5
Series Solutions of Linear Equations
Exercises 5.1
1. lim
n
an+1 2n+1 xn+1 /(n + 1) 2n = lim = lim |x| = 2|x| n n n + 1 an 2n xn /n (-1)n converges by n n=1
The series is absolutely convergent for 2|x| < 1 or |x| < 1/2. At x = -1/2, the series
16
Numerical Solutions of Partial Differential Equations
Exercises 16.1
1. The figure shows the values of u(x, y) along the boundary. We need to determine u11 and u21 . The system is u21 + 2 + 0 + 0 - 4u11 = 0 1 + 2 + u11 + 0 - 4u21 = 0 or -4u11 +
13
Boundary-Value Problems in Rectangular Coordinates
Exercises 13.1
1. If u = XY then ux = X Y, uy = XY , X Y = XY , and X Y = = 2 . X Y Then X 2 X = 0 so that X = A1 e x ,
2
and Y 2 Y = 0
Y = A2 e y ,
2
and u = XY = c1 ec2 (x+y) . 2. If u =
11
Systems of Nonlinear Differential Equations
Exercises 11.1
1. The corresponding plane autonomous system is x = y, y = -9 sin x.
If (x, y) is a critical point, y = 0 and -9 sin x = 0. Therefore x = n and so the critical points are (n, 0) for n =
10
Systems of Linear Differential Equations
Exercises 10.1
1. Let X =
x . Then y X = x . Then y X = 4 5 -7 0 X. 3 4 -5 8 X.
2. Let X =
x 3. Let X = y . Then z
-3 X = 6 10 x 4. Let X = y . Then z
4 -1 4
-9 0 X. 3
1 X = 1 -1
20
Conformal Mappings and Applications
Exercises 20.1
1. For w =
1 -y 1 1 x 1 1 and v = 2 . If y = x, u = ,u= 2 ,v=- , and so v = -u. The image is the 2 2 z x +y x +y 2 x 2 x
line v = -u (with the origin (0, 0) excluded.) 2. If y = 1, u = x 1 -1
6
Numerical Solutions of Ordinary Differential Equations
Exercises 6.1
All tables in this chapter were constructed in a spreadsheet program which does not support subscripts. Consequently, xn and yn will be indicated as x(n) and y(n), respectively.
8
Matrices
Exercises 8.1
1. 2 4 6. 8 1
2. 3 2 7. Not equal
3. 3 3 8. Not equal
4. 1 3 9. Not equal
5. 3 4 10. Not equal
11. Solving x = y - 2, y = 3x - 2 we obtain x = 2, y = 4. 12. Solving x2 = 9, y = 4x we obtain x = 3, y = 12 and x =
14
Boundary-Value Problems in Other Coordinate Systems
Exercises 14.1
1. We have A0 = 1 2
0 0
u0 d =
u0 2
1 An = Bn = and so 1
u0 cos n d = 0 u0 sin n d =
0
u0 [1 - (-1)n ] n
u0 u0 u(r, ) = + 2 2. We have A0 = An = Bn = and so u(r, )
3
Higher-Order Differential Equations
Exercises 3.1
1. From y = c1 ex + c2 e-x we find y = c1 ex - c2 e-x . Then y(0) = c1 + c2 = 0, y (0) = c1 - c2 = 1 so that c1 = 1/2 and c2 = -1/2. The solution is y = 1 ex - 1 e-x . 2 2 2. From y = c1 e4x + c2
2
First-Order Differential Equations
Exercises 2.1
1.
y
2.
y y
x t
x t
3.
y
4.
y y
x
x t
5.
y
6.
y
x
x
7.
y
8.
y
x
x
17
Exercises 2.1
9.
y
10.
y
x
x
11.
y
12.
y
x
x
13.
y
14.
y
x
x
15. Writing the different
1
Introduction to Differential Equations
Exercises 1.1
1. Second-order; linear. 2. Third-order; nonlinear because of (dy/dx)4 . 3. The differential equation is first-order. Writing it in the form x(dy/dx) + y 2 = 1, we see that it is nonlinear in y
9
1.
Vector Calculus
Exercises 9.1
2. 3.
4.
5.
6.
7.
8.
9.
Note: the scale is distorted in this graph. For t = 0, the graph starts at (1, 0, 1). The upper loop shown intersects the xz-plane at about (286751, 0, 286751). 10.
375
Exercises 9.
15
Integral Transform Method
Exercises 15.1
1. (a) The result follows by letting = u or u = {t-1/2 } =
2
2 in erf( t ) =
t
e-u du.
2
0
(b) Using
s1/2
and the first translation theorem, it follows from the convolution theorem that