E v1 2v2 2 1 as our eigenvector thus weve determined

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 5t . Now, to find a particular solution, we assume that it can be written in the form 1 2 x p = u1 + u2 e 5t . 2 1 As shown in the course notes, this requires us to solve the system )= 1 t (2) )=4+ 2 t (3) u01 (1) + u02 ( 2e u01 (2) + u02 (e 5t 5t , AMATH 350 Page 3 Assignment #7 Solutions - Fall 2013 Multiplying (3) by 2, and then subtracting it from (4), we obtain 5e 5t 0 u2 4 Therefore u02 = e5t , so 5 u2 = From (3) we also know that u01 = = 4. 4 5t e. 25 1 + 2e t 5t 0 u2 u1 = ln |t| + = 18 + , so t5 8t . 5 Therefore a particular solution is ✓ ◆ 8t 4 1 2 xp = ln |t| + + 2 1 5 25 8 8 ln |t| + 5t 25 = , 16t 4 2 ln |t| + 5 + 25 and the full general solution is 1 2 x = C1 + C2 e 2 1 5t ✓ 8t + ln |t| + 5 ◆ 1 2 4 + 25 2 1 . AMATH 350 Page 4 Assignment #7 Solutions - Fall 2013 3. This is simply a different way of expressing the same kinds of problems we’ve been solving; we have 11 1 0 0 ~= x ~+ x 2t, ~ (0) = x . 11 1 0 The characteristic equation is 1 1 det =( 1 1 1) 2 1= 2 2= ( 2) = 0 , so the eigenvalues are 0 and 2. To find an eigenvector for 0, we write 11 v1 11 v2 = 0 0 and conclude that v1 + v2 = 0, so we may use ~ = v 1 . 1 For the eigenvector 2, we have instead 11 v1 0 = 1 1 v2 0 1 and so v1 v2 = 0, and we may use ~ = v . 1 We can now see that the complementary function is 1 1 ~ h = c1 x + c2 e 2t . 1 1 Therefore, to find the general solution, we let 1 1 ~ = u1 x + u2 1 1 e 2t and to find u1 and u2 we must solve the system of equations u01 + u02 e2t = 2t u01 + u02 = 2t. Adding these, we find that 2u02 e2t = 0 =) u02 = 0 =) u2 = c2 . Subtracting them instead gives us 2u01 = 4t =) u01 = 2t =) u1 = t2 + c1 . Thus we’ve found that ~ = c1 x 1 1 + c2 1 1 2t e+ 1 1 t2 . Applying the initial condition, we require that c1 + c2 = 0 c1 + c2 = 0, so c1 = c2 = 0, and the desired solution is simply x (t) = t2 , y (t) = t2 . AMATH 350 Page 5 Assignment #7 Solutions - Fall 2013 4. Commodity Problem: 7. (a) The model is dD = 2D 5p + 10, dt dp = D S = D 50, dt or y0 = Ay + F where D p y=...
View Full Document

Ask a homework question - tutors are online