Ordinary Differential Equations (ODEs)

Transcription

1 Chapter 13 Ordinary Differential Equations (ODEs) We briefly review how to solve some of the most standard ODEs First Order Equations Separable Equations A first-order ordinary differential equation is separable if it can be written in the form = f (x) g (y) Where y = y (x) is the function we are trying to find. Theory To solve a separable equation, we move all the expressions involving y on one side, and those involving x on the other side. We get = f (x) g (y) Integrating both sides gives g (y) = f (x) How we proceed depends on f and g. Example 159 The simplest separable equation is the growth-decay equation = λy 161

2 162 CHAPTER 13. ORDINARY DIFFERENTIAL EQUATIONS (ODES) Proceeding as explained above, we get y = λ Integrating both sides gives that is y = λ ln y = λx + C Where C is a constant. Since y represents a certain quantity, we can assume y > 0. So, solving for y gives: y = e λx+c = e C e λx = Ke λx Where K = e C is a constant. When λ > 0, this represents exponential growth. When λ < 0, this represents exponential decay. The constant K can be found given initial conditions. Problems In the problems below, y is a function of x. 1. Solve y = 3y 2. Solve y = 20y knowing that y (0) = Linear Equations Theory We review how to solve first-order linear ODEs by using the integrating factor technique. Definition 160 A first-order linear ODE is of the form where y = y (x) is the function we are solving for. y + p (x) y = q (x) (13.1) We outline the technique used in the general case, then illustrate it with some examples. Algorithm 161 To solve equation 13.1, follow the steps below.

3 13.1. FIRST ORDER EQUATIONS 163 step 1 Multiply each side of the equation by g (x) = e p(x). This is called the integrating factor. The reason for this will be obvious in the next step. Using the chain rule, we see that g (x) = d p (x) e p(x) Using the fundamental theorem of Calculus, we see that Thus, we get The left side of this equation is Thus the equation becomes g (x) = p (x) e p(x) = p (x) g (x) y g (x) + p (x) g (x) y = q (x) g (x) d (yg (x)) d (yg (x)) = q (x) g (x) step 2 We integrate both sides with respect to x to get d (yg (x)) = q (x) g (x) or yg (x) = q (x) g (x) + C Solving for y gives y = 1 ( x ) q (x) g (x) + C g (x) a Remark 162 Of course, we have the solution symbolically. To have an exact expression implies we can find the integrals involved. We now show how this works with an example. Example 163 Solve y + y = x First, we note that this is indeed a first-order linear ODE with p (x) = 1 and q (x) = x. The integrating factor is e = e x. Thus, we multiply our equation by e x to get y e x + ye x = xe x that is (ye x ) = xe x

4 164 CHAPTER 13. ORDINARY DIFFERENTIAL EQUATIONS (ODES) Next, we integrate each side to get ye x = xe x = e x (x 1) + C Solving for y gives y = x 1 + Ce x Example 164 Solve y + 3y = e x First, we note that this is indeed a first-order linear ODE with p (x) = 3 and q (x) = e x. The integrating factor is e 3 = e 3x. Thus, we multiply our equation by e 3x to get y e 3x + 3ye 3x = e 4x that is ( ye 3x ) = e 4x Next, we integrate with respect to x to get ye 3x = e 4x = 1 4 e4x + C Thus y = 1 4 ex + Ce 3x Problems 1. Solve y + 2y = e x, (Exact solution is: { Ce 2x + 1 } e x ) 2. Solve y = 3y, (Exact solution is: { Ce 3x} ) 3. Solve y + 2y = sin x, (Exact solution is: { 2 5 sin x 1 } 5 cos x + C 7e 2x ) 13.2 Second Order Equations Linear, Constant-Coeffi cient Homogeneous Equations A linear, second-order constant-coeffi cient homogeneous equation is of the form ay + by + cy = 0

5 13.2. SECOND ORDER EQUATIONS 165 A general solution of such an equation is a linear combination of two linearly independent solutions. In other words, if y 1 (x) and y 2 (x) are two linearly independent solutions, then y (x) = Ay 1 (x) + By 2 (x) where A and B are constant, determined from additional conditions such as initial conditions. To find linearly independent solutions, we solve the quadratic equation ar 2 + br + c = 0 Let r 1 and r 2 be the solutions of this quadratic equation. There are three possibilities: 1. r 1 and r 2 are distinct real solutions. The two linearly independent solutions are e r1x and e r2x. A general solution of our linear, second-order constant-coeffi cient homogeneous equation will be y (x) = Ae r1x + Be r2x 2. r 1 and r 2 are real and equal, that is r 1 = r 2 = r. Then, two linearly independent solutions are e rx and xe rx. A general solution of our linear, second-order constant-coeffi cient homogeneous equation will be y (x) = Ae rx + Bxe rx 3. r 1 and r 2 are complex conjugates, that is r 1 = α+iβ and r 2 = α iβ for some real numbers α and β. Then, two linearly independent solutions are e αx sin βx and e αx cos βx. A general solution of our linear, second-order constant-coeffi cient homogeneous equation will be y (x) = (A sin βx + B cos βx) e αx Remark 165 The formulas in the third case are derived using Euler s formula e ix = cos x + i sin x Example 166 Let a be a constant. Find the general solution of y + a 2 y = 0. First, we find linearly independent solutions by solving r 2 +a 2 = 0. The solutions are r = ±ia. From the third case above, we see that the general solution is y (x) = A cos ax + B sin ax Example 167 Let a be a constant. Find the general solution of y a 2 y = 0. First, we find linearly independent solutions by solving r 2 a 2 = 0. The solutions are r = ±a. From the first case above, we see that the general solution is y (x) = Ae ax + Be ax

6 166 CHAPTER 13. ORDINARY DIFFERENTIAL EQUATIONS (ODES) Remembering that We can rewrite the solution as cosh x = ex + e x 2 sinh x = ex e x y (x) = A 1 cosh ax + B 1 sinh ax Linear, Constant-Coeffi cient Nonhomogeneous Equations Proposition 168 A general solution of the nonhomogeneous ODE is of the form 2 y + ay + by = r (x) (13.2) y (x) = y h (x) + y p (x) where y h (x) is a solution of the corresponding homogeneous problem (when r (x) = 0) as described in the previous section and y p (x) is any solution of equation The table below shows how to select y p (x) depending on the form of r (x). Term in r (x) Choice for y p (x) ke γx Ce γx kx n for n = 0, 1, 2,... K n x n + K n 1 x n K 1 x + K 0 k cos ωx K cos ωx + M sin ωx k sin ωx K cos ωx + M sin ωx ke ax cos ωx e ax (K cos ωx + M sin ωx) ke ax sin ωx e ax (K cos ωx + M sin ωx) Remark 169 If r (x) is a combination of functions shown in the table above, follow the following rules: 1. Modification Rule: If a term in the choice for y p (x) happens to solve the corresponding homogeneous ODE, multiply it by x (or x 2 if the solution corresponds to a double root). 2. Sum Rule: If r (x) is a sum of function in the first column of the table above, choose for y p (x) the sum of the functions in the corresponding function of the second column. Example 170 Solve y + y = 0.001x 2 y (0) = 0 y (0) = 1.5

7 13.3. PROBLEMS 167 Therefore, the answer is y = cos x sin x x y x Problems Graph of the solution and r (x) (in blue) Find two independent solutions and the general solution of the equations below. 1. y + y + y = 0 2. y 6y = 0 3. y + 6y = 0