# Match the differential equations with the solution graphs labeled I–IV. Give reasons for your choices. (a) y ′ = 1 + x 2 + y 2 (b) y ′ = x e − x 2 − y 2 (a) y ′ = 1 1 + e x 2 + y 2 (a) y ′ = sin ( x y ) cos ( x y )

### Calculus (MindTap Course List)

8th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781285740621

Chapter
Section

### Calculus (MindTap Course List)

8th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781285740621
Chapter 9.1, Problem 13E
Textbook Problem
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## Match the differential equations with the solution graphs labeled I–IV. Give reasons for your choices.(a) y ′ = 1 + x 2 + y 2 (b) y ′ = x e − x 2 − y 2 (a) y ′ = 1 1 + e x 2 + y 2 (a) y ′ = sin ( x y ) cos ( x y )

To determine

To match:

The differential equation with the solution graphs and to give the reasons for choices.

### Explanation of Solution

1) Concept:

i. From the graph of function see the nature of function.

ii. P is increasing if dPdt>0 and decreasing if dPdt<0

2) Given:

The differential equation y'=1+x2+y2

And the solution graphs are,

3) Calculation:

The given differential equation is

y'=1+x2+y2

And the graph of functions is

Consider y'=1+x2+y2

Notice that, y'=1+x2+y21 for all values of x & y and as x increases y' is also increases, as x, y'

From the above graph see that only graph with labelled III is satisfactory with this condition.

Therefore, the differential equation y'=1+x2+y2 matches with the solution graph labelled  III, because it is reasonable with the condition, as x, y'

Conclusion:

The differential equation y'=1+x2+y2 matches with the solution graph labelled  III, because it is reasonable with the condition, as x, y'

To match:

The differential equation with the solution graphs and to give the reasons for choices.

Solution:

The differential equation y'=xe-x2-y2 matches with the solution graph labelled  I, because it is only graph reasonable with the condition, y'>0 for x>0 and y'<0 if  x<0

1) Concept:

i. From the graph of function see the nature of function.

ii. P is increasing if dPdt>0 and decreasing if dPdt<0

2) Given:

The differential equation y'=xe-x2-y2

And the solution graphs

3) Calculation:

The given differential equation is

y'=xe-x2-y2

And the graphs of functions are

Consider y'=xe-x2-y2

Notice that, y'=xe-x2-y2>0 for x>0 and y'=xe-x2-y2<0 if  x<0

From the above graph see that only function with negative tangent slope ( that is decreasing) when x<0 and positive tangent slope (increasing) when x>0 is function in graph labelled I

Therefore, the graph with labelled I is satisfactory with this condition.

Therefore, the differential equation y'=xe-x2-y2 matches with the solution graph labelled  I, because it is only graph reasonable with the condition, y'>0 for x>0 and y'<0 if  x<0

Conclusion:

The differential equation y'=xe-x2-y2 matches with the solution graph labelled  I, because it is only graph reasonable with the condition, y'>0 for x>0 and y'<0 if  x<0

To match:

The differential equation with the solution graphs and to give the reasons for choices.

Solution:

The differential equation y'=11+ex2+y2  matches with the solution graph labelled  IV, because it is only graph reasonable with the condition, y'>0 for x and y'0 as  x

1) Concept:

i. From the graph of function see the nature of function.

ii. P is increasing if  dPdt>0 and decreasing if  dPdt<0

2) Given:

The differential equation y'=11+ex2+y2

And the solution graphs are

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