If the solution curve y=f(x) of the differential equation (x^ 2 -4 ) y^ -2 x y+2 x (4-x^ 2 )^ 2 =0, x2, passes through the point (3,15), then the local maximum value of f is \_\_\_\_.

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If the solution curve y=f(x) of the differential equation (x^ 2 -4 ) y^ -2 x y+2 x (4-x^ 2 )^ 2 =0, x>2, passes through the point (3,15), then the local maximum value of f is \_\_\_\_.

Quantrex Academy · Accepted Answer

Correct answer: A. A

Rewriting: dy dx - 2x x^2-4 y = -2x(x^2-4). IF = e^ - 2x x^2-4 dx = 1 x^2-4 . d dx ( y x^2-4 ) = -2x y x^2-4 = -x^2 + C. y = (x^2-4)(C-x^2). At (3,15): 15 = 5(C-9) C = 12. y = (x^2-4)(12-x^2). y' = 4x(8-x^2) = 0 x = 2 2 (for x > 2). y'' = 32 - 12x^2 = -64 f(2 2 ) = (8-4)(12-8) = 16.

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