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If one of the fundamental functions is t²cost, the original equation must be at least of these many degrees: 6 4 3 this fundamental function is not possible. y(t)= Ctcos2t is a solution for a differential equation. Select all equations that could have y(t) a solution. C is a constant. ot equal to 0. y(4) +8y"+16y=0 y"+4y= 3 cos2t y"+y = 3 sin2t (y6) - 64y = 0

If one of the fundamental functions is t²cost, the original equation must be at least of these many degrees: 6 4 3 this fundamental function is not possible. y(t)= Ctcos2t is a solution for a differential equation. Select all equations that could have y(t) a solution. C is a constant. ot equal to 0. y(4) +8y"+16y=0 y"+4y= 3 cos2t y"+y = 3 sin2t (y6) - 64y = 0

Advanced Engineering Mathematics
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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Po If one of the fundamental functions is t²cost, the original equation must be at least of these many degrees: 6 4 3 this fundamental function is not possible. y(t)= Ctcos2t is a solution for a differential equation. Select all equations that could have y(t) a solution. C is a constant. ot equal to 0. y(4) +8y"+16y = 0 y"+4y= 3 cos2t y"+y = 3 sin2t y6) - 64y = 0
Transcribed Image Text:Po If one of the fundamental functions is t²cost, the original equation must be at least of these many degrees: 6 4 3 this fundamental function is not possible. y(t)= Ctcos2t is a solution for a differential equation. Select all equations that could have y(t) a solution. C is a constant. ot equal to 0. y(4) +8y"+16y = 0 y"+4y= 3 cos2t y"+y = 3 sin2t y6) - 64y = 0
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