Prove that the general solution of the differential equation √√1+cos(t) d0=(√1 + cos(t) — t sin(t) 0(t))dt is - -2t√√1+cos(t)+4√√1-cos(t) Se dt + C 0(t) = C 2t√√11 cos(t) 4√1 cos(t)

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Prove that the general solution of the differential equation
√1 + cos(t) d0=(√1 + cos(t) — t sin(t) 0(t))dt is
-2t√1+cos(t)+4√√1-cos(t)
So
с
'dt + C
0(t):
-
C
2t √11 cos(t) 14√1 cos(t)
Transcribed Image Text:Prove that the general solution of the differential equation √1 + cos(t) d0=(√1 + cos(t) — t sin(t) 0(t))dt is -2t√1+cos(t)+4√√1-cos(t) So с 'dt + C 0(t): - C 2t √11 cos(t) 14√1 cos(t)
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