Homogeneous Equations with Constant Coefficients MATH 365 Ordinary Differential Equations J. Robert Buchanan Department of Mathematics Spring 2018
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).
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).
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.
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.
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.
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.
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.
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).
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
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
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.
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
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.
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
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.
Illustration y y = 0 y(0) = 2 y (0) = 1 2.0 1.5 1.0 y(t) y (t) 0.5 0.2 0.4 0.6 0.8 1.0 t -0.5-1.0
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?
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.
Example Solve the following second order linear constant coefficient homogeneous IVP. y + 3y + 2y = 0 y(0) = 2 y (0) = 0
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.
Graph y(t) = 4e t 2e 2t 2.0 1.5 1.0 y(t) y (t) 0.5 0.5 1.0 1.5 2.0 2.5 3.0 t -0.5-1.0
Example Solve the following second order linear constant coefficient homogeneous IVP. y + y 20y = 0 y(2) = 1 y (2) = 0
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 10 + 4c 2 e 8 which implies c 1 = 4 9 e10 and c 2 = 5 9 e 8.
Graph y(t) = 4 9 e10 5t + 5 9 e4t 8 15 10 5 y(t) y (t) 1.8 2.0 2.2 2.4 t -5-10 -15
Example Solve the following second order linear constant coefficient homogeneous IVP. 3y 12y 36y = 0 y(0) = 2 y (0) = 1
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.
Graph y(t) = 11 8 e 2t + 5 8 e6t 10 8 y(t) y (t) 6 4 2 0.00 0.05 0.10 0.15 0.20 t
Homework Read Section 3.1 Exercises: 1 27 odd