This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ( / 3) = 2 , y ( / 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)....
View Full
Document
 Spring '10
 PHAN
 Math

Click to edit the document details