prelim2_1 - = A = 1 y (0) = 3 Ae 3(0) 3 B e 3(0) + Be 3(0)...

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293 Test 2 Problem 1 Geoffrey Recktenwald Oct. 31, 2005 Solve the following initial value problems A) y ′′ + 9 y = 0 ICs y (0) = 1 and y (0) = 3 Assume a solution of the form y ( x ) = e rx to get the characteristic equation r 2 + 9 = 0. Solve the characteristic equation to get r = ± 3 i . The solution has the form y ( x ) = A cos3 x + B sin 3 x or equivalently y ( x ) = ae 3 ix + be 3 ix . We will take our solution in the form y ( x ) = A cos3 x + B sin 3 x and the derivative y ( x ) = 3 A sin 3 x + 3 B cos3 x Use the initial conditions to solve for A and B . y (0) = A cos3(0) + B sin 3(0) = A = 1 y (0) = 3 A sin 3(0) + 3 B cos3(0) = 3 B = 3 Thus, the final solution is y ( x ) = cos3 x + sin 3 x B) y ′′ + 6 y + 9 y = 0 ICs y (0) = 1 and y (0) = 3 Assume a solution of the form y ( x ) = e rx to get the characteristic equation r 2 + 6 r + 9 = 0. Solve the characteristic equation to get a repeated root at r = 3. The solution has the form y ( x ) = Ae 3 x + Bxe 3 x and the derivative is y ( x ) = 3 Ae 3 x 3 Bxe 3 x + Be 3 x Use the initial conditions to solve for A and B . y (0) = Ae 3(0) + B 0 e 3(0)
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Unformatted text preview: = A = 1 y (0) = 3 Ae 3(0) 3 B e 3(0) + Be 3(0) = 3 A + B = 3 Thus, A = 1 and B = 6. The final solution is y ( x ) = e 3 x + 6 xe 3 x C) y + 10 y + 9 y = 0 ICs y (0) = 1 and y (0) = 3 Assume a solution of the form y ( x ) = e rx to get the characteristic equation r 2 + 10 r + 9 = 0. Solve the characteristic equation to get r 1 = 1 and r 2 = 9. The solution has the form y ( x ) = Ae x + Be 9 x and the derivative is y ( x ) = Ae x 9 Be 9 x Use the initial conditions to solve for A and B . y (0) = Ae + Be 9(0) = A + B = 1 y (0) = Ae 9 Be 9(0) = A 9 B = 3 Thus, A = 3 / 2 and B = 1 / 2. The final solution is y ( x ) = 3 2 e x 1 2 e 9 x...
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