The general solution of the equation: t?y" + 2ty' = 0 is y = c1+ c2/t. You use the variation of parameters method to look for a solution of ty"+2ty' = sin(t) in the form y = u1 (t)+u2(t)/t. Then u1 and u2 should satisfy the system: (a) u1 + u2/t = 0, u – uz/t² = sin(t) (b) u + uz/t = sin(t), u – uz/t² = 0 (c) u + u2/t = 0, u2 = – sin(t) (d) u + u/t = 0, u = -t sin(t)
The general solution of the equation: t?y" + 2ty' = 0 is y = c1+ c2/t. You use the variation of parameters method to look for a solution of ty"+2ty' = sin(t) in the form y = u1 (t)+u2(t)/t. Then u1 and u2 should satisfy the system: (a) u1 + u2/t = 0, u – uz/t² = sin(t) (b) u + uz/t = sin(t), u – uz/t² = 0 (c) u + u2/t = 0, u2 = – sin(t) (d) u + u/t = 0, u = -t sin(t)
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:The general solution of the equation: t'y" + 2ty' = 0 is y = c1+c2/t. You use the variation of
parameters method to look for a solution of t'y" +2ty' = sin(t) in the form y
Then u1 and u2 should satisfy the system:
= u1 (t)+u2(t)/t.
(a) u1 + u2/t = 0, u – uz/t? = sin(t)
(b) u + u/t = sin(t), u – us/t? = 0
|
(c) u + u/t = 0, uz = – sin(t)
(d) u + u½/t = 0, u' = -t² sin(t)
b)
a)
c)
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