Given the ordinary differential equationd2y+dy= 0dx2dxwith(0)

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Given the ordinary differential equation
d²y + dy = 0
dx² dx

with(0) = 0 and dy (0) = 1, the value of y(1) is
dx

_______ (correct to two decimal places).

1. 1.4678
2. 1.4628
3. 1.4698
4. 1.46

Correct Option: A

(D² + D – 6) y = 0
y (0) = 0,
y' (0) = 1
(D + 3) (D – 2) y = 0
D = 2, – 3
C .F. = C₁ e^2x + C₂ e^–3x
y = c₁ e^2x + C₂ e^–3x
y (0) = 0
So, 0 = C₁ + C₂ ----------------------(i)

y = (1) =	e² - e^{- 3}	= 1.4678
	5

y (0) = 1
1 = 2 C₁ – 3 C₂ ___(ii)
From equations (i) & (ii), we get

C₁ =	1	, C₂ =	- 1
	5		5

y =	e^2x	-	e^{- 3x}
	5		5

when, x = 1

y = (1) =	e² - e^{- 3}	= 1.4678
	5

	d²y	+	dy	= 0
	dx²		dx

with(0) = 0 and	dy	(0) = 1, the value of y(1) is
	dx