Chapter 2. First-Order Ordinary Differential Equations
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Chapter 2 First-Order Ordinary Differential Equations
containing the point x0 , y0 , there exists an interval x x0 < h centered at x0 on which
there exists one and only one solution to the differential equation that satisﬁes the initial
condition.
Often, we can use the command
DSolve[{y’[x]==f[x,y[x]],y[x0]==y0},y[x],x]
to solve the initial-value problem (2.1);
DSolve[y’[x]==f[x,y[x]],y[x],x]
f x, y .
attempts to ﬁnd a general solution of y
EXAMPLE 2.1.1: Solve the initial-value problem
dy/ dx
y 0
x/ y
0.
Does this result contradict the Existence and Uniqueness Theorem?
SOLUTION: This equation is solved with DSolve to determine the
family of solutions y2 x2 C.
In[74]:= Clear x, y
DSolve y x
Out[74]=
y x
x
,y x ,x
y x
x2
2C 1
, y x
x2
2C 1
We note that the graph of y2 x2
C for various values of C is the
same as the graph of the level curves of f x, y
y2 x2 . Members of
this family corresponding to C
40, 38, . . . , 38, 40 are graphed with
ContourPlot in Figure 2-1.
In[75]:= cvals
Table i, i, 40, 40, 2
In[76]:= ContourPlot y2 x2 , x, 6, 6 , y, 6, 6 ,
PlotPoints 120, Contours > cvals,
ContourShading False
Application of the initial condition yields 02 02 C, so C 0. There0, so there
fore, solutions that pass through 0, 0 , satisfy y2 x2
2.1 Theory of First-Order Equations: A Brief Discussion
43
6
4
2
0
-2
-4
-6
-6
-4
-2
Figure 2-1 Plot of f x, y
0
2
4
6
C for various values of C
are two solutions, y x and y
x, that satisfy the differential equation and the initial condition. Although more than one solution satisﬁes
this initial-value problem, the Existence and Uniqueness Theorem is not
contradicted because the function f x, y
x/ y is not continuous at the
point 0, 0 ; the requirements of the theorem are not met.
EXAMPLE 2.1.2: Verify that the initial-value problem dy/ dx
1 has a unique solution.
y, y 0
SOLUTION: In this case, f x, y
y, x0
0, and y0
1. Hence, both
f and f / y 1 are continuous on all rectangular regions containing
0, 1 . Therefore by the Existence and Uniqueness
the point x0 , y0
Theorem, there exists a unique solution to the differential equation that
satisﬁes the initial condition y 0
1.
We can verify this by solving the initial-value problem. The unique
solution is y ex , which is computed with DSolve and then graphed
44
Chapter 2 First-Order Ordinary Differential Equations
2.5
2
1.5
1
0.5
-1
-0.5
0.5
Figure 2-2 Plot of y
1
ex
with Plot in Figure 2-2. Notice that the graph passes through the point
0, 1 , as required by the initial condition.
In[77]:= Clear x, y, sol
Out[77]=
sol DSolve
x
y x
y x
y x ,y 0
1 ,y x ,x
In[78]:= Plot y x /. sol, x, 1, 1
EXAMPLE 2.1.3: Show that the initial-value problem
dy
dx
y 0
x
y
x2 cos x
0
has inﬁnitely many solutions.
SOLUTION: Writing xy
y
dy
dx
x2 cos x in the form y
x2 cos x
x
f x, y results in
y
and because f x, y
x2 cos x y / x is not continuous on an interval
containing x 0, the Existence and Uniqueness Theorem does not guarantee the existence or uniqueness of a solution. In fact, using DSolve
we see that a general solution of the equation is y x sin x Cx and for
every value of C, y 0
0.
2.1 Theory of First-Order Equations: A Brief Discussion
45
10
7.5
5
2.5
-10
-5
5
10
-2.5
-5
-7.5
-10
Figure 2-3
Every solution satisﬁes y 0
0
In[79]:= Clear y
Out[79]=
y x
sol DSolve x y x
y x
xC 1
x Sin x
xˆ2 Cos x , y x , x
We conﬁrm this graphically by graphing several solutions. First, we use
Table to deﬁne toplot to be a set of functions obtained by replacing
the arbitrary constant in y x by 4, 3, . . . , 3, 4.
In[80]:= toplot Table sol 1, 1, 2 /.C 1
i, 4, 4
Out[80]=
4 x x Sin x , 3 x x Sin x , 2 x
x x Sin x , x Sin x , x x Sin x
2 x x Sin x , 3 x x Sin x , 4 x x
> i,
x Sin x ,
,
Sin x
These functions are then graphed with Plot in Figure 2-3.
In[81]:= Plot Evaluate toplot , x, 10, 10 ,
PlotRange > 10, 10 , AspectRatio > 1
46
Chapter 2 First-Order Ordinary Differential Equations
2.2 Separation of Variables
Deﬁnition 5 (Separable Differential Equation). A differential equation that can be
f x or g y dy
f x dx is called a separable differential
written in the form g y y
equation.
Separable differential equations are solved by collecting all the terms involving
y on one side of the equation, all the terms involving x on the other side of the
equation, and integrating:
g y dy
f x dx
g y dy
f x dx
C,
where C is a constant.
EXAMPLE 2.2.1: Show that the equation
dy
dx
2 y 2y
x
is separable, and solve by separation of variables.
SOLUTION: The equation y
be written in the form
2 y
1
dy
2 y 2y
2y / x is separable because it can
1
dx.
x
To solve the equation, we integrate both sides and simplify. Observe
that we can write this equation as
1
1
dy
2 y1
y
1
dx
x
C.
To evaluate the integral on the left-hand side, let u
1
dy. We then obtain
2 y
1
du
u
so that
ln u
ln x
1
dx
x
C1 . Recall that
ln
1
u
ln x
ln u
C1 .
1
y so du
C1
ln u
1
, so we have
2.2 Separation of Variables
47
Using Mathematica, we have
1
y
2 y 2y
In[82]:=
2
1
y
2
Out[82]=
y Log
y 2y
1
y
which we then simplify with Simplify.
2
In[83]:= Simplify
Out[83]=
Log
y
1
y Log
2
1
y
1
y
2y
Note that Log[x]
represents the natural
logarithm function,
y ln x.
y
The integral on the right-hand side of the equation is computed in the
same way.
1
x
x
Out[84]= Log x
In[84]:=
Simpliﬁcation yields
1
u
where C2
eln x
C1
C2 x
eC1 . Resubstituting we ﬁnd that
1
1
y
or
C2 x
x
C3
1
y
.
Solving for y shows us that
y
1
y
y
C
x
x C
x
x C
x
2
is a general solution of the equation y
same results with Mathematica,
In[85]:= Solve
Log
1
y
cons, y //Simplify
2 cons
cons
1
x
y
Out[85]=
2
x
2 y
2y / x. We obtain the
Log x
2
where E cons represents the arbitrary constant in the solution. We obtain
an equivalent result with DSolve. Entering
We use cons to represent
the arbitrary constant C to
avoid ambiguity with the
built-in symbol C.
48
Chapter 2 First-Order Ordinary Differential Equations
In[86]:= Clear x, y
y x
2
DSolve y x
gensol
2y x
,
x
y x ,x
Solve
ifun
Inverse functions are being used by Solve, so some solutions may not be
found.
2
C 1
2
Out[86]=
y x
x
x2
ﬁnds a general solution of the equation which is equivalent to the one
we obtained by hand and names the result gensol. The formula for
the solution, which is the second part of the ﬁrst part of the ﬁrst part
of gensol, is extracted from gensol with gensol[[1,1,2]]. Alternatively, if you are using Version 5, you can select, copy, and paste the
result to any location in the notebook.
In[87]:= gensol 1, 1, 2 //Simplify
2
C 1
2
Out[87]=
x
x2
To graph the solution for various values of C[1], which represents
the arbitrary constant in the formula for the solution, we use Table
together with ReplaceAll (/.) to generate a set of functions obtained
by replacing C[1] in the formula for the solution by i for i
3, 2.50,
. . . , 2.50, and 3, naming the resulting set of functions toplot. We view
an abbreviation of toplot with Short.
In[88]:= toplot Table gensol
i , i , 3, 3, 1/ 2
1, 1 , 2
/.C 1
Short toplot, 4
1
Out[88]=
2
x
3/2
1
,
x2
x
5/4
2
1
,
x2
x
2
1
,
x2
x
3/4
2
,
x2
2
1
x
1
,
x2
x
x2
x
2
x2
2
,
x2
3/ 4
,
x2
3/ 2
2
x
1/4
x
1
x
x2
2
,
x
x2
1/ 4
2
,
,
x2
2
,
x2
5/ 4
2
x
x
2
,
2.2 Separation of Variables
49
10
8
6
4
2
-4
Figure 2-4
-2
Various solutions of y
4
2
2 y
2y / x
We then graph the set of functions toplot with Plot in Figure 2-4.
In[89]:= Plot Evaluate toplot , x, 5, 5 ,
PlotRange > 0, 10 , AspectRatio
1
An initial-value problem involving a separable equation is solved through the
following steps.
1. Find a general solution of the differential equation using separation of
variables.
2. Use the initial condition to determine the unknown constant in the general
solution.
EXAMPLE 2.2.2: Solve (a) y cos x dx 1 y2 dy 0 and (b) the initial1.
value problem y cos x dx 1 y2 dy 0, y 0
SOLUTION: (a) Note that this equation can be rewritten as dy/ dx
y cos x / 1 y2 . We ﬁrst use DSolve to solve the equation.
50
Chapter 2 First-Order Ordinary Differential Equations
In[90]:= sol1 DSolve y x
y x ,x
y x Cos x / 1
y x ˆ2 ,
InverseFunction
ifun
Inverse functions are being used. Values may be lost for multivalued
inverses.
Solve
ifun
Inverse functions are being used by Solve, so some solutions may not be
found.
Out[90]=
y x
ProductLog
y x
ProductLog
2C 1
2C 1
2 Sin x
,
2 Sin x
In this case, we see that DSolve is able to solve the nonlinear equation,
although the result contains the ProductLog function. Given z, the
Product Log function returns the principal value of w that satisﬁes z
wew . A more familiar form of the solution is found using traditional
techniques. Separating and integrating gives us
y2
1
dy
cos x dx
1
y
y dy
cos x dx
ln y
1 2
y
2
y
sin x
C.
We can also use Mathematica to implement the steps necessary to solve
the equation by hand. To solve the equation, we must integrate both
the left and right-hand sides which we do with Integrate, naming
the resulting output lhs and rhs, respectively.
In[91]:= lhs
rhs
Integrate
1
yˆ2 /y, y
Integrate Cos x , x
2
y
Log y
2
Out[91]= Sin x
Out[91]=
1 2
Therefore, a general solution to the equation is ln y
sin x C.
2y
1 2
sin x C in Figure 2-5
We now use ContourPlot to graph ln y 2 y
for various values of C by observing that the level curves of f x, y
1 2
1 2
sin x correspond to the graph of ln y
sin x C for
ln y
2y
2y
various values of C.
2.2 Separation of Variables
51
10
8
6
4
2
0
0
4
2
1 2
y
2
Figure 2-5 Plot of ln y
6
sin x
8
10
C for various values of C
In[92]:= ContourPlot lhs rhs, x, 0, 10 , y, 0, 10 ,
Contours > 30, PlotPoints > 150,
ContourShading > False
By substituting y 0
1 into this equation, we ﬁnd that C
implicit solution is given by ln y 12 y2 sin x 1/ 2.
1/ 2, so the
In[93]:= Clear x, y, c
In[94]:= Solve Evaluate lhs
x > 0, y > 1 , c
1
Out[94]=
c
2
rhs
c /.
We can also use DSolve to solve the initial value problem as well. The
solution is then graphed in Figure 2-6 with Plot.
In[95]:= sol2
Out[95]=
y x
DSolve
y 0
y x
y x Cos x / 1
1 ,y x ,x
ProductLog
1 2 Sin x
In[96]:= Plot y x /.sol2, x, 0, 10
y x ˆ2 ,
52
Chapter 2 First-Order Ordinary Differential Equations
1.4
1.2
4
2
6
8
10
0.8
0.6
Figure 2-6 Plot of the solution that satisﬁes y 0
1
EXAMPLE 2.2.3: Solve each of the following equations: (a) y
0; (b) y Αy 1 K1 y , K, Α > 0 constant.
y2 sin t
SOLUTION: (a) The equation is separable:
1
dy
y2
1
dy
y2
1
y
y
sin t dt
sin t dt
cos t
C
1
.
cos t C
We check our result with DSolve.
In[97]:= sola
Out[97]=
DSolve y t
y t ˆ2 Sin t
1
y t
C 1
Cos t
0, y t , t
Observe that the result is given as a list. The formula for the solution is
the second part of the ﬁrst part of the ﬁrst part of sola.
In[98]:= sola
Out[98]=
C 1
1, 1, 2
1
Cos t
We then graph the solution for various values of C with Plot in
Figure 2-7.