(b) Draw the graphs of the solution and of the forcing function; explain how they are related. [1, 0 27 3. y" +4y = sint - U2n(1) sin(t – 27); y(0) = 0, y'(0) = 0 %3D 24. y"+4y = sin t + u, (t) sin(t – 1); y(0) = 0, y'(0) = 0 |1, 0st< 10 f(1) = |0, t> 10 2 5. y" + 3y' + 2y = f(t); y(0) = 0, y'(0) = 0; 6. y" +3y' + 2y = u2(t); 2 7. y" +y= U3,(1); 2 8. y" +y' + y =t- uz/2(1)(t – 1/2); y(0) = 0, y'(0) = 1 y(0) = 1, y'(0) = 0 %3D y(0) = 0, y'(0) = 0 S1/2, 0 6 sin t, 0st< g(t) = 0, 10. y" + y' + y = g(1); y(0) = 0, y'(0) = 0; %3D 11. y" + 4y = u, () – u3, (1); y(0) = 0, y(0) = 0 %3D
(b) Draw the graphs of the solution and of the forcing function; explain how they are related. [1, 0 27 3. y" +4y = sint - U2n(1) sin(t – 27); y(0) = 0, y'(0) = 0 %3D 24. y"+4y = sin t + u, (t) sin(t – 1); y(0) = 0, y'(0) = 0 |1, 0st< 10 f(1) = |0, t> 10 2 5. y" + 3y' + 2y = f(t); y(0) = 0, y'(0) = 0; 6. y" +3y' + 2y = u2(t); 2 7. y" +y= U3,(1); 2 8. y" +y' + y =t- uz/2(1)(t – 1/2); y(0) = 0, y'(0) = 1 y(0) = 1, y'(0) = 0 %3D y(0) = 0, y'(0) = 0 S1/2, 0 6 sin t, 0st< g(t) = 0, 10. y" + y' + y = g(1); y(0) = 0, y'(0) = 0; %3D 11. y" + 4y = u, () – u3, (1); y(0) = 0, y(0) = 0 %3D
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Number 2 please and use laplace transforms to solve
Transcribed Image Text:PROBLEMS
In each of Problems 1 through 13:
(a) Find the solution of the given initial value problem.
(b) Draw the graphs of the solution and of the forcing function; explain how they are related.
2 1. y" +y= f(1);
1, 0st < 3n
|0, 37 <t < ∞
y(0) = 0, y'(0) = 1;
f(t) =
1, n<t < 2n
0, 0<t<n and t> 27
2 2. y" + 2y' + 2y = h(t);
y(0) = 0, y'(0) = 1;
h(t) =
2 3. y" + 4y = sint – u2, (t) sin(t – 27);
y(0) = 0, y'(0) = 0
y" + 4y = sin t + u„(t) sin(t – n);
y(0) = 0, y'(0) = 0
1, 0<t< 10
f(1t) =
|0, t> 10
25. y" +3y' + 2y = f (t);
y(0) = 0, y'(0) = 0;
2 6. y" +3y' + 2y = u2(t);
7. у" + у%3D из" ();
2 8. y" +y' + y =t – un/2(t)(t – 1/2);
У (0) %3 0, у(0) %3D 1
y(0) = 1, y'(0) = 0
y(0) = 0, y'(0) = 0
t/2, 0<t < 6
g(t) =
| 3,
2 9. y" +y = g(t);
y(0) = 0, y'(0) = 1;
t > 6
sin t, 0<t < A
|0,
10. y" +y' + y = g(t);
У0) 3 0, у(0) - 0;B
g(t) =
11. y" + 4y = u(t) – U3m (t);
y(0) = 0, y'(0) = 0
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