Finding Solutions. Many differential equations have solutions of the form y(t) = eat, where a is some constant. For example, y(t) = e³t is a solution to the equation y' = 3y. Find all values of a such that y(t) = eat is a solution to the given equation in Exercise Group 1.1.9.10-15. 10. y'= -2y 11. y' + 7y=0
Finding Solutions. Many differential equations have solutions of the form y(t) = eat, where a is some constant. For example, y(t) = e³t is a solution to the equation y' = 3y. Find all values of a such that y(t) = eat is a solution to the given equation in Exercise Group 1.1.9.10-15. 10. y'= -2y 11. y' + 7y=0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Answer only odd number questions
Transcribed Image Text:ns
= of
S
Finding Solutions. Many differential equations have solutions of the form
y(t) = eat, where a is some constant. For example, y(t) = e³t is a solution to
the equation y' = 3y. Find all values of a such that y(t) = eat is a solution to
the given equation in Exercise Group 1.1.9.10-15.
11. y' + 7y=0
13. y" + 4y 12y = 0
15. y(5)- 5y"" + 4y' = 0
10. y'=-2y
12. y" - 3y +2y=0
14. y" - y" - 4y' + 4y = 0
Initial Value Problems. Use direct substitution to verify that y(t) is a
solution of the given differential equation in Exercise Group 1.1.9.16-21. Then
use the initial conditions to determine the constants C or c₁ and c₂.
4
16. y' = 4y, y(0) = 2. y(t) = Cet
17. y' + 7y= 0, y(0) = 2. y(t) = Ce 7t
18. y" + 4y 0. y(0) = 1, y'(0) = 0, y(t)- ci cos 2t + c₂ sin 2t
19. y" - 5y' + 4y = 0, y(0)= 1, y'(0) 0, y(t) =cie + c₂e¹t
At
20. y" + 4y + 13y 0, y(0) = 1, y'(0)
0.
2t
y(t)
cos 3t+
cje
2t
C₂e
sin 3t
21. y" 4y + 4y = 0, y(0)
1. y' (0) - 0, y(t) - ce²t + c₂te²t
Transcribed Image Text:1. Baking a Potato (Part 1). Suppose that you bake a potato in your
microwave. When you remove the potato from the microwave oven, you record
the temperature every two minutes (see Table 1.1.11). If room temperature is
73 degrees, write a differential equation that models the temperature of the
potato, T(t), at time t.
Table 1.1.11. Temperatures for a Baked Potato
t (minutes) T(t) (temperature in degrees Fahrenheit)
0
197.2
2
191.6
4
187.1
6
182.8
8
178.7
Verifying Solutions.
Use direct substitution to verify that y(t) is a solution
of the given differential equation in Exercise Group 1.1.9.2-9.
2. y(t) = et y' = 4y
4. y(t) = 3e5t; y' - 5y = 0
1
2'
6. y(t):
y = 2ty + t
7et²
8. y(t) =ty"-ty' y=0
3. y(t) = 3e 2t: y'
-2y
5. y(t) e³-2; y = 3y + 6
7. y(t) = (t8 – +¹)¹/4;
y'
2y¹ + t¹
ty³
9. y(t)
y" - 4y' | 4y
et + et;
p²
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