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- Let ² (²+2r + 5) = 0 be the characteristic equation of 4th order D.E. with constant coefficients. Then the general linear homogeneous solution is........

- Let ² (²+2r + 5) = 0 be the characteristic equation of 4th order D.E. with constant coefficients. Then the general linear homogeneous solution is........

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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differential equation, please solve question 5

Q5. Let ² (²+2r + 5) = 0 be the characteristic equation of 4th order linear homogeneous D.E. with constant coefficients. Then the general solution is ² + 2r ²³² +5r²²=0 / r ² (r ) (r ) -bt rzor=0 Q6. L-1 ¹ {3²+45+6} = 5²+us = (5 + A) ³²-A)² (5+2=2 ³ (5+2)² +2 (8+2) -L- (5+2)²+2 (5+2) ²74 315 *)-(² 1²_ u +6 |L'S6+ 7 = -1 + 21 ^2 = -1 -2 (r₁ +1+2i)(√₂+1-2²)
Transcribed Image Text:Q5. Let ² (²+2r + 5) = 0 be the characteristic equation of 4th order linear homogeneous D.E. with constant coefficients. Then the general solution is ² + 2r ²³² +5r²²=0 / r ² (r ) (r ) -bt rzor=0 Q6. L-1 ¹ {3²+45+6} = 5²+us = (5 + A) ³²-A)² (5+2=2 ³ (5+2)² +2 (8+2) -L- (5+2)²+2 (5+2) ²74 315 *)-(² 1²_ u +6 |L'S6+ 7 = -1 + 21 ^2 = -1 -2 (r₁ +1+2i)(√₂+1-2²)
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