The solution of the differential equation d y d x 4 - d y d x 2 - 2 = 0 is y = ± λ x + C (where, C is an arbitrary constant). Then, λ 2 is equal to

Question

The solution of the differential equation d y d x 4 - d y d x 2 - 2 = 0 is y = ± λ x + C (where, C is an arbitrary constant). Then, λ 2 is equal to, 2, 4, 8, 16

MARKS App · Accepted Answer

Let d y d x 2 = t ∴ t 2 - t - 2 = 0 ⇒ t - 2 t + 1 = 0 ⇒ d y d x 2 = 2 or - 1 (rejected) ⇒ d y d x = ± 2 or d y = ± 2 d x ⇒ ∫ d y = ± 2 ∫ 1 d x ⇒ y = ± 2 x + C ∴ λ = 2 ⇒ λ 2 = 4

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