Solutions to Homework Assignment 2
MATH 256-01 Section 1.2, Page 14 To Keep: 1, 2, 6, 11, 12, 13, 15 1. (a) Since dy = y + 5, we have dt dy/dt =1 y + 5 d [ln | y + 5|] = 1 dt ln | y + 5| = t + C ln | y + 5| = t + C | y + 5| = et+C y + 5 = Cet y = 5 Cet .
Solutions to Homework Assignment 3
MATH 256-01 Section 1.3, Page 22 To Keep: 1-21, 23, 28 1. Second order linear. 2. Second order linear. 3. Fourth order linear. 4. First order nonlinear. 5. Second order nonlinear. 6. Third order linear. d 7. Since y1 = e
Solutions to Homework Assignment 8
MATH 256-01 Section 2.6, Page 95 Problems: 1-14, 18-22, 25-30 1. M = 2x + 3, My = 0; N = 2y 2, Nx = 0. Since My = Nx , this equation is exact. We have x = M x = 2x + 3 = x2 + 3x + h(y). This gives N = y = h (y) = 2y 2, s
Solutions to Homework Assignment 12
MATH 256-01 Section 3.3, Page 152 Problems: 1-10, 15-18, 21-25, 28 1. W (f, g) = pendent. 2. cos 3 = cos 2 cos - sin 2 sin = (cos3 - cos sin2 ) - (2 sin2 cos ) = cos3 - cos + cos3 - 2 cos - 2 cos2 = 4 cos3 - 3 cos . Sin
Solutions to Homework Assignment 13
MATH 256-01 Section 3.4, Page 158 Problems: 1-6, 7-21 odd, 24, 28, 29, 33, 35, 36, 37 1. e1+2i = e(cos 2 + i sin 2) 3. ei = cos + i sin = -1. 2. e2-3i = e2 (cos(-3) + i sin(-3) = e2 (cos 3 - i sin 3).
1 (cos(ln 2) + i s
Solutions to Homework Assignment 9
MATH 256-01 Section 2.7, Page 103 Problems: 1, 3, 4, 5, 7, 11, 13 I will use the MAPLE code we created in class for most of these exercises. I modified the code a little since the example we did was autonomous, but the e
Solutions to Homework Assignment 10
MATH 256-01 Section 3.1, Page 136 Problems: 1-17 odd, 20, 21, 29-43 odd 1. The characteristic equation is r2 + 2r - 3 = 0, so (r + 3)(r - 1) = 0. Thus r = -3 or r = 1, and the general solution is c1 e-3t + c2 et . 3. 6r
Solutions to Homework Assignment 18
MATH 256-01 Section 3.7, Page 183 Problems: 1-19 odd 1. The solution to the corresponding homogeneous equation is c1 e2t + c2 e3t , so let Y (t) = u1 e2t + u2 e3t . Then Y = u1 e2t +2u1 e2t +u2 e3t +3u2 e3t = 2u1 e2t +3