { t s (A 0 t n +A 1 t n 1 + +A n )e αt sinβt P n (t)e αt t s [(A 0 t n +A 1 t n 1 + +A n )e αt cosβt cosβt +(B 0 t n +B 1 t n 1 + +B n )e αt sinβt] Note: Here s is the smallest nonnegative integer (s = 0,1,or 2) that will ensure that no term in the guess Y(t) is a solution of the corresponding homogeneous equation. 15

16 Section 3.6 Variation of Parameters. The undetermined coefficients method can handle certain types of equations, such as the right hand side has to be the combination of polynomials, trig. functions, or exponential functions. Other than those functions, we cannot directly apply this method. In this section, we will study a much powerful and efficient method; Variation of parameters. The main feature of this method is that it is very general in the sense that as long as we know the fundamental set of solutions for the corresponding homogeneous equation, we can get the solution of the homogeneous equation instantly for arbitrary given right hand side function g(t). Consider the general form of a homogeneous linear second order ODE; Its corresponding homogeneous equation is y +p(t)y +q(t)y = g(t) (5) y +p(t)y +q(t)y = 0 (6) Assumption Let us assume that we know y 1, y 2 are the fundamental set of solutions for the homogeneous equation (6). In order to find all solutions for the nonhomogeneous equation (5), we need to find a special solution Y(t). Similar as the reduced order technique, we seek the solution Y(t) in a special form: Y(t) = u 1 (t)y 1 (t)+u 2 (t)y 2 (t), (7) where u 1 and u 2 are functions of t which are unknown at the moment. We need to set up some equation(s) to solve for u 1 and u 2. 16

17 Theorem 0.3. If in the equation (5), p, q, and g are continuous on an open interval I and t 0 is some point in I. Assume that y 1, y 2 are the fundamental set of solutions for the corresponding homogeneous equation (6), then a particular solution of equation (5) is t Y(t) = y 1 (t) t 0 y 2 (s)g(s) t W(y 1,y 2 )(s) ds+y 2(t) t 0 where t 0 is any conveniently chosen point in I. The general solution is y(t) = c 1 y 1 (t)+c 2 y 2 (t)+y(t). y 1 (s)g(s) W(y 1,y 2 )(s) ds, Ex11) Find a particular solution of y +4y = 3csct. 17

18 Ex12) Find a particular solution of t 2 y t(t+2)y +(t+2)y = 2t 3, t > 0, y 1 = t, y 2 = te t. 18

19 Summary The variation of parameters method is a relatively general method for nonhomogeneous linear second order ODE. It can be applied for the equations when p(t) and q(t) are not constant. However, the key assumption is that we know a fundamental set of solutions y 1 and y 2 for the corresponding homogeneous equation. This suggests us that to solve a nonhomogeneous equation, it suffices to solve the corresponding homogeneous equation. This principle is also true for many other types of ODE/PDE. 19