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Chapter 3. Applications of First-Order Ordinary Differential Equations

Chapter 3. Applications of First-Order Ordinary Differential Equations

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120



Chapter 3 Applications of First-Order Equations



EXAMPLE 3.1.1: Use the definition of orthogonality to verify that the

1 x2 are orthogonal at the point

curves given by y

x and y

2/ 2, 2/ 2 .



SOLUTION: First note that the point

graph of y

given by y



x and y

1 and y



1

x/ 1



2/ 2, 2/ 2 lies on both the



x2 .



The derivatives of the functions are

x2 , respectively.



In[260]:= Clear x, y

y1 x



x



y2 x



1



x2



In[261]:= y1 x

Out[261]= 1

In[262]:= y2 x

x

Out[262]=

1 x2



Hence, the slope of the tangent line to y

stitution of x

tangent line at



2/ 2 into y

2/ 2, 2/ 2 .



In[263]:= y2

Out[263]=



x/ 1



x2



x at



2/ 2, 2/ 2 is 1. Sub-



yields 1 as the slope of the



2

2



1



Thus, the curves are orthogonal at the point



2/ 2, 2/ 2 because the



slopes of the lines tangent to the graphs of y

x and y

1 x2 at

the point

2/ 2, 2/ 2 are negative reciprocals. We graph these two

1 x2 at

2/ 2, 2/ 2 in

curves along with the tangent line to y

Figure 3-1 to illustrate that the two are orthogonal. Note that the graphs

are displayed correctly even though several error messages, which are

1 x2 is undefined

not all displayed here, are generated because y



3.1 Orthogonal Trajectories



121



1.5



1



0.5



-1.5



-1



-0.5



0.5



1



1.5



-0.5



-1



-1.5

Figure 3-1



The curves are orthogonal at the point



2/ 2, 2/ 2



if x < 1 or x > 1. The option AspectRatio->1 specifies that the ratio

of lengths of the x-axis to the y-axis in the resulting graphics object be 1.

3 3

,

,

2 2

3 3

3 3

PlotRange

,

,

,

,

2 2

2 2

AspectRatio 1, PlotStyle

Dashing

GrayLevel 0 , GrayLevel 0.5



In[264]:= Plot



x,



1



x2 , x



1 x2

Plot

plnr

is not a machine size real number at x



2 , x,



0.01



,



1.5.



The next step in our discussion of orthogonal curves is to determine the set of

orthogonal curves to a given family of curves. We refer to this set of orthogonal

curves as the family of orthogonal trajectories. Suppose that a family of curves is

defined as F x, y

C and that the slope of the tangent line at any point on these

curves is dy/ dx

f x, y . Then, the slope of the tangent line on the orthogonal

trajectory is dy/ dx

1/ f x, y so the family of orthogonal trajectories is found by

solving the first-order equation dy/ dx

1/ f x, y .



122



Chapter 3 Applications of First-Order Equations



EXAMPLE 3.1.2: Determine the family of orthogonal trajectories to the

family of curves y cx2 .



SOLUTION: First, we must find the slope of the tangent line at any

point on the parabola y cx2 . Differentiating with respect to x results in

dy/ dx 2cx. However from y cx2 , we have that c y/ x2 . Substitution

into dy/ dx 2cx then yields dy/ dx 2 y/ x2 x 2y/ x on the parabolas.

Hence, we must solve dy/ dx

x/ 2y to determine the orthogonal trajectories. This equation is separable, so we write it as 2y dy

x dx, and

2

2

x

k, where k is a constant,

then integrating both sides gives us 2y

which we recognize as a family of ellipses. Note that an equivalent

result is obtained with DSolve.

In[265]:= sol

Out[265]=



x

,y x ,x

2y x

4C 1

x2

, y x

2



DSolve y x



y x



x2



4C 1

2



We graph several members of the family of parabolas y cx2 , the family

k, and the two families of curves together. First,

of ellipses 2y2 x2

we define parabs to be the list of functions obtained by replacing c in

y cx2 by nine equally spaced values of c between 3/ 2 and 3/ 2.

In[266]:= parabs



Table c x2 , c,



3 3 1

, ,

2 2 8



Next, we graph the list of functions parabs for 3 x 3 with Plot

and name the result p1. The graphs in p1 are not displayed because the

option DisplayFunction->Identity is included in the Plot command. We graph several ellipses 2y2 x2 k by using ContourPlot to

graph several level curves of f x, y

y2 12 x2 and name the result p2.

Including the option PlotPoints->120 helps assure that the ellipses

appear smooth in the result. Including the option ContourStyle->

GrayLevel[.4] specifies that the contours be drawn in a light gray.

(This will help us distinguish between the ellipses and the parabolas

when we show the graphs together.) As with p1, p2 is not displayed.

Finally, p1 and p2 are displayed together with Show in Figure 3-2.

Notice that these two families appear orthogonal, confirming the results

we obtained.

In[267]:= p1



Plot Evaluate parabs , x, 3, 3 ,

DisplayFunction Identity



3.1 Orthogonal Trajectories



123



3



2



1



-3



-2



-1



1



2



3



-1



-2



-3

Figure 3-2 The two sets of curves are orthogonal to each other



In[268]:= p2



x2

, x, 3, 3 , y, 3 , 3 ,

2

Contours 30, ContourStyle GrayLevel 0.5 ,

ContourShading False, PlotPoints 120,

DisplayFunction Identity



ContourPlot y2



In[269]:= Show p1, p2, PlotRange

3 , 3 , 3, 3

AspectRatio 1,

DisplayFunction $DisplayFunction



,



EXAMPLE 3.1.3 (Temperature): Let T x, y represent the temperature at

the point x, y . The curves given by T x, y

c (where c is constant) are

called isotherms. The orthogonal trajectories are curves along which

heat will flow. Determine the isotherms if the curves of heat flow are

given by y2 2xy x2 c.



SOLUTION: We begin by finding the slope of the tangent line at each

point on the heat flow curves y2 2xy x2 c using implicit differentiation.



124



Chapter 3 Applications of First-Order Equations



In[270]:= eq1

Out[270]=



x



2



y x



2xy x



In[271]:= step1

Out[271]=



2



2x



x2



2x y x

y x



2



c



c



Dt eq1, x

2y x



2xy x



2y x y x



Dt c, x



Because c represents a constant, d/ dx c

0. We interpret step2 to be

equivalent to the equation 2yy 2y 2xy 2x 0, where y dy/ dx.

In[272]:= step2

Out[272]=



2x



step1/.Dt c, x

2y x



2xy x



0

2y x y x



0



We calculate y

dy/ dx by solving step2 for y’[x] with Solve and

name the result imderiv.

In[273]:= imderiv

Out[273]=



This equation is also

homogeneous of degree one.



y x



Solve step2, y x

x y x

x y x



Thus, dy/ dx

x y / x y so the orthogonal trajectories satisfy the

differential equation dy/ dx

x y/ x y.

Writing this equation in differential form as x y dx x y dy 0,

we see that this equation is exact because / y x y

1

/ x x y.

Thus, we solve the equation by integrating x y with respect to x to yield

1 2

xy g y . Differentiating f with respect to y then gives us

f x, y

2x

x g y . Then, because the equation is exact, x g y

x y.

fy x, y

1 2

y which implies that g y

y

.

This

means

that

Therefore, g y

2

the family of orthogonal trajectories (isotherms) is given by 12 x2 xy

1 2

k.

2y

Note that DSolve is able to solve this differential equation.

In[274]:= DSolve y x

Out[274]=



y x

y x



x

x



x y x

, y x , x //Simplify

x y x

2C 1

2C 1



2 x2 ,

2 x2



To graph y2 2xy x2 c and 12 x2 xy 12 y2 k for various values of

c and k to see that the curves are orthogonal, we use ContourPlot.

First, we graph several level curves of y2 2xy x2 c on the rectangle 4, 4

4, 4 and name the result cp1. The option Contours->40



3.1 Orthogonal Trajectories



125



instructs Mathematica to graph 40 contours instead of the default

of ten.

In[275]:= cp1



ContourPlot y2 2xy x2 , x, 4, 4 ,

y, 4, 4 , ContourShading False,

Axes Automatic, Contours 40,

PlotPoints 120, AxesOrigin

0, 0 ,

Frame False,

DisplayFunction Identity



Next we graph several level curves of 12 x2 xy 12 y2

k on the same

rectangle and name the result cp2. In this case, the option

ContourStyle->{{GrayLevel[0.4],Dashing[{0.01}]}}

specifies that the contours are to be dashed in a medium gray.

In[276]:= cp2



x2 y2

, x, 4, 4 ,

2

2

y, 4, 4 , ContourShading False,

Axes Automatic, Contours 40,

PlotPoints 120,

ContourStyle

GrayLevel 0.4 ,

Dashing 0.01

, AxesOrigin

0, 0 ,

Frame False,

DisplayFunction Identity



ContourPlot xy



The graphs are then displayed side-by-side using Show and Graphics

Array in Figure 3-3 and together using Show in Figure 3-4.



-4



4



4



2



2



-2



2



4



-4



-2



2



-2



-2



-4



-4



Figure 3-3



Several members of each family of curves



4



126



Chapter 3 Applications of First-Order Equations



4



2



-4



-2



2



4



-2



-4

Figure 3-4 The two sets of curves are orthogonal to each other



In[277]:= Show GraphicsArray

In[278]:= Show cp1, cp2,

DisplayFunction



cp1, cp2



$DisplayFunction



EXAMPLE 3.1.4: Determine the orthogonal trajectories of the family of

curves given by y2 2cx c2 . Graph several members of both families

of curves on the same set of axes.



SOLUTION: After defining eq to be the equation y2

implicitly differentiate.

In[279]:= eq



Out[279]=



yˆ2



2c x



2cx



c2 , we



cˆ2



step1 Dt eq

2 x Dt c

2 c Dt x



2 y Dt y



2 c Dt c



As in the previous examples, we interpret Dt[x] to be 1, Dt[c] to be

0, and Dt[y] to represent dy/ dx.



3.1 Orthogonal Trajectories



127



In[280]:= step2

Out[280]=



The equation y2



2c



step1 /. Dt c

2 y Dt y



0, Dt x



1



0



c2 is a quadratic in c. Solving for c,



2cx



In[281]:= cval



Solve eq, c



Out[281]=



x



c



x2



y2 , c



x



x2



y2



we choose to substitute the first value into the equation y dy/ dx

In[282]:= imderiv

Out[282]=



Solve step2, Dt y

x

x2 y2

y



Dt y



dy

dx



Then, we must solve



y

x2



x



y2



/.cval



c.



1



.



In[283]:= de

y x

Evaluate 1/imderiv

y y x

y x

Out[283]= y x

x

x2 y x 2



1, 1 , 2



/.



Note that Mathematica is able to solve this equation.

In[284]:= DSolve de, y x , x //Simplify

Out[284]=



y x

y x



C 1

2

C 1

2



C 1

C 1



2x ,

2x



Thus, y2 4C2 4Cx and replacing 2C with C yields y2 C2 2Cx or

y2 2Cx c2 , which means that this family of curves is self-orthogonal.

We confirm that the family is self-orthogonal with ContourPlot in

Figures 3-5 and 3-6.

In[285]:= cp1



ContourPlot cval 1, 1, 2 , x, 10, 10 ,

y, 10, 10 , ContourShading False,

Frame False, Axes Automatic,

AxesOrigin

0, 0 , Contours 30,

PlotPoints 120,

DisplayFunction Identity



In[286]:= cp2



ContourPlot cval 2, 1, 2 , x, 10, 10 ,

y, 10, 10 , ContourShading False,

Frame False, Axes Automatic,

AxesOrigin

0, 0 , Contours 30,

PlotPoints 120,

DisplayFunction Identity



128



Chapter 3 Applications of First-Order Equations



10



10



5



5



-10 -5



5



-10 -5



10



5



-5



-5



-10



-10



Figure 3-5



The plots are symmetric about the y-axis



10



5



-10



-5



5



10



-5



-10

Figure 3-6 The family of curves is self-orthogonal



In[287]:= Show GraphicsArray



In[288]:= Show cp1, cp2,

DisplayFunction



cp1, cp2



$DisplayFunction



10



3.1 Orthogonal Trajectories



129



Application: Oblique Trajectories

If we are given a family of curves that satisfies the differential equation dy/ dx

f x, y and we want to find a family of curves that intersects this family at a constant angle Θ, we must solve the differential equation

f x, y tan Θ

.

1 f x, y tan Θ



dy

dx



For example, to find a family of curves that intersects the family of curves x2 y2

c2 at an angle of Π/ 6, we first implicitly differentiate the equation to obtain

2x

Because tan Θ



tan Π/ 6



2y



dy

dx



dy

dx



0



x

y



f x, y .



1/ 3, we solve

dy

dx



x/ y

1



x 3



1/ 3



x/ y 1/ 3



y 3



y

x



,



which is a first-order homogeneous equation. With the substitution x

obtain the separable equation

1 v 3

dv

1 v2



3

dy.

y



Integrating yields

3

ln 1

2

so

3

ln 1

2



v2



tan



1



v



3 ln y



k1



x2

y2



tan



1



x

y



3 ln y



k1 .



Mathematica finds an equivalent implicit solution.

In[289]:= sol1



3x



DSolve y x



y x



3y x



, y x , x //



x



FullSimplify

y x

Out[289]= Solve 2 ArcTan

x

3 2 Log x



Log 1



y x

x2



2



2 C 1 ,y x



Similarly, for

dy

dx



x/ y

1



1/ 3



x/ y 1/ 3



x 3

y 3



y

x



,



vy, we



130



Chapter 3 Applications of First-Order Equations



we obtain

1 v 3

dv

1 v2



3

dy

y



so that the trajectories are

3

ln 1

2

In[290]:= sol2



x2

y2



tan



1



x

y



3 ln y



3x



DSolve y x



y x



3y x



Out[290]= Solve

C 1



y x

1

x

2

3 Log x , y x



ArcTan



k1 .



,y x ,x



x



3 Log 1



y x

x2



2



To confirm the result graphically, we graph several members of each family of

curves in Figure 3-7

In[291]:= toplot2

Out[291]=



Solve sol2 1 , C 1

1

y

C 1

2 ArcTan

2

x

y2

3 Log 1

2 3 Log x

x2



In[292]:= cp1



/.y x



y



ContourPlot x2 y2 , x, 10, 10 ,

y, 10, 10 , Frame False, Contours 20,

ContourStyle GrayLevel 0.5 ,

Axes Automatic, ContourShading False,

PlotPoints 120, AxesOrigin

0, 0 ,

DisplayFunction Identity



10



10



10



5



5



5



-10 -5

-5



5 10 -10 -5

-5



5 10 -10 -5

-5



-10



-10



-10



Figure 3-7



Several members of each family of curves



5 10



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