Sixth Order Differential Equation
f satisfies
f ( x ) = − 1 0 1 x 3 ( 1 0 1 x 3 f ( 6 ) ( x ) + 1 0 9 x 2 f ( 5 ) ( x ) + 5 9 x f ( 4 ) ( x ) + 5 3 f ( 3 ) ( x ) ) ,
f ( 1 ) = 0 , f ′ ( 1 ) = 0 , f ′ ′ ( 1 ) = 0 , f ( 3 ) ( 1 ) = 0 , f ( 4 ) ( 1 ) = 0 , and f ( 5 ) ( 1 ) = 3 5 2 0
If f ( e π ) = e a π − e − π , where a is an integer, find a .
The answer is 3.
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Looking for solutions of the form x u , we see that we must satisfy the equation u ( u − 1 ) ( u − 2 ) ( u − 3 ) ( u − 4 ) ( u − 5 ) + 9 u ( u − 1 ) ( u − 2 ) ( u − 3 ) ( u − 4 ) + 1 8 u ( u − 1 ) ( u − 2 ) ( u − 3 ) + 6 u ( u − 1 ) ( u − 2 ) u 2 ( u − 1 ) 2 ( u − 2 ) 2 + 1 0 0 = − 1 0 0 = 0 which has solutions u = − 1 ± i , u = 1 ± 2 i , 3 ± i . Thus the general solution of the differential equation is f ( x ) = x A cos ( ln x ) + B sin ( ln x ) + x ( C cos ( 2 ln x ) + D sin ( 2 ln x ) ) + x 3 ( E cos ( ln x ) + F sin ( ln x ) ) x > 0 for constants A , B , C , D , E , F . Matching the initial conditions, we deduce that f ( x ) = 3 x 3 cos ( ln x ) + 2 sin ( ln x ) + 3 4 x sin ( 2 ln x ) − 3 1 x 3 ( 3 cos ( ln x ) − 2 sin ( ln x ) ) x > 0 and this gives f ( e π ) = e 3 π − e − π , making the answer 3 .