### 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:

(D^{2} + 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 = (C_{1} + C_{2}x)e^{ax}

In this case, a = –2

Solution y = (C_{1} + C_{2}x) e^{–2x}

y = C_{1}e^{–2x} + C_{2}xe^{–2x}

This is similar to y = f_{1}(x)e^{–2x} + f_{2}(x)e^{–2x}