What is the solution of given initial value problem? =y+ sin t dt2 d'y dx + cos t dt %3D dt2 (0) = 1, z'(0) = 0, y(0) = -1, y'(0) = -1 %3D %3D %3D O a. z(t) = cos t, y(t) = Cos t - sin t O b. z(t) = cos t, y(t) = - cost +sint %3D Oc. (t) = sin t, y(t) cos t + sin t %3D %3D O d. z(t) = cos = cos t, y(t) = - cost- sin t | %3D O e. r(t) = sin t, y(t) = -cost – sin t %3D

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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What is the solution of given initial value problem?
=y + sin t
dt2
d²y
dx
+ cos t
dt2
dt
I(0) = 1, r'(0) = 0, y(0) = -1, y'(0) = -1
O a. r(t)
= cos t, y(t)
cos t - sint
O b. a(t) = cos t, y(t)
cos t + sint
%3D
O c. r(t) = sin t, y(t)
cost + sint
%3D
O d. z(t) = cos t, y(t) =
Cos t – sint
%3D
O e. r(t) = sint, y(t)
cos t
sin t
Transcribed Image Text:What is the solution of given initial value problem? =y + sin t dt2 d²y dx + cos t dt2 dt I(0) = 1, r'(0) = 0, y(0) = -1, y'(0) = -1 O a. r(t) = cos t, y(t) cos t - sint O b. a(t) = cos t, y(t) cos t + sint %3D O c. r(t) = sin t, y(t) cost + sint %3D O d. z(t) = cos t, y(t) = Cos t – sint %3D O e. r(t) = sint, y(t) cos t sin t
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