SOLH_Const_Coef - 16 Second-Order Homogeneous Linear...

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16 Second-Order Homogeneous Linear Equations with Constant Coefficients A very important class of second-order homogeneous linear equations consists of those with constant coefficients; that is, those that can be written as ay ′′ + by + cy = 0 where a , b and c are real-valued constants (with a negationslash= 0 ). Some examples are y ′′ 5 y + 6 y = 0 y ′′ 6 y + 9 y = 0 and y ′′ 6 y + 13 y = 0 . There are two reasons these sorts of differential equations are important: First of all, they often arise in applications. Secondly, as we will see, it is relatively easy to find fundamental sets of solutions for these equations. Do note that, because the coefficients are constants, they are, trivially, continuous functions on the entire real line. Consequently, we can take the entire real line as the interval of interest, and be confident that any solutions derived will be valid on all of ( −∞ , ) . IMPORTANT: What we will derive and define here (e.g., “the characteristic equation”) is based on the assumption that the coefficients in our differential equation— the a , b and c above — are constants. Some of the results will even require that these constants be real valued. Do not, later, try to blindly apply what we develop here to differential equations in which the coefficients are not real-valued constants. 16.1 Deriving the Basic Approach Seeking Inspiration Let us look for clues on how to solve our second-order equations by first looking at solving a first-order, homogeneous linear differential equation with constant coefficients, say, 2 dy dx + 6 y = 0 . 337
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338 Second-Order Homogeneous Linear Equations with Constant Coefficients Since we are considering ‘linear’ equations, let’s solve it using the method developed for first- order linear equations: First divide through by the first coefficient, 2 , to get dy dx + 3 y = 0 . The integrating factor is then ρ = e integraltext 3 dx = e 3 x . Multiplying through and proceeding as usual with first-order linear equations: e 3 x bracketleftbigg dy dx + 3 y bracketrightbigg = e 3 x · 0 equal1⇒ e 3 x dy dx + 3 e 3 x y = 0 equal1⇒ d dx bracketleftbig e 3 x y bracketrightbig = 0 equal1⇒ e 3 x y = c equal1⇒ y = ce 3 x . So a general solution to 2 dy dx + 6 y = 0 is y = ce 3 x . Clearly, there is nothing special about the numbers used here. Replacing 2 and 6 with constants a and b in the above would just as easily have given us the fact that a general solution to a dy dx + by = 0 is y = ce rx where r = − b a . Thus we see that all solutions to first-order homogeneous linear equations with constant coeffi- cients are given by constant multiples of exponential functions. Exponential Solutions with Second-Order Equations Now consider the second-order case. For convenience, we will use y ′′ 5 y + 6 y = 0 as an example, keeping in mind that our main interest is in finding all possible solutions to an arbitrary second-order homogeneous differential equation ay ′′ + by + cy = 0 where a , b and c are constants.
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Deriving the Basic Approach 339
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