Let y = y(x) be a differentiable function in the interval (0, ) such that y(1) = 2, and t x ( t2y(x) - x2y(t)x - t ) = 3 for each x > 0. Then 2y(2) is equal to :
- 27
- 18
- 23
- 12
Answer
(3) 23
Solution
t2y(x) - x2y(t) = y(x)(t2 - x2) + x2 (y(x) - y(t) )
⇒ t2y(x) - x2y(t)x - t = -(t + x) y(x) + x2 y(x) - y(t)x - t
As t x: -2x y + x2 dydx = 3
⇒ dydx - 2xy = 3x2 (linear)
I.F. = e- 2xdx = 1x2
⇒ yx2 = 3x4dx = - 1x3 + C
y(1) = 2 ⇒ 2 = -1 + C ⇒ C = 3
⇒ y = 3x2 - 1x
∴ y(2) = 12 - 12 = 232 ⇒ 2y(2) = 23
Answer: (3)