GATE2008(2): Higher Order Differential Equation

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Question: GATE2008 (2 Marks)

It is given that \(y” + 2y’+ y = 0, y(0) = 0, y(1) = 0 \). What is y(0.5) $

(A) 0                     (B) 0.37
(C) 0.62 (D) 1.13

Solution:

(D2 + 2D + 1) y = 0

(D + 1)2 y = 0

D = –1, –1 (real repeated roots; similar to (D – a)2 y = 0; having solution y = (C1 + C2x)eax

In this case, a = –1

Solution y = (C1 + C2x)e–x

Applying boundary conditions:

y(0) = 0

0 = C1

y(1) = 0

0 = C1 + C2

C2 = 0

Equation becomes y = 0

Therefore, y (0.5) = 0

Answer: (A)