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In Exercises 1-6, find the general solution (in scalar form) of the given second-order equation. 1. d'y dt² 7. 6- dt -7y=0 d1² +9y=0 dt 3. + dt² dt d²y 5. +8 +25y = 0 di² dt In Exercises 7-12, find the solution of the given initial-value problem. +2-3y=0 dt -Sy=0 dt² di y (0) = 6, y' (0) = -2 y (0) = 11, y'(0) = -7 d'y dy d'y dy 12y = 0 +4y=0 d1² d²y -4 +29y=0 dt² dt dt

In Exercises 1-6, find the general solution (in scalar form) of the given second-order equation. 1. d'y dt² 7. 6- dt -7y=0 d1² +9y=0 dt 3. + dt² dt d²y 5. +8 +25y = 0 di² dt In Exercises 7-12, find the solution of the given initial-value problem. +2-3y=0 dt -Sy=0 dt² di y (0) = 6, y' (0) = -2 y (0) = 11, y'(0) = -7 d'y dy d'y dy 12y = 0 +4y=0 d1² d²y -4 +29y=0 dt² dt dt

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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The four cases just described completely classify the various long-term behav- iors of all harmonic oscillators. In the next section we will derive a geometric way to classify these behaviors. EXERCISES FOR SECTION 3.6 In Exercises 1-6, find the general solution (in scalar form) of the given second-order equation. 1. 3. 5. d²y dt2 7. dy 6 dt -7y=0 d²y +6 dt2 dt d²y dy +8 +25y = 0 dt² dt d²y dy 4 +9y=0 2. +13-0 4. 6. 8. dy dt² dt 10 d²y dt² 4 dt In Exercises 7-12, find the solution of the given initial-value problem. d²y dy +2- - 3y = 0 dt2 dt y (0) = 6, y'(0) = -2 12y = 0 = +4y=0 d²y 4. +29y = 0 dt² dt dy +4 - 5y = 0 dt² dt y (0) = 11, y'(0) = -7 d²y dy +4 +2012-0
Transcribed Image Text:The four cases just described completely classify the various long-term behav- iors of all harmonic oscillators. In the next section we will derive a geometric way to classify these behaviors. EXERCISES FOR SECTION 3.6 In Exercises 1-6, find the general solution (in scalar form) of the given second-order equation. 1. 3. 5. d²y dt2 7. dy 6 dt -7y=0 d²y +6 dt2 dt d²y dy +8 +25y = 0 dt² dt d²y dy 4 +9y=0 2. +13-0 4. 6. 8. dy dt² dt 10 d²y dt² 4 dt In Exercises 7-12, find the solution of the given initial-value problem. d²y dy +2- - 3y = 0 dt2 dt y (0) = 6, y'(0) = -2 12y = 0 = +4y=0 d²y 4. +29y = 0 dt² dt dy +4 - 5y = 0 dt² dt y (0) = 11, y'(0) = -7 d²y dy +4 +2012-0
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