### Question: GATE2005 (2 Marks)

The complete solution of the ordinary differential equation \(\displaystyle \frac{d^2y}{dx^2} + p\frac{dy}{dx} + qy = 0 \text{ is } y = C_1e^{-x} + C_2e^{-3x} \)

Then p and q are

(A) p = 3, q = 3 | (B) p = 3, q = 4 |

(C) p = 4, q = 3 | (D) p = 4, q = 4 |

Which of the following is a solution of the differential equation \(\displaystyle \frac{d^2y}{dx^2} + p\frac{dy}{dx} + (q +1) y = 0 \)

\((A) e^{-3x} \) | \((B) xe^{-x} \) |

\((C) xe^{-2x} \) | \((D) x^2e^{-2x} \) |

### Solution:

(D^{2} + pD + q) y = 0

If D^{2} + pD + q = 0 can be simplified as (D – a) (D – b) = 0; where a and b are roots;

The solution becomes

y = C_{1}e^{ax} + C_{2}e^{bx}

Given that the solution is y = C_{1}e^{-x} + C_{2}e^{-3x} (a = –1 and b = –3)

Equation becomes (D + 1)(D + 3) = 0

D^{2} + 4D + 3 = 0

Comparing with original equation, we get p = 4, q = 3

**Answer: (C)**

(D^{2} + pD + q+1) y = 0; where p = 4 and q = 3

(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}

Checking with the options, we can easily identify that xe^{-2x} matches with the solution when C_{1} = 0 and C_{2} = 1