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...
View
Full Document
 Fall '07
 TERRELL,R
 Differential Equations, Equations

Click to edit the document details