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Unformatted text preview: Math 267 – Assignment 1  Solutions 1. Find the general solution of the following DEs. (a) y 00 + 5 y = 0, (b) y 00 + 6 y + 13 y = 0, (c) 25 y 00 20 y + 4 y = 0. Solution. (a) The characteristic equation of the given ODE is r 2 + 5 r = r ( r + 5) = 0 ⇒ r 1 = 0 , r 2 = 5 . So, the general solution is y ( t ) = c 1 e t + c 2 e 5 t = c 1 + c 2 e 5 t where c 1 and c 2 are arbirary constants. (b) The characteristic equation is r 2 + 6 r + 13 = 0, which has the roots r 1 = 3 + 2 i and r 2 = 3 2 i . Therefore, the general solution is given by y ( t ) = c 1 e 3 t cos(2 t ) + c 2 e 3 t sin(2 t ) where c 1 and c 2 are arbirary constants. (c) The characteristic equation is 25 r 2 20 r +4 = 0, which has the roots r 1 = r 2 = 2 / 5. Then the general solution is y ( t ) = c 1 e (2 / 5) t + c 2 te (2 / 5) t where c 1 and c 2 are arbirary constants. 2. Solve the following IVPs and sketch the solutions. (a) y 00 6 y + 9 y = 0 , y (0) = 0 , y (0) = 2. (b) y 00 + y = 0 , y ( π/ 3) = 2 , y ( π/ 3) = 4. (c) y 00 + 4 y + 3 y = 0 , y (0) = 2 , y (0) = 1. Solution. (a) First, we find the general solution. The corresponding characteristic equation is r 2 6 r + 9 = 0, which has the roots r 1 = r 2 = 3. Thus, the general solution is y ( t ) = c 1 e 3 t + c 2 te 3 t . 1 Using the initial conditions, we get: y (0) = c 1 = 0 . Next, we compute y ( t ) after setting c 1 = 0 which yields y ( t ) = c 2 e 3 t + 3 c 2 te 3 t . Then, using the second initial condition, we get y (0) = c 2 = 2 . We conclude that the solution of our IVP is y ( t ) = 2 te 3 t . As seen in Figure 1, y ( t ) → ∞ as t → ∞ . 5 10 15 20 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 x 10 27 t y Figure 1: The graph of y ( t ) = 2 te 3 t (Problem 2a)....
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 Spring '10
 PHAN
 Math, Boundary value problem, general solution, Boundary conditions

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