12y Consider the differential equation xy" – y' = (1) Making the changes of variable x=e', y = u(t)e 2" in the differential equation (1), we obtain the differential equation, u depending on t, given by u"-6u'-4u = 0 The solution of the differential equation (1), depending on the parameter t, with constant c1 and C2, is given by: a) { z(1) y(t) et e* (cı cos(tv/13) + c2 sen(t/13)) S ¤(t) et b) | y(t) et (cietvi3 + cze-tV13 S 2(t) = et c) y(t) e3t -t/13 + c2e Į æ(t) et d) l y(t) e' (c1 cos(tv/13) + C2 sen(tv/13))
12y Consider the differential equation xy" – y' = (1) Making the changes of variable x=e', y = u(t)e 2" in the differential equation (1), we obtain the differential equation, u depending on t, given by u"-6u'-4u = 0 The solution of the differential equation (1), depending on the parameter t, with constant c1 and C2, is given by: a) { z(1) y(t) et e* (cı cos(tv/13) + c2 sen(t/13)) S ¤(t) et b) | y(t) et (cietvi3 + cze-tV13 S 2(t) = et c) y(t) e3t -t/13 + c2e Į æ(t) et d) l y(t) e' (c1 cos(tv/13) + C2 sen(tv/13))
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 by hand)

Transcribed Image Text:12y
Consider the differential equation xy" – y' =
(1)
Making the changes of variable x=e', y = u(t)e-2t in the differential equation (1), we obtain the
%3D
differential equation, u depending on t, given by u"-6u'-4u = 0
The solution of the differential equation (1), depending on the parameter t, with constant ci and
C2, is given by:
a)
Į ¤(t)
et
y(t)
ešt (cı cos(t/13) + c2
sen(tv/13))
S #(t)
b)
et
| y(t)
e (cretVB
+ Cze¯tV13
S¤(t)
et
c)
y(t)
3t
/13
+ C2e¬tV13
e°
Į ¤(t)
et
d)
y(t)
e' (c1 cos(tv/13) + C2 sen(tv/13))
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