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Ordinary Differential Equations

# Ordinary Differential Equations - 25 Linear Differential...

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25 Linear Differential Equations Ordinary Differential Equations may be divided into two large classes: the so- called linear equations and the nonlinear equations. Linear equations are much simpler in many respects, because various properties of their solutions can be characterised in a general way, and standard methods are available for solving many of these equations. We have previously seen linear first order ODES, which can be written as ) x ( Q y ) x ( P dx dy A second order linear ODE is an ODE that can be written in the form ) x ( R y ) x ( Q dx dy ) x ( P dx y d 2 2 The characteristic feature of this equation is that it is linear in the unknown function y and its derivatives, whereas P, Q and R can be any given functions of x. P and Q are called the coefficients of the equation. Any second order ODE that cannot be written in this form is said to be nonlinear . 26 Homogeneous and Non-Homogeneous ODES If the function R(x) 0, then the ODE becomes 0 y ) x ( Q dx dy ) x ( P dx y d 2 2 in which case the equation is said to be homogeneous . If this is not the case, then the equation is said to be non-homogeneous or inhomogeneous . Examples 0 y 6 dx dy x 2 dx y d x 1 2 2 2 is a second order homogeneous linear ODE x sin e y 4 y x is a second order non-homogeneous linear ODE 0 y y y is a second order nonlinear ODE 1 dx dy dx y d 2 2 2 is a second order nonlinear ODE

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27 Homogeneous ODES with Constant Coefficients We are going to concentrate on second order equations here, but these can be generalised to ODES of higher orders. A second order homogeneous linear ODE with constant coefficients has the form 0 by dx dy a dx y d 2 2 To solve this recall that the solution to a first order homogeneous linear ODE with constant coefficient, i.e. 0 ky dx dy is an exponential function, kx Ce y We can use the same form to find solutions to our second order problem by trying the solution x e y , giving x e y and x 2 e y . 28 Substituting this we have 0 e b a x 2 . So this is a solution given that the characteristic equation (or auxillary equation) 0 b a 2 has solutions. The solutions are given by 2 b 4 a a 2 1 2 b 4 a a 2 2 These solutions for are of one of three forms: 1. two distinct real roots 2. two complex conjugate roots 3. a real double root
29 General Solution Consider x 1 1 e y and x 2 2 e y . If y 1 (x) and y 2 (x) are linearly independent on some interval I then a general solution to our second order homogeneous linear ODE is x 2 x 1 2 1 e c e c y where c 1 and c 2 are arbitrary constants. (Conversely, If y 1 and y 2 are not linearly independent, then y is not a general solution to the ODE). In the cases of two distinct real roots or two complex conjugate roots of the characteristic equation, y 1 and y 2 are linearly independent and y is a general solution to the ODE. In the case of double roots, we need to obtain a basis given that the single value of gives us one y 1 that satisfies the ODE.

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