As a specific example we consider the non-homogeneous problem y" + 25y = sec² (5x) (1) The general solution of the homogeneous problem (called the complementary solution, Yc = ay₁ + by2 ) is given in terms of a pair of linearly independent solutions, y₁, y2. Here a and b are arbitrary constants. Find a fundamental set for y" + 25y = 0 and enter your results as a comma separated list cos(5x), sin(5x) * BEWARE Notice that the above set does not require you to decide which function is to be called y₁ or y2 and normally the order you name them is irrelevant. But for the method of variation of parameters an order must be chosen and you need to stick to that order. In order to more easily allow WeBWork to grade your work I have selected a particular order for y₁ and y2. In order to ascertain the order you need to use please enter a choice for y₁ = -5sin(5x) and if your answer is marked as incorrect simply enter the other function from the complementary set. Once you get this box marked as correct then y2 sin(2x) With this appropriate order we are now ready to apply the method of variation of parameters. (2) For our particular problem we have W(x) U1 = [-Y2(x)f(x) W(x) U2 = Y = Yc + Yp - [ Ур Y₁(x)f(x) W(x) And combining these results we arrive at = dx = dx = = 2 S J dx dx = (3) Finally, using a and b for the arbitrary constants in yc, the general solution can then be written as
As a specific example we consider the non-homogeneous problem y" + 25y = sec² (5x) (1) The general solution of the homogeneous problem (called the complementary solution, Yc = ay₁ + by2 ) is given in terms of a pair of linearly independent solutions, y₁, y2. Here a and b are arbitrary constants. Find a fundamental set for y" + 25y = 0 and enter your results as a comma separated list cos(5x), sin(5x) * BEWARE Notice that the above set does not require you to decide which function is to be called y₁ or y2 and normally the order you name them is irrelevant. But for the method of variation of parameters an order must be chosen and you need to stick to that order. In order to more easily allow WeBWork to grade your work I have selected a particular order for y₁ and y2. In order to ascertain the order you need to use please enter a choice for y₁ = -5sin(5x) and if your answer is marked as incorrect simply enter the other function from the complementary set. Once you get this box marked as correct then y2 sin(2x) With this appropriate order we are now ready to apply the method of variation of parameters. (2) For our particular problem we have W(x) U1 = [-Y2(x)f(x) W(x) U2 = Y = Yc + Yp - [ Ур Y₁(x)f(x) W(x) And combining these results we arrive at = dx = dx = = 2 S J dx dx = (3) Finally, using a and b for the arbitrary constants in yc, the general solution can then be written as
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Transcribed Image Text:As a specific example we consider the non-homogeneous problem y" + 25y = sec² (5x)
(1) The general solution of the homogeneous problem (called the complementary solution,
=
Yc
= ay₁ + by2 ) is given in terms of a pair of linearly independent solutions, y₁, Y2. Here a and
bare arbitrary constants.
Find a fundamental set for y" + 25y = 0 and enter your results as a comma separated list
cos(5x), sin(5x)
*
BEWARE Notice that the above set does not require you to decide which function is to be called y₁
or y2 and normally the order you name them is irrelevant. But for the method of variation of
parameters an order must be chosen and you need to stick to that order. In order to more easily
allow WeBWork to grade your work I have selected a particular order for y₁ and y2. In order to
ascertain the order you need to use please enter a choice for y₁ -5sin(5x)
and if
your answer is marked as incorrect simply enter the other function from the complementary set.
Once you get this box marked as correct then y₂ sin(2x)
U 1 =
With this appropriate order we are now ready to apply the method of variation of parameters.
(2) For our particular problem we have W(x)
[-Y₂(x) f(x)
W(x)
U2
=
y = Yc + Yp
Ур
Y₁(x)f(x)
W(x)
[ 31 (2)
=
And combining these results we arrive at
=
dx
dx
=
=
=
S
J
= 2
=
dx
dx
||
=
(3) Finally, using a and b for the arbitrary constants in yc, the general solution can then be written
as
=
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