# GATE2005: Higher Order Differential Equation

### 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:

(D2 + pD + q) y = 0

If D2 + pD + q = 0 can be simplified as (D – a) (D – b) = 0; where a and b are roots;

The solution becomes

y = C1eax + C2ebx

Given that the solution is y = C1e-x + C2e-3x (a = –1 and b = –3)

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

D2 + 4D + 3 = 0

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

(D2 + pD + q+1) y = 0; where p = 4 and q = 3

(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

Checking with the options, we can easily identify that xe-2x matches with the solution when C1 = 0 and C2 = 1