The solution of the differential equation y ″ + 3 y ′ + 2 y = u 2 ( t ) with initial condition y ( 0 ) = 0 and y ′ ( 0 ) = 1 .

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Answer 1

Question

Chapter 6.4, Problem 6P

(a)

To determine

The solution of the differential equation y″+3y′+2y=u2(t) with initial condition y(0)=0 and y′(0)=1.

(b)

To determine

To sketch: The solution of the differential equation solution of the differential equation (e−t−e−2t)+12u2(t)(1−2e−(t−2)+e−2(t−2)) with forcing function and explain the relation between them.

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Answer 2

Question

Chapter 6.4, Problem 6P

(a)

To determine

The solution of the differential equation y″+3y′+2y=u2(t) with initial condition y(0)=0 and y′(0)=1.

(b)

To determine

To sketch: The solution of the differential equation solution of the differential equation (e−t−e−2t)+12u2(t)(1−2e−(t−2)+e−2(t−2)) with forcing function and explain the relation between them.

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Answer 3

Question

Chapter 6.4, Problem 6P

(a)

To determine

The solution of the differential equation y″+3y′+2y=u2(t) with initial condition y(0)=0 and y′(0)=1.

(b)

To determine

To sketch: The solution of the differential equation solution of the differential equation (e−t−e−2t)+12u2(t)(1−2e−(t−2)+e−2(t−2)) with forcing function and explain the relation between them.

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Answer 4

Question

Chapter 6.4, Problem 6P

(a)

To determine

The solution of the differential equation y″+3y′+2y=u2(t) with initial condition y(0)=0 and y′(0)=1.

(b)

To determine

To sketch: The solution of the differential equation solution of the differential equation (e−t−e−2t)+12u2(t)(1−2e−(t−2)+e−2(t−2)) with forcing function and explain the relation between them.

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Answer 5

Chapter 6.4, Problem 6P

(a)

To determine

The solution of the differential equation y″+3y′+2y=u2(t) with initial condition y(0)=0 and y′(0)=1.

(b)

To determine

To sketch: The solution of the differential equation solution of the differential equation (e−t−e−2t)+12u2(t)(1−2e−(t−2)+e−2(t−2)) with forcing function and explain the relation between them.

Answer 6

Chapter 6.4, Problem 6P

(a)

To determine

The solution of the differential equation y″+3y′+2y=u2(t) with initial condition y(0)=0 and y′(0)=1.

Answer 7

Chapter 6.4, Problem 6P

(a)

To determine

The solution of the differential equation y″+3y′+2y=u2(t) with initial condition y(0)=0 and y′(0)=1.

Answer 8

(b)

To determine

To sketch: The solution of the differential equation solution of the differential equation (e−t−e−2t)+12u2(t)(1−2e−(t−2)+e−2(t−2)) with forcing function and explain the relation between them.

Answer 9

(b)

To determine

To sketch: The solution of the differential equation solution of the differential equation (e−t−e−2t)+12u2(t)(1−2e−(t−2)+e−2(t−2)) with forcing function and explain the relation between them.

Answer 10

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Answer 11

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Answer 12

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Answer 13

Ch. 6.1 - In each of Problems 1 through 4, sketch the graph...Ch. 6.1 - In each of Problems 1 through 4, sketch the graph...Ch. 6.1 - In each of Problems 1 through 4, sketch the graph...Ch. 6.1 - Prob. 4P Ch. 6.1 - Find the Laplace transform of each of the...Ch. 6.1 - Find the Laplace transform of f (t) = cos at,...Ch. 6.1 - Recall that cosh bt = (ebt + e−bt)/2 and sinh bt =...Ch. 6.1 - Prob. 8P Ch. 6.1 - Prob. 9P Ch. 6.1 - Recall that cosh bt = (ebt + e−bt)/2 and sinh bt =...

The solution of the differential equation y ″ + 3 y ′ + 2 y = u 2 ( t ) with initial condition y ( 0 ) = 0 and y ′ ( 0 ) = 1 .

The solution of the differential equation y″+3y′+2y=u2(t) with initial condition y(0)=0 and y′(0)=1.

To sketch: The solution of the differential equation solution of the differential equation (e−t−e−2t)+12u2(t)(1−2e−(t−2)+e−2(t−2)) with forcing function and explain the relation between them.

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