h is linear with constant coefficients. d²y dx² = we will work on solving the corresponding homogeneous equation. The characteristic equation g m as your variable) is = 0 which has root = dy 10 dx ause this is a repeated root, we don't have much choice but to use the exponential function esponding to this root: to do reduction of order. = ue5ª (using the prime notation for the derivatives) +25y = x efore y₂ = plugging y2 into the left side of the differential equation, and reducing, we get y2 - 10y2 + 25y2 = -5x ow our equation is e5xu" = x. To solve for u we need only integrate xe irst constant of integration and b as the second we get " the general solution. twice, using a as
h is linear with constant coefficients. d²y dx² = we will work on solving the corresponding homogeneous equation. The characteristic equation g m as your variable) is = 0 which has root = dy 10 dx ause this is a repeated root, we don't have much choice but to use the exponential function esponding to this root: to do reduction of order. = ue5ª (using the prime notation for the derivatives) +25y = x efore y₂ = plugging y2 into the left side of the differential equation, and reducing, we get y2 - 10y2 + 25y2 = -5x ow our equation is e5xu" = x. To solve for u we need only integrate xe irst constant of integration and b as the second we get " the general solution. twice, using a as
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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please solve it on paper
Transcribed Image Text:Consider the DE
which is linear with constant coefficients.
Y2
=
d²y
dx²
=
First we will work on solving the corresponding homogeneous equation. The characteristic 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:
to do reduction of order.
ue 5x
Y2
Then (using the prime notation for the derivatives)
Y/2
dy
dx
U =
- 10-
Therefore y2 =
+ 25y
=
= x
So, plugging y2 into the left side of the differential equation, and reducing, we get
y2 - 10y2 + 25y2
-5x
So now our equation is e5u" = x. To solve for u we need only integrate xe
our first constant of integration and b as the second we get
the general solution.
twice, using a as
We knew from the beginning that e5 was a solution. We have worked out is that xe5* is another
solution to the homogeneous equation, which is generally the case when we have multiple roots.
Then is the particular solution to the nonhomogeneous equation, and the general solution we
derived is pieced together using superposition.
2+5x
125
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