SOLH_Const_Coef

# SOLH_Const_Coef - 16 Second-Order Homogeneous Linear...

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
Deriving the Basic Approach 339

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern