Let y = y(x) be the solution of the differential equation (x + y + 2)2 dx = dy, y(0) = -2. Let the maximum and minimum values of the function y = y(x) in [0, 3 ] be and , respectively. If (3 + )2 + 2 = + 3, , Z, then + equals .
Answer
31
Solution
Put v = x + y + 2 dvdx = 1 + dydx = 1 + v2
dv1 + v2 = -1(x + y + 2) = x + C
y(0) = -2 C = 0 y = x - x - 2
y' = 2 x - 1 = 2 x 0 ⇒ y is increasing on [0, 3 ]
= y(0) = -2, = y ( 3 ) = 3 - 3 - 2 3 + = 3 3 - 6
(3 3 - 6)2 + 4 = 27 + 36 - 36 3 + 4 = 67 - 36 3
= 67, = -36 + = 31
Answer: 31