If the solution curve y = f(x) of the differential equation (x2 - 4)y' - 2xy + 2x(4 - x2)2 = 0, x > 2, passes through the point (3, 15), then the local maximum value of f is
Answer
16
Solution
Dividing by (x2 - 4): dydx - 2xx2 - 4y = -2x(x2 - 4)
Linear equation; I.F. = e- 2xx2 - 4dx = e- e(x2 - 4) = 1x2 - 4
ddx ( yx2 - 4 ) = -2x
yx2 - 4 = -x2 + C
At (3, 15): 155 = -9 + C C = 12
f(x) = (x2 - 4)(12 - x2)
f'(x) = 2x(12 - x2) - 2x(x2 - 4) = 4x(8 - x2)
f'(x) = 0 at x = 2 2 (x > 2); f' changes from + to - local maximum
f(2 2) = (8 - 4)(12 - 8) = 16
Answer: 16