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