Let y = y ( x ) be the solution of the differential equation d y d x = 2 y + 2 sin x - 5 x - 2 cos x such that y ( 0 ) = 7 . Then y ( π ) is equal to

Question

Let y = y ( x ) be the solution of the differential equation d y d x = 2 y + 2 sin x - 5 x - 2 cos x such that y ( 0 ) = 7 . Then y ( π ) is equal to, 7 e π 2 + 5, e π 2 + 5, 2 e π 2 + 5, 3 e π 2 + 5

MARKS App · Accepted Answer

d y d x - 2 x y = 2 2 sin x - 5 x - 2 cos x I F = e - x 2 So, y e - x 2 = ∫ e - x 2 2 x 2 sin x - 5 - 2 cos x d x ⇒ y · e - x 2 = e - x 2 5 - 2 sin x + C y = 5 - 2 sin x + C · e x 2 Given at x = 0 , y = 7 ⇒ 7 = 5 + C ⇒ C = 2 So, y = 5 - 2 sin x + 2 e x 2 Now at x = π y = 5 + 2 e π 2

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