Homogeneous Equations with Constant Coefficients

Transcription

1 Homogeneous Equations with Constant Coefficients MATH 365 Ordinary Differential Equations J. Robert Buchanan Department of Mathematics Spring 2018

2 General Second Order ODE Second order ODEs have the form d 2 y dt 2 If the ODE is linear then ( f t, y, dy ) dt so that we may write = f ( t, y, dy dt ). = g(t) p(t) dy dt q(t)y y + p(t)y + q(t)y = g(t).

3 Another Linear Form Sometimes authors write a second order linear ODE as If P(t) 0 then we may define P(t)y + Q(t)y + R(t)y = G(t). p(t) = Q(t) P(t) q(t) = R(t) P(t) g(t) = G(t) P(t) and then the equation above is equivalent to y + p(t)y + q(t)y = g(t).

4 Another Linear Form Sometimes authors write a second order linear ODE as If P(t) 0 then we may define P(t)y + Q(t)y + R(t)y = G(t). p(t) = Q(t) P(t) q(t) = R(t) P(t) g(t) = G(t) P(t) and then the equation above is equivalent to y + p(t)y + q(t)y = g(t). Remarks: If a second order equation cannot be written in one of these two forms, it is nonlinear. We will assume p(t), q(t), and g(t) are continuous on some interval in common.

5 Initial Value Problems A second order initial value problem (IVP) consists of a second order ODE and two initial conditions. d 2 ( y dt 2 = f t, y, dy ) dt y(t 0 ) = y 0 y (t 0 ) = y 0 where y 0 and y 0 are the initial values of y(t) and y (t) respectively at t = t 0.

6 Homogeneous Equations A second order linear ODE of the form P(t)y + Q(t)y + R(t)y = G(t) or y + p(t)y + q(t)y = g(t) is called homogeneous if G(t) = 0 or g(t) = 0 for all t.

7 Homogeneous Equations A second order linear ODE of the form P(t)y + Q(t)y + R(t)y = G(t) or y + p(t)y + q(t)y = g(t) is called homogeneous if G(t) = 0 or g(t) = 0 for all t. Remarks: Otherwise the equation is said to be nonhomogeneous. We will often solve the homogeneous version of a nonhomogeneous equation before turning to the nonhomogeneous equation.

8 Constant Coefficient Equations Consider P(t)y + Q(t)y + R(t)y = 0 If P(t), Q(t), and R(t) are the constants a, b, and c respectively then the ODE can be written as ay + by + cy = 0 and is called a second order linear constant coefficient homogeneous ordinary differential equation.

9 Applications The general mathematical model of the motion of a mass connected to a Hooke s Law spring, subject to viscous damping, and an external force is m y + γ y + k y = F(t).

10 Applications The general mathematical model of the motion of a mass connected to a Hooke s Law spring, subject to viscous damping, and an external force is m y + γ y + k y = F(t). Bessel s equation must be solved in order to model the motion of a vibrating circular membrane (for example a drum). t 2 y + ty + (t 2 ν 2 )y = 0

11 Applications The general mathematical model of the motion of a mass connected to a Hooke s Law spring, subject to viscous damping, and an external force is m y + γ y + k y = F(t). Bessel s equation must be solved in order to model the motion of a vibrating circular membrane (for example a drum). t 2 y + ty + (t 2 ν 2 )y = 0 Legendre s equation and its solution are often encountered in quantum mechanics. (1 t 2 )y 2t y + α(α + 1)y = 0

12 Example (1 of 2) Consider the second order linear, constant coefficient, homogeneous ODE: y y = 0 Check by direct substitution that y(t) = c 1 e t + c 2 e t is a solution to this ODE when c 1 and c 2 are constants.

13 Example (1 of 2) Consider the second order linear, constant coefficient, homogeneous ODE: y y = 0 Check by direct substitution that y(t) = c 1 e t + c 2 e t is a solution to this ODE when c 1 and c 2 are constants. y(t) = c 1 e t + c 2 e t y (t) = c 1 e t c 2 e t y (t) = c 1 e t + c 2 e t

14 Example (1 of 2) Consider the second order linear, constant coefficient, homogeneous ODE: y y = 0 Check by direct substitution that y(t) = c 1 e t + c 2 e t is a solution to this ODE when c 1 and c 2 are constants. y(t) = c 1 e t + c 2 e t y (t) = c 1 e t c 2 e t y (t) = c 1 e t + c 2 e t and y (t) y(t) = c 1 e t + c 2 e t (c 1 e t + c 2 e t ) = 0.

15 Example (2 of 2) Consider the second order linear constant coefficient homogeneous IVP: Find the solution to this IVP. y y = 0 y(0) = 2 y (0) = 1

16 Example (2 of 2) Consider the second order linear constant coefficient homogeneous IVP: Find the solution to this IVP. y y = 0 y(0) = 2 y (0) = 1 If we assume y(t) = c 1 e t + c 2 e t then 2 = c 1 + c 2 1 = c 1 c 2 which implies c 1 = 1/2 and c 2 = 3/2.

17 Illustration y y = 0 y(0) = 2 y (0) = y(t) y (t) t

18 General Solution (1 of 2) We observed that exponential functions were the solution to a second order linear homogeneous ODE considered earlier. If we now consider the general constant coefficient equation a y + b y + c y = 0 we can hypothesize that solutions take the form y(t) = e rt where r is a constant. Question: how do we determine r?

19 General Solution (2 of 2) a y + b y + c y = 0 a(r 2 e rt ) + b(re rt ) + c e rt = 0 a r 2 + b r + c = 0 Thus r must be the solution to a quadratic equation called the characteristic equation.

20 Example Solve the following second order linear constant coefficient homogeneous IVP. y + 3y + 2y = 0 y(0) = 2 y (0) = 0

21 Solution Characteristic equation: r 2 + 3r + 2 = 0 (r + 1)(r + 2) = 0 r = 1 or r = 2 General solution: y(t) = c 1 e t + c 2 e 2t Evaluate the constants: 2 = c 1 + c 2 0 = c 1 2c 2 which implies c 1 = 4 and c 2 = 2.

22 Graph y(t) = 4e t 2e 2t y(t) y (t) t

23 Example Solve the following second order linear constant coefficient homogeneous IVP. y + y 20y = 0 y(2) = 1 y (2) = 0

24 Solution Characteristic equation: r 2 + r 20 = 0 (r + 5)(r 4) = 0 r = 5 or r = 4 General solution: y(t) = c 1 e 5t + c 2 e 4t Evaluate the constants: 1 = c 1 e 10 + c 2 e 8 0 = 5c 1 e c 2 e 8 which implies c 1 = 4 9 e10 and c 2 = 5 9 e 8.

25 Graph y(t) = 4 9 e10 5t e4t y(t) y (t) t

26 Example Solve the following second order linear constant coefficient homogeneous IVP. 3y 12y 36y = 0 y(0) = 2 y (0) = 1

27 Solution Characteristic equation: 3r 2 12r 36 = 0 3(r + 2)(r 6) = 0 r = 2 or r = 6 General solution: y(t) = c 1 e 2t + c 2 e 6t Evaluate the constants: 2 = c 1 + c 2 1 = 2c 1 + 6c 2 which implies c 1 = 11/8 and c 2 = 5/8.

28 Graph y(t) = 11 8 e 2t e6t 10 8 y(t) y (t) t

29 Homework Read Section 3.1 Exercises: 1 27 odd