Explanation Given information: The equation is d y d t = 1 y Definition used: Isoclines is a way of eliminating a variable. Instead of calculating y ' , we set it equal to a constant then we solve for y. That is if the differential equation is d y d x = f ( x ) then set d y d x = k ⇒ f ( x ) = k . Calculation: Consider the given differential equation. d y d t = 1 y The isoclines are given by the equation 1 y = k y = 1 k y = C where C is a constant

4th Edition

ISBN: 9781319308858

Author: Rogawski

Publisher: MAC HIGHER

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Differential Equations

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Chapter 9.3, Problem 6E

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The isoclines of dydt=1y are horizontal lines and to sketch the slope field for −2≤t≤2,−2≤y≤2 and to plot the solutions with initial conditions y(0)=0 and y(0)=1.

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Chapter 9.3, Problem 5E

Chapter 9.3, Problem 7E

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CALCULUS:EARLY TRANS.(LL)-W/ACCESS

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The isoclines of d y d t = 1 y are horizontal lines and to sketch the slope field for − 2 ≤ t ≤ 2 , − 2 ≤ y ≤ 2 and to plot the solutions with initial conditions y ( 0 ) = 0 and y ( 0 ) = 1 .

Solution Summary: The author explains that the isoclines of dy dt are horizontal lines and plot the solutions with initial conditions.

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To show:

The isoclines of dydt=1y are horizontal lines and to sketch the slope field for −2≤t≤2,−2≤y≤2 and to plot the solutions with initial conditions y(0)=0 and y(0)=1.

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Chapter 9 Solutions

The isoclines of d y d t = 1 y are horizontal lines and to sketch the slope field for − 2 ≤ t ≤ 2 , − 2 ≤ y ≤ 2 and to plot the solutions with initial conditions y ( 0 ) = 0 and y ( 0 ) = 1 .

Solution Summary: The author explains that the isoclines of dy dt are horizontal lines and plot the solutions with initial conditions.

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To show: The isoclines of dydt=1y are horizontal lines and to sketch the slope field for −2≤t≤2,−2≤y≤2 and to plot the solutions with initial conditions y(0)=0 and y(0)=1.

Want to see the full answer?

Chapter 9 Solutions

To show:

The isoclines of dydt=1y are horizontal lines and to sketch the slope field for −2≤t≤2,−2≤y≤2 and to plot the solutions with initial conditions y(0)=0 and y(0)=1.