8-27 HOMEWORK FIGURES FOR MATH 242
3
2
y(t)
1
0
0
1
2
3
4
t
-1
-2
-3
Figure 1. Slope field for y 0 + 3y = t + e−2t
1
2
2
t
y(t)
0
0.4
0.8
1.2
0
-2
-4
-6
Figure 2. Slope field for y 0 − 2y = t2 e2t
3
4
y(t)
2
0
0
1
2
3
4
t
-2
-4
Figure 3. Slope field for y 0 + (1/t)y = 3 cos(2t)
5
4
4
y(t)
2
0
0
1
2
3
4
t
-2
-4
Figure 4. Slope field for ty 0 + 2y = sin t
5
5
4
y(t)
2
0
0
1
2
3
4
t
-2
-4
Figure 5. Slope field for (1 + t2 )y 0 + 4ty = (1 + t2 )−2
5
6
4
y(t)
2
0
0
1
2
3
4
t
-2
-4
Figure 6. Slope field for ty 0 − y = t2 e−t
5
7
10
y(t)
5
0
0
5
10
15
20
t
-5
-10
Figure 7. Slope field for y 0 − 12 y = 2 cos t. The solutions have the initial conditions y(0) = a, with a =
−0.5, −0.7, −0.79, −0.8, −1.
8
10
t
y(t)
0
2
4
6
8
10
12
0
-10
-20
-30
Figure 8. Slope field for 2y 0 − y = et/3 . The solutions have the initial conditions y(0) = a, with a =
−2.5, −2.6, −2.7, −2.8, −3.
9
10
y(t)
5
0
0
1
2
3
4
5
t
-5
-10
Figure 9. Slope field for 3y 0 − 2y = e−πt/2 . The solutions have the initial conditions y(0) = a, with a =
−1, −0.5, 0, 0.5, 1.
6
10
4
y(t)
2
0
1
2
3
t
-2
-4
Figure 10. Slope field for ty 0 + (t + 1)y = 2te−t . The
solutions have the initial conditions y(1) = a, with a =
−1, −0.5, 0, 0.5, 1.
4
11
4
y(t)
2
0
-4
-3
-2
-1
0
t
-2
-4
Figure 11. Slope field for ty 0 + 2y = (sin t)/t. The
solutions have the initial conditions y(−π/2) = a, with
a = −1, −0.5, 0, 0.5, 1.