Correct Option: C

	d²y	= y ⇒ (D² - 1)y = 0
	dx²

D² – 1 = 0
⇒ D = ± 1
y = c₁ e^x + c₂ e^–x
142,

Passes through (0, 0) and			3		(0,0)
			4

⇒ 0 = C₁ + C₂ ...(1)
142,

			3		(0,0)
			4

	3	= C1 + C2e-142 = C1 2 +	C2
	4		2

⇒ 2C1 +	1	C2 =	3
	2		4

solving (1) and (2)
⇒ C₁ = 1/2
C₂ = – 1/2

∴ y =	1	ex -	1	e-x =	1	(ex - e-x)
	2		2		2