# GATE2008(1): Higher Order Differential Equation

### 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 = $$\displaystyle \frac{180}{\pi} \times 1 = 57.3°$$

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