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.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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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