11.1.4 Direction Fields

Question 1

Direction Fields

Answer 1

• Given a first-order differential equation, you can use direction fields to analyze the behavior of its solution curves.
• The direction field illustrates the slopes of the lines tangent to the various solution curves. From this information you can deduce the shapes of particular solutions.

Question 2

note

Answer 2

• This differential equation is harder to solve because it is not separable. However, you can plot it.
• It only involves a first derivative, which represents slope. At any point (x, y) on the graph, you can determine the slope of the line tangent to one of the solution curves.
• Make a table of x- and y-values. Then calculate dy/dx for each ordered pair. Draw a hash mark at each point tilted to correspond to the slope.
• The result is a direction field. You can see patterns of
  movement in the direction field. They correspond to the
  solution curves.
• Here is a separable equation. You can generate a direction field for it, too.
• Consider the point (2, 1). Plug x = 2 and y = 1 into the
  differential equation to find the slope of the tangent line at that point. The circled mark has a slope of one.
• To draw a solution curve, first choose a starting point. Use the hash marks to determine what the curve does to the right of the point and to the left of the point.
• The direction field indicates that the solution curves will be hyperbolas. These curves are just a few examples.
• First-order differential equations are like “slope functions.” Given x and y, the equation returns the slope at that point.
• To draw this solution curve, start at the origin. On the right side of the origin, the marks are tilted down. The graph must follow them. On the left side, the marks indicate that the graph must curve to the upper left.
• Remember, there is a solution curve for each value of the constant C.

Question 3

note 2

Answer 3

• Euler suggested a numerical method to approximate a particular solution to a differential equation (dy/dx = F(x, y)). To use this method, begin with a known point (x 0 , y 0 ) on the particular solution. Extrapolate along the tangent to the curve at that point a short distance h in the x-direction (and a corresponding distance in the y-direction). The coordinates of this new point (x 1 , y 1 ) are (x 0 + h, y 0 + hF(x 0 , y 0 )). The actual values of this new point can then be used to advance the extrapolation another interval h to the next point, and so
  on. A smaller step interval (h) improves the accuracy of the approximation.
• The graph shows a hypothetical example with the result of the approximations next to the actual solution to the differential equation.
• Here is a worked example of the method using the equation dy/dx = 3x/(5y + 1).
• Consider the curve starting from the known point (0.5, 0.5). Beginning here, develop a table of values for each of the approximated points. The first entry is worked out in detail.
• The graph shows the curve of the actual solution and the plotted points of the approximation.
• Here is an example of Euler’s method using the equation dy/dx = x + 2y. This equation cannot be solved by separation.
• The solution starts at the point (0, 2) and proceeds in steps of h = 0.1. The approximate coordinates appear in the table. These are plotted in the graph.
• Notice the congruence of the solution using Euler’s method with the solution using the direction field method.

Question 4

A solution curve y=y(x) of dydx=sinx−y passes through (3,2). At this point the function y=y(x)____________.

Answer 4

is decreasing

Question 5

Which differential equation produces a series of parallel lines as its solution curve?

Answer 5

dy/dx=5

Question 6

A solution curve of dy/dx=xyx2+y2passes through (5,5). At this point,what is the equation of the line tangent to this solution curve?

Answer 6

y−5=1/2(x−5)

Question 7

A solution curve y=y(x) of dy/dx=ex+ey passes through (−1,−2). At this point the function y=y(x)____________.

Answer 7

is increasing

Question 8

A solution curve of dy/dx=x2−y2passes through (−1,4). At this point,what is the equation of the line tangent to this solution curve?

Answer 8

y − 4 = −15 (x + 1)

Question 9

A solution curve y=y(x) of dydx=x2−y+1 passes through (3,2). At this point the function y=y(x)____________.

Answer 9

is increasing.

Question 10

A solution curve of dy/dx=xy+ex passes through (ln2,3). At this point,what is the equation of the line tangent to this solution curve?

Answer 10

y − 3 = (3 ln 2 + 2) (x − ln 2)

Question 11

A solution curve of dydx=x+ypasses through (2,3). At this point,what is the slope of the line tangent to this solution curve?

Answer 11

5

Question 12

A solution curve of dydx=x2y passes through (3,6). At this point,what is the slope of the line tangent to this solution curve?

Answer 12

54

Question 13

A solution curve y=y(x) of dydx=2x2−3y2 passes through (2,3). At this point the function y=y(x)____________.

Answer 13

is decreasing.

Question 14

Which differential equation produces a series of parallel lines as its solution curve?

Answer 14

dydx=2

Question 15

A solution curve of dydx=ycosxpasses through (π,2). At this point,what is the equation of the line tangent to this solution curve?

Answer 15

y − 2 = −2 (x − π )