GATE2008(1): Higher Order Differential Equation

Reading Time: 1 minute

Question: GATE2008(1 Mark)

Given that \(\ddot{x} + 3x = 0 \) and \(x(0) = 1, \dot{x}(0) = 0,  \) what is x(1) ?

(A) -0.99                     (B) -0.16
(C) 0.16 (D) 0.99

Solution:

Let x = f(t)

(D2 + 3)x= 0

\((D – \sqrt{3}i)(D + \sqrt{3}i)x = 0 \)

This is of the form (D – a)(D – b)x = 0;

Where a = α + iβ and b = α – iβ which has solution x = eαt (C1cosβt + C2sinβt)

In this case, α = 0 and β =\(\sqrt{3} \)

Solution: x = C1cos \(\sqrt{3} \) t + C2sin\(\sqrt{3} \) t

 \(\dot{x}= -\sqrt{3} C_1 sin\sqrt{3}t+ \sqrt{3} C_2 cos \sqrt{3} t \)

Boundary conditions:

x(0) = 1

1 = C1

\(\dot{x}(0) = 0 \)

\(0 = \sqrt{3} C_2 \)

Hence, C1 = 1 and C2 = 0

Equation: x = cos \(\sqrt{3} \) t

When x = 1 radian, the angle in degrees = \(\displaystyle \frac{180}{\pi} \times 1 = 57.3° \)

\(x (1) = cos(\sqrt{3} \times 57.3) =  –0.16 \)

Answer: (B)