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Consider the DE which is linear with constant coefficients. d²y dr² First we will work on solving the corresponding homogeneous equation. The auxiliary equation (using m as your variable) is dy - 16- +64y=2 da Because this is a repeated root, we don't have much choice but to use the exponential function corresponding to this root: Y2 = ue&r Then (using the prime notation for the derivatives) 3/₂ = 3/2 = So, plugging y2 into the left side of the differential equation, and reducing, we get Therefore y₂ = the general solution. y2-16y2 +64y2 = = 0 which has root So now our equation is esu" = x. To solve for u we need only integrate ae-8 twice, using a as our first constant of integration and b as the second we get U = to do reduction of order. We knew from the beginning that es was a solution. We have worked out is that æe is another solution to the homogeneous equation, which is generally the case when we have multiple roots. Then 218 is the particular solution to the nonhomogeneous equation, and the general solution we derived is pieced together using superposition.

Consider the DE which is linear with constant coefficients. d²y dr² First we will work on solving the corresponding homogeneous equation. The auxiliary equation (using m as your variable) is dy - 16- +64y=2 da Because this is a repeated root, we don't have much choice but to use the exponential function corresponding to this root: Y2 = ue&r Then (using the prime notation for the derivatives) 3/₂ = 3/2 = So, plugging y2 into the left side of the differential equation, and reducing, we get Therefore y₂ = the general solution. y2-16y2 +64y2 = = 0 which has root So now our equation is esu" = x. To solve for u we need only integrate ae-8 twice, using a as our first constant of integration and b as the second we get U = to do reduction of order. We knew from the beginning that es was a solution. We have worked out is that æe is another solution to the homogeneous equation, which is generally the case when we have multiple roots. Then 218 is the particular solution to the nonhomogeneous equation, and the general solution we derived is pieced together using superposition.

Algebra: Structure And Method, Book 1
Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter4: Polynomials
Section4.8: Rate-time-distance Problems
Problem 1OE
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Consider the DE which is linear with constant coefficients. d²y dr² First we will work on solving the corresponding homogeneous equation. The auxiliary equation (using m as your variable) is dy - 16- +64y=2 da Because this is a repeated root, we don't have much choice but to use the exponential function corresponding to this root: Y2 = ue&r Then (using the prime notation for the derivatives) 3/₂ = 3/2 = So, plugging y2 into the left side of the differential equation, and reducing, we get Therefore y₂ = the general solution. y2-16y2 +64y2 = = 0 which has root So now our equation is esu" = x. To solve for u we need only integrate ae-8 twice, using a as our first constant of integration and b as the second we get U = to do reduction of order. We knew from the beginning that es was a solution. We have worked out is that æe is another solution to the homogeneous equation, which is generally the case when we have multiple roots. Then 218 is the particular solution to the nonhomogeneous equation, and the general solution we derived is pieced together using superposition.
Transcribed Image Text:Consider the DE which is linear with constant coefficients. d²y dr² First we will work on solving the corresponding homogeneous equation. The auxiliary equation (using m as your variable) is dy - 16- +64y=2 da Because this is a repeated root, we don't have much choice but to use the exponential function corresponding to this root: Y2 = ue&r Then (using the prime notation for the derivatives) 3/₂ = 3/2 = So, plugging y2 into the left side of the differential equation, and reducing, we get Therefore y₂ = the general solution. y2-16y2 +64y2 = = 0 which has root So now our equation is esu" = x. To solve for u we need only integrate ae-8 twice, using a as our first constant of integration and b as the second we get U = to do reduction of order. We knew from the beginning that es was a solution. We have worked out is that æe is another solution to the homogeneous equation, which is generally the case when we have multiple roots. Then 218 is the particular solution to the nonhomogeneous equation, and the general solution we derived is pieced together using superposition.
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