AMath350.F13.A7.Sol

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

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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=...
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