Explanation Given information: The slope field for d y d t = sin y sin t is given as follows: Calculation: Given differential equation can be solved as follows: d y d t = sin y sin t d y sin y = sin t d t ∫ d y sin y = ∫ sin t d t ln | csc y − cot y | = − cos t + C Now apply the initial conditions y ( 0 ) = 1 and y ( 0 ) = − 1 as shown. ln | csc ( 1 ) − cot ( 1 ) | = − cos 0 + C C = 0

4th Edition

ISBN: 9781319055905

Author: FRANZOSA

Publisher: VST

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Chapter 9.3, Problem 1E

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The graph of the solution with initial conditions y(0)=1 and y(0)=−1, to show y(t)=0 is a solution and add it to the graph to the plot.

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Chapter 9.3, Problem 4PQ

Chapter 9.3, Problem 2E

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Chapter 9 Solutions

EBK CALCULUS: EARLY TRANSCENDENTALS

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The graph of the solution with initial conditions y ( 0 ) = 1 and y ( 0 ) = − 1 , to show y ( t ) = 0 is a solution and add it to the graph to the plot.

Solution Summary: The author illustrates the graph of the solution with initial conditions y(0)=1 and (t)=0.

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To sketch:

The graph of the solution with initial conditions y(0)=1 and y(0)=−1, to show y(t)=0 is a solution and add it to the graph to the plot.

Want to see the full answer?

Chapter 9 Solutions

The graph of the solution with initial conditions y ( 0 ) = 1 and y ( 0 ) = − 1 , to show y ( t ) = 0 is a solution and add it to the graph to the plot.

Solution Summary: The author illustrates the graph of the solution with initial conditions y(0)=1 and (t)=0.

Concept explainers

Videos

To sketch: The graph of the solution with initial conditions y(0)=1 and y(0)=−1, to show y(t)=0 is a solution and add it to the graph to the plot.

Want to see the full answer?

Chapter 9 Solutions

To sketch:

The graph of the solution with initial conditions y(0)=1 and y(0)=−1, to show y(t)=0 is a solution and add it to the graph to the plot.