- 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........
- 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........
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)
With integration, one of the major concepts of calculus. Differentiation is the derivative or rate of change of a function with respect to the independent variable.
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