GATE1995: Higher Order Differential Equation

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Question: GATE1995 (2 Marks)

The solution to the differential equation \(f”(x) +4f’ (x) + 4f (x) = 0 \) is

\((a) f_1 (x) = e^{-2x} \)
\((b) f_1 (x) = e^{2x}, f_2 (x) = e^{-2x} \)
\((c) f_1 (x) = e^{-2x}, f_2 (x) = e^{-2x} \)
\((d) f_1(x) = e^{-2x}, f_2(x) = e^{-x} \)

Solution:

(D2 + 4D + 4) y = 0

(D + 2)2 y = 0

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

In this case, a = –2

Solution y = (C1 + C2x) e–2x

y = C1e–2x + C2xe–2x

This is similar to y = f1(x)e–2x + f2(x)e–2x

Answer: (C)