ans:
M351 Study Guide 1 (S. Zhang) .
f=
fy =
1
2
Verify that the piecewise dened function
1.45
x2 , x < 0
x2 ,
x0
For any point (x0 , y0 ) in the rst and the third quadrant,
excluding y = 0 line, we have locally a unique solution for IV
y(x0 ) = y0 .
is
ans:
M351 H11 (S. Zhang) 8.7-10.
1. 8.7: 5-8
1. (8.7:6)
det(A I) =
Solve Ax = b by Cramers rule
1
5
2
,
1
= 3i
A=
5
4
, b=
10 6
1
5
= 3i
ans:
A I =
xk =
Bk
det A
where Bk is obtained by replacing k-th column of A.
x=C
14
35
=
1/5
1/2
A I =
Solve Ax =
M351 H12 (S. Zhang) 8.12.
(Note: Only x1 is free, which does not appear in Ax = 0.)
1. 8.12: 9-17, 39-40
For = 2, solve (A I)x = 0:
1 3 1
3
4 x = c 1
1
0
1. (8.12:10)
Factor the matrix A into a product XDX 1 ,
where D is diagonal.
1
1/2
(A =
2
.
1
(Note
M351 H5 (S. Zhang) 3.4-6.
1.
3.5: 1, 2, 9-12, 25
22.99
1.
3.4: 7-8, 12-14, 17, a1-a2
1.
20.99
1.
(3.5:2)
Solve
y + y = tan x
22.15
(3.4:12)
Solve
ans: For yc = yH ,
20.31
y 16y = 2e4x
r2 + 1 = 0, r = i
ans:
For yH , char equation
yH = c1 cos x + c2 sin
M351 H8 (S. Zhang) 8.1-2.
1. (8.2:3)
Solve the linear system by
(1) Gaussian elimination
(2) Gauss-Jordan elimination
1. 8.1: 11-12, 16-18,21,23,26-30, 35,38
1. (8.1:18)
Find AB and BA
1 4
A = 5 10 , B =
8 12
9x1
2x1
4 6 3
1 3 2
ans: (1) Row-echelon form
M351 H6 (S. Zhang) 3.7-8.
(C) u = y , but
1. 3.7: a1-a2,3-4,13-15
1. (3.7:14)
y =u
Taylor series solution of IVP:
y + y2 = 1
y(0) = 2,
y (0) = 3
3u
ans:
du
dy
du
2u = 0
dy
Divided by u (may check additional solution u = 0 below.)
2
2
y = y + 1
y (0) = 2
ans:
M351 H3 2.6-8 (S. Zhang) .
1.
2.6: 2-4
T (0) = 70
A = 10
T (0.5) = 50
dT
= k(A T ) = k(10 T )
dt
dT
= kdt
10 T
ln(10 T ) = kt + C,
10.99
1.
10.26
(2.6:2)
Approximate y(0.2) by the Euler method with
h = 0.1 and again with h = 0.05:
y = x + y2 ,
y(0)
R
M351 H4 3.1-3 (S. Zhang) .
1.
y2 = y1
3.1: 17-19, a1
1.
2
e
= x2
1
x2 x4 dx = x2 ( x5 + C)
5
2
y1 = 1 + x , y2 = x x, y3 = x + 3x + 4
x4
We can let C = 0 above as it is included in the general
solution below:
Method 1 (nonzero solutions)
c1 y1 + c2 y2 +
M351 Study Guide Final (S. Zhang) .
1.
y = 4y y 2 = y(4 y)
y = 4y 2yy = (4 2y)y
Solve
2.55
1
x<1
1 x 1
y +y =
y(0) = 1
By y = 0, we have critical points y = 0 and y = 4.
For y = 0, we have an additional point, y = 2.
ans: The solution is done in two step
M351 Study Guide 2 (S. Zhang) .
4. Find the general solution of the homogeneous equation yc
and write down the form for undetermined coecients for
yp . (Do not solve for yp .)
1. Solve
y 16y = 2e4x
(D2 9)(D + 3)(D2 + 4)y =ex sin x e3x + cos 2x.
ans:
For
2. (1.2:18)
Without solving the equation, nd and plot the
region in xy plane for the following equation such that there
is a unique solution through each given initial point in this
region
y = xy
M351 H1 1.1, 1.2, 2.1 (S. Zhang) .
1. (1.1:25)
Verify that