Math 293 solution set 10

Math 293 solution set 10 - Math 293 Solutions to Problem...

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Math 293 Solutions to Problem Set 10. § 10.2 #4 . y 00 + 9 y = 0 has general solution y = c 1 sin 3 x + c 2 cos 3 x . Plugging in the condition y (0) = 0 gives c 2 = 0, so y = c 1 sin 3 x . To apply the condition y 0 ( π ) = - 6 we compute y 0 = 3 c 1 cos 3 x , and then plugging in y 0 ( π ) = - 6 yields c 1 = 2. The answer is therefore y = 2 sin 3 x . #9 . To solve y 00 + λy = 0 we examine the 3 possible cases: Case (1): λ > 0. Write λ = α 2 with α > 0. The quadratic equation is r 2 + α 2 = 0, with roots r = ± α i , so the general solution is y = c 1 sin αx + c 2 cos αx . The boundary condition y (0) = 0 translates into c 2 = 0, so we have y = c 1 sin αx . The derivative is y 0 = αc 1 cos αx , so the boundary condition y 0 ( π ) = 0 implies that cos απ = 0. (We are assuming that α > 0, and if c 1 were 0 then y would be the trivial solution 0.) The equation cos απ = 0 implies that απ = 2 n - 1 2 π for some positive integer n , hence α = 2 n - 1 2 . Thus we have the solutions y = c 1 sin 2 n - 1 2 x corresponding to the eigenvalues λ = α 2 = ( 2 n - 1 2 ) 2 .

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