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Solve [D^4+D^2+1]y=0 |
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Answer» Answer: Step-by-step explanation: (D^4 + 1)y = 0 -(given) Now, from the auxillary equation F(m) = 0 , we get: (m^4 + 1)=0 m^4 + 2m^2 - 2m^2 + 1 = 0 (m^2 + 1)^2 - 2m^2 = 0 (m^2 + 1 - 2m^2) (m^2 + 1 + 2m^2) =0 -m^2 + 1 = 0 or 3m^2 + 1 =0 m^2 = 1 or m^2 = -1/3 m = -1,+1 or m= +_√-1/3 = 0 +- i√1/3 [ since i=√-1] Therefore, COMPLEMENTARY function(C.F.) would be: C1 e^-x + C2 e^x + C3 cos √1/3 x + C4 SIN √1/3 x ✓✓ ___________ii method __________ (D4 - 4D3 + 6D2 - 4D + 1)y = 0 (D + 1Q)4y = 0 Auxiliary equation is (m + 1)4 = 0 m = - 1, - 1, - 1, - 1 Therefore, y = C1 + C2xe-x + C3x2e-x + C4x3e-x ____________________________ Hope it helps you ❤ |
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