Let y = y x be the solution of the differential equation 1 - x 2 d y = x y + x 3 + 2 1 - x 2 d x , - 1 x 1 and y 0 = 0 . If ∫ - 1 2 1 2 1 - x 2 y x d x = k then k - 1 is equal to

Question

Let y = y x be the solution of the differential equation 1 - x 2 d y = x y + x 3 + 2 1 - x 2 d x , - 1 x 1 and y 0 = 0 . If ∫ - 1 2 1 2 1 - x 2 y x d x = k then k - 1 is equal to,

MARKS App · Accepted Answer

Given, 1 - x 2 d y d x = x y + x 3 + 2 1 - x 2 ⇒ d y d x + - x 1 - x 2 y = x 3 + 2 1 - x 2 IF = e ∫ - x 1 - x 2 d x = 1 - x 2 y x · 1 - x 2 = x 4 4 + 2 x + c y 0 = 0 ⇒ c = 0 1 - x 2 y x = x 4 4 + 2 x So,required value = ∫ - 1 2 1 2 x 4 4 + 2 x d x - 1 4 · 2 ∫ 0 1 2 x 4 d x = 1 10 x 5 0 1 2 = 1 320 k - 1 = 320

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