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