Section 1.1 Some basic math models and direction fields 15-20, 25
15. (j) y=2-y, since y=2 is equilibrium, and when y>2, y<0, and when y<2, y>0,
16. ( c) y=y-2, since y=2 is equilibrium, and when y>2, y>0, and when y<2, y<0.
17. (g) y=-y-2, since y=-2 is
Section 3.6: 7, 10, 13
7.
y 00 + 4y 0 + 4y = t
2
e
2t
;t > 0
Solution 1 Characteristic equation for the corresponding homogeneous equation is
r2 + 4r + 4 = 0;
(r + 2)2 = 0;
r 1 = r2 =
2:
so the fundamental set of the solution are
y1 = e
The Wronskian is
2
Section 3.3: 35
35.
t2 y " + ty 0 + y = 0:
Solution 1 Assume that the solution has the form of y = tr ; plug it into ODE, we have
(r(r
1) + r + 1)tr = 0;
r 2 + 1 = 0;
r1;2 = i:
So two independent solutions are
y1 = ti = ei ln t = cos(ln t) + i sin(ln t);
Section 3.1: 12, 13, 20, 22
12.
y" 3y 0, y 0 2, y 0 3.
Solution
The characteristic equation is
r 2 3r 0,
r r 3 0,
r 1 0, r 2 3.
So the general solution is
y t c 1 c 2 e 3t .
The derivative is
y t 3 c 2 e 3t .
set t 0, we have
c 1 c 2 2,
3c 2 3.
So we have
Section 2.6: 26, 28
26. Find an integrating factor and solve it
y 0 = e2x + y
1:
Solution 1 The original equation could be rewritten as
(e2x + y
1)dx
dy = 0:
where
M (x; y ) = e2x + y 1; My = 1;
N (x; y ) = 1; Nx = 0:
we have
My
Nx
1
=
N
0
=
1
1
so the in
Section 2.2: 5, 6, 12, 14
5:
y 0 = cos2 x cos2 2y:
Soln:
First, by inspection that if cos 2y = 0; then 2y = n + 2 ; y =
(We call it trivial because it is a constant.)
Next, solve the equation by separation of variables:
n
2
(1)
+ 4 ; for every x is a triv
Section 2.1: 13, 14, 18, 20, 32
13:
y0
y = 2te2t ; y (0) = 1:
(1)
dy
dt
(2)
Soln:Step1:
(t)
Step 2: Compare lhs with
d( y )
dt
=
dy
dt
y = 2 te2t :
d
=
at
d
at y;
we choose
)
+
= e t , (we choose c = 1)
(3)
Step 3: equation in step 1 becomes
d(e t y )
= 2