Consider the DE d²y dy - 18- +81y = x dx² dx which is linear with constant coefficients. First we will work on solving the corresponding homogeneous equation. The auxiliary equation (using m as your variable) is = 0 which has root Because this is a repeated root, we don't have much choice but to use the exponential function corresponding to this root: Y2 = to do reduction of order. Then (using the prime notation for the derivatives) = 3½ = So, plugging y2 into the left side of the differential equation, and reducing, we get y — 18, + 81y2 = So now our equation is eu" = x. To solve for u we need only integrate ace first constant of integration and b as the second we get 9x twice, using a as our u = Therefore y2 = , the general solution. We knew from the beginning that ex was a solution. We have worked out is that re⁹ is another solution to the homogeneous equation, which is generally the case when we have multiple roots. Then 2+9x is the particular solution to the nonhomogeneous equation, and the general solution we derived is pieced together using superposition. 729
Consider the DE d²y dy - 18- +81y = x dx² dx which is linear with constant coefficients. First we will work on solving the corresponding homogeneous equation. The auxiliary equation (using m as your variable) is = 0 which has root Because this is a repeated root, we don't have much choice but to use the exponential function corresponding to this root: Y2 = to do reduction of order. Then (using the prime notation for the derivatives) = 3½ = So, plugging y2 into the left side of the differential equation, and reducing, we get y — 18, + 81y2 = So now our equation is eu" = x. To solve for u we need only integrate ace first constant of integration and b as the second we get 9x twice, using a as our u = Therefore y2 = , the general solution. We knew from the beginning that ex was a solution. We have worked out is that re⁹ is another solution to the homogeneous equation, which is generally the case when we have multiple roots. Then 2+9x is the particular solution to the nonhomogeneous equation, and the general solution we derived is pieced together using superposition. 729
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![Consider the DE
d²y
dy
- 18-
+81y = x
dx²
dx
which is linear with constant coefficients.
First we will work on solving the corresponding homogeneous equation. The auxiliary equation (using m
as your variable) is
= 0 which has root
Because this is a repeated root, we don't have much choice but to use the exponential function
corresponding to this root:
Y2 =
to do reduction of order.
Then (using the prime notation for the derivatives)
=
3½
=
So, plugging y2 into the left side of the differential equation, and reducing, we get
y — 18, + 81y2 =
So now our equation is eu" = x. To solve for u we need only integrate ace
first constant of integration and b as the second we get
9x
twice, using a as our
u =
Therefore y2 =
, the general solution.
We knew from the beginning that ex was a solution. We have worked out is that re⁹ is another
solution to the homogeneous equation, which is generally the case when we have multiple roots. Then
2+9x is the particular solution to the nonhomogeneous equation, and the general solution we derived is
pieced together using superposition.
729](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F85836af1-4c0d-435d-82f2-1b92079701bf%2Fc42f0f6f-79b9-4201-b974-a5db5984e15f%2Fgki1r2b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Consider the DE
d²y
dy
- 18-
+81y = x
dx²
dx
which is linear with constant coefficients.
First we will work on solving the corresponding homogeneous equation. The auxiliary equation (using m
as your variable) is
= 0 which has root
Because this is a repeated root, we don't have much choice but to use the exponential function
corresponding to this root:
Y2 =
to do reduction of order.
Then (using the prime notation for the derivatives)
=
3½
=
So, plugging y2 into the left side of the differential equation, and reducing, we get
y — 18, + 81y2 =
So now our equation is eu" = x. To solve for u we need only integrate ace
first constant of integration and b as the second we get
9x
twice, using a as our
u =
Therefore y2 =
, the general solution.
We knew from the beginning that ex was a solution. We have worked out is that re⁹ is another
solution to the homogeneous equation, which is generally the case when we have multiple roots. Then
2+9x is the particular solution to the nonhomogeneous equation, and the general solution we derived is
pieced together using superposition.
729
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