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Differential
Chapter
15
equations
15.1 Basic concepts, separable and
homogeneous equations
15.2 First-order linear equations
15.3 Exact equations
15.4 Strategy for solving first-order equations
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15.5 Second-Order Linear Equations
A second-order linear differential equation has the form
2
d
y
dy
(1)
P( x) 2  Q( x)  R( x) y  G( x)
dx
dx
where P, Q, R, and G are continuous functions.
If G(x) = 0 for all x, such equations are called secondorder homogeneous linear equations. (This use of the
word homogeneous has nothing to do with the meaning
given in Section 15.1.)
(2)
d2y
dy
P ( x ) 2  Q( x )  R ( x ) y  0
dx
dx
If
G ( x)  0
for some x, Equation 1 is
nonhomogeneous.
Two basic facts enable us to solve homogeneous linear
equations. The first of these says that if we know two
solutions y1 and y 2 of such an equation, then the linear
combination 2y 2c  1y 1c  y
is also a solution.
(3)Theorem If y1 ( x ) and y 2 ( x ) are both solutions of
the linear equation (2) and c1 and c2 are any
constants, then the function )x ( 2y 2c  )x ( 1y 1c  )x (y
is also a solution of Equation 2.
Proof
have
Since y1 and y 2 are solutions of Equation 2, we
0  1y )x (R  1y )x (Q  1y )x (P
and
0  2y )x (R  2y )x (Q  2y )x (P
Therefore
y )x (R  y )x (Q  y )x (P
) 2y 2c  1y 1c()x (R  ) 2y 2c  1y 1c()x (Q  ) 2y 2c  1y 1c()x (P 
) 2y 2c  1y 1c()x (R  ) 2y 2c  1y 1c()x (Q  )2y 2c  1y 1c()x (P 
] 2y )x (R  2y )x (Q  2y )x (P [ 2c  ] 1y )x (R  1y )x (Q  1y )x (P [1c 
0  2c  0  1c 
0
Thus
2 y 2c  1y 1c 
y
is a solution of Equation 2.
Let x and y are two variables, if neither x nor y is a
constant multiple of the other, we say x and y are two
linearly independent variables. For instance, the function
f ( x)  x 2 and
g ( x)  5 x 2
are linearly dependent, but
f ( x)  e x and g ( x )  xe x are linearly independent.
The second theorem says that the general solution of a
homogeneous linear equation is a linear combination of
two linearly independent solutions.
(4)Theorem If y1 and y 2 are linearly independent
solutions of Equation 2 , then the general solution is
given by
)x ( 2y 2c  )x ( 1y 1c  )x ( y
where c1
and c2 are arbitrary constants.
In general, it is not easy to discover particular
solutions to a second-order linear equation. But it is always
possible to do so if the coefficient functions P, Q and R are
constant functions, that is, if the differential equation has
the form
(5)
0  yc   yb  ya
It is not hard to think of some likely candidates for
particular solutions of Equation 5. For example, the
exponential function y  erx because its derivatives are
rx
2 rx
constants multiple of itself: y   re , y   r e
.
Substitute these expression into Equation 5
ar 2 e rx  brerx  cerx  0
(ar 2  br  c)e rx  0
Notice erx is never 0 so y  e rx is a solution of
Equation 5 if r is a root of the equation
(6)
ar 2  br  c  0
which is called the auxiliary equation(or characteristic
equation) of Equation 5. Using the quadratic formula, the
root r1 and r2 of the auxiliary equation can be found:
(7)
 b  b 2  4ac
r1 
2a
 b  b 2  4ac
r2 
2a
We distinguish three cases according to the sign of
the discriminant b 2  4ac .
Case 1
b 2  4ac  0
In this case the roots r1 and r2 of the auxiliary
equation are real and distinct, so y1  e r1 x and y 2  e r2 x
are two linearly independent solutions of Equation 5.
(8) If the roots r1 and r2 of the auxiliary equation
` ar 2  br  c  0 are real and unequal, then the
general solution of ay   by   cy  0 is
y  c1e r1 x  c2 e r2 x
Example 1 Solve the equation
y   y   6 y  0
d 2 y dy
Example 2 Solve the equation 3 2 
 y  0.
dx
dx
Case 2
b 2  4ac  0
In this case r1  r2 ; that is, the root of the auxiliary
equation are real and equal. Denote r as the common value
of r1 and r2 , we have
(9)
b
r
2a
so
2ar  b  0
We know that y1  e rx is one solution of Equation 5. We
now verify that y2  xerx is also a solution:
ay2  by2  cy2
 a ( 2re rx  r 2 xerx )  b(e rx  rxerx )  cxerx
 ( 2ar  b)e rx  ( ar 2  br  c) xerx
 0(e rx )  0( xerx )  0
Since y1  e and y2  xe are linearly
independent solutions, Theorem 4 provides us with the
general solution:
rx
rx
(10) If the auxiliary equation ar 2  br  c  0
has only one real root r, then the general solution of
` ay   by   cy  0 is
y  c1e rx  c2 xerx  (c1  c2 x)e rx
Example 3 Solve the equation 4 y  12 y  9 y  0
Case 3
b 2  4ac  0
In this case the roots r1 and r2 of the auxiliary
equation are complex numbers, we can write
r1    i
r2    i
where    b (2a ),
  4ac  b 2 (2a)
Using Euler’s equation
ei  cos  i sin 
we write the solution of the differential equation as
y  C1e r1 x  C2 e r2 x
 C1e ( i ) x  C2 e ( i ) x
 C1ex (cos x  i sin x)  C2 ex (cos x  i sin x)
 ex [(C1  C2 ) cos x  i (C1  C2 ) sin x]
 ex (c1 cos x  c2 sin x)
where
c1  C1  C2 ,
c2  C1  C2
We summarize the discussion as follows:
(11) If the roots of auxiliary equation ar 2  br  c  0
are the complex numbers r1    i , r2    i
,
then the general solution of ay   by   cy  0
is
y  ex (c1 cos x  c2 sin x)
Example 4 Solve the equation
y  6 y  13 y  0.
Initial-value and boundary-value problems
An initial-value problem for the second-order Equation 1 or
2 consists of finding a solution y of the differential equation
that also satisfies initial conditions of the form
y( x0 )  y0
y( x0 )  y1
where y 0 and y1 are given constants. If P, Q, R, and G are
continuous on an interval and P ( x )  0 there, then a
theorem found in more advanced books guarantees the
existence and uniqueness of a solution to this initial-value
problem.
A boundary-value problem for Equation 1 consists of
finding a solution of the differential equation that also
satisfies boundary conditions of the form
y( x0 )  y0
y( x1 )  y1
In contrast with the situation for initial-value problems, a
boundary-value problem does not always have a
solution.
Example 5 Solve the initial-value problem
y  y  6 y  0
y (0)  1 y(0)  0
Example 6 Solve the initial-value problem
y  y  0
y (0)  2 y(0)  3
Example 7 Solve the boundary-value problem
y  2 y  y  0
y (0)  1
y (1)  3
15.1 Basic concepts, separable and
homogeneous equations
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Basic concepts
In Sections 8，
We have known the concept of the differential
equations.
10 Two kinds of equations
Def:
An ordinary differential equation:
involves an unknown function of a single variable
and some of its derivatives.
Such as
dy
x
 e xy
dx
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A partial differential equation:
involves an unknown function of two or more
variables and some of its partial derivatives.
Such as
 2u
 2u  2u
 2 xy 2  2  0
2
x
x y
20 The order of the equation
The order of a differential equation is the order of the
highest derivative that appears in the equation.
dy
x
 e xy
dx
Is an ordinary differential equation of order 1
y  2 xy  y  sin x
Is a third-order differential equation
 2u
 2u  2u
 2 xy 2  2  0
2
x
x
y
Is a second-order partial differential equation
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12.1.2 Separable equations
Note: We study only ordinary differential equations
mainly
In general,a first-order differential equation has
dy
the form :
 F ( x, y )
dx
where F is some function of the two variables x and y
10 Define the separable equations
When F can be factored as a function of x times a
function
F ( x, y )  g ( x ) f ( y )
then
dy
 g ( x) f ( y )
dx
Is called a separable equation
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Form
(1)
dy
 g ( x) f ( y )
dx
Also can be written as:
(2)
dy g ( x)

dx h( y )
h( y )dy  g ( x)dx
We integrate both sides：
 h( y)dy   g ( x)dx
H ( y )  G ( x)  C
This defines y implicitly as a function of x.
Sometimes ,we can solve for y in terms of x ，
thus , y=f(x) are called general solutions of the equation
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20 An initial-value problem
The separable equation
dy g ( x)

dx h( y )
Has a solution satisfies an initial condition of the form
(3)
y( x0 )  y0
We say this is an initial problem。
In general, there is a unique solution to the initial
problem given by equations (2),(3)
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Example 1
(a) Solve the differential equation
(b) Solve the initial-value problem
dy
 e  y cos x.
dx
y  e y cos x
y (0)  1
dy
y

e
cos x.
Solution (a)
dx
y
e
dy  cos xdx.
Write it as
y
e
 dy   cos xdx.
e y  sin x  C
y  ln(sin x  C )
This is the general solution, involves an arbitrary constant
C
Solution (b)
y(0) =1 tell us x=o,y=1
So
y  ln(sin x  e)
1  ln(sin 0  C )
1  ln C , C  e
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Example 2
Solution:
Solve the differential equation
ln x
y 
xy  xy 3
ln x
( y  y )dy 
dx
Write it as
x
We integrate both sides
3
ln x
 ( y  y )dy   x dx
3
So
y2 y4 1 2
1
  ln x  C
2 4 2
2
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Example 3
Solve the differential
equation
2
x
y 
Solution：Write it as
3x  e
4 y3
( y  y 3 )dy 
ln x
dx
x
We integrate both sides
3
(
y

y
)dy  

ln x
dx
x
So
y2 y4 1 2
1
  ln x  C
2 4 2
2
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Homogeneous equations
A first-order differential equation
If F(x,y) can be written as
The form (4)
is called homogeneous equations
For instance:
Turn it as :
dy
 F ( x, y )
dx
y
F ( x, y )  g ( )
x
dy
y
 g( )
dx
x
dy x 2  xy  y 2
x y

 ln x  ln y 
2
2
dx
x y
x  2y
y
y 2
y
1 ( )  ( )
1 ( )
dy
y
x
x  ln( ) 
x

y 2
y
dx
x
1 ( )
1  2( )
x
x
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y
y  g ( )
x
We see
make the change of variable
then
thus
y  xv
v  xv  g (v)
and
so
y
x
y  1v  xv  v  xv
v
y  xv
g (v )  v
v 
x
or
This is a separable differential equation ,
it’s general solution v ( x )
So, the original differential equation has the general
solution
y  xv ( x )
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xy  y 2
y 
x2
Example 5 Solve the differential equation
Solution: write it as
make the substitution
v
y
y
y  ( )  ( ) 2
x
x
y
x
,
which gives,
y  v  xv
2
v
so
v  xv  v  v 2
v 
x
this is a separable differential equation,
in a usual way solve it,
1
  ln x  C
v
x
y
ln x  C
aslo
vo


dv

2
v
v  
yo

v0
dx
x
1
ln x  C
is another solution
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Example 6 Solve the homogeneous differential equation
y
y
xy sin
 y sin  x
x
x
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15.1Homework
P 955
1.3.5.
15.2 First-order linear
equations
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The general solutions of the first-order linear
equations
dy
 P( x) y  Q( x)
dx
(1)
where P and Q are contunuous functions on agiven interval.
It’s solution is
ye 
 P ( x )dx
(2)
 Q( x)e P( x)dx dx  C 
 

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When Q(x)=0, form (1) become
dy
 P( x) y  0
dx
(3)
This is a separate equations ,solve it
dy
 y   P( x)dx,
ln y    P( x)dx  C ,
y  e  P ( x ) dx C
Thus
y  Ce  P ( x ) dx
(4)
Is the solution of the equation form (3),
C is an arbitrary constant
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Equation
(3)
dy
 P( x) y  0
dx
The solution is written as
y  Ce  P ( x ) dx
Let us replace the constant C in it by arbitrary
We look for the solution of the equation of the form
dy
 P( x) y  Q(x)
dx
y( x)  C ( x)e  P ( x ) dx
(5)
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Differentiating Equation (5),we get
  P ( x ) dx
  P ( x ) dx
dC(x)   P ( x ) dx
e
 C ( x)(1) P( x)e
 P( x)C ( x)e
 Q( x)
dx
dC(x)   P ( x ) dx
e
 Q( x)
dx
P ( x ) dx
dC(x)

 Q ( x )e
dx
P ( x ) dx

C(x)  Q( x)e
dx  C

y( x)  C ( x)e  P ( x ) dx
Put it into (5),
thus: The solution of the form (1) is
ye 
 P ( x ) dx
 Q( x)e P ( x ) dx dx  C 
 

(6)
Another method see page 997
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Example 1
Solve the differential equation
dy
 3x 2 y  6 x 2
dx
Solution:
this is a linear equation
By the general solution form (6) P( x)  3x 2
First step:
The second step:
e
  P ( x ) dx
e
  3 x 2 dx
 P ( x ) dx dx 
Q
(
x
)
e

e
 6x
Q( x )  6 x 2
 x3
2
 e dx
x3
  2  e dx  2e
x3
Last step:
ye
 x3
 2e x  C 


3
3
 2  Ce
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 x3
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Example 2 Find the solution of the initial-value problem
x 2 y  xy  1
x0
y (1)  2
Solution : We must first put the differential equation into
standard form:
1
1
y  y  2
x
x
First:
Second:
e
  P ( x ) dx
e
1
  dx
x
e
x0
 ln x
e
ln x1  1
x
1
1
 x2  xdx  xdx
1
y   ln x  C 
x

P ( x ) dx

dx 
 Q( x)e
Last:
Since y(1)=2, We have
2
 ln x
1
 ln1  C   C
1
Therefore, the solution to the initial-value problem is
y
1
 ln x  2 
x
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Example 3
find the solution of the initial-value problem
dy ty 3t

,
dt t 2 1
y (2) 2
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A Bernoulli differential equation
dy
 P( x) y  Q( x) y n
dx
Observe that ,if n=0 or 1,it is linear.
n
the substitution
uy
transforms the Bernoulli equation into the linear
equation
du
 (1  n) P( x)u  Q( x)
dx
Example
1
Solve the differential equation
dy
y
 6  xy 2
dx
x
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15.2 homework
15.3 Exact equations
Definition of the exact
f ( x, y )  C
Suppose the equation
(1)
Then y=f(x) satisfies a first-order differential
equation obtained by using the Chain Rule to
differentiate both sides of equation with respect to x:
f x( x, y)  f y( x, y) y  0
(2)
A differential equation of the form of Equation
(2)is called exact。
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Definition： A first –order differential equation of the
form
dy
P ( x, y )  Q ( x, y )  0
dx
(3)
is called exact if there is a function f(x,y) such that
f x( x, y)  P( x, y)
f y( x, y )  Q( x, y )
If f(x,y) is known,thus the solution is given implicitly by
f ( x, y )  C
(4)
We may be able to solve Equation 4 for y as an explicit
function of x.
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Theorem
We have the following convenient method for
the exactness of a differential equation.
Theorem:
Q ( x, y )
Suppose P ( x, y ) and
have continuous partial derivatives on a
simply –connected domain.
The the differential equation
dy
P ( x, y )  Q ( x, y )  0
dx
is exact if and only if
P
Q

Y
x
(5)
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Find solutions of exact equation
Example 1 Solve the differential equation
4 x  3 y  3( x  y 2 ) y  0
Solution
Here
2
Q
(
x
,
y
)

3
x

3
y
P( x, y)  4 x  3 y and
have continuous partial derivatives on R2
Also
P
Q
3
y
x
So the differential equation is exact by theorem .
Thus there exists a function f such that
f x( x, y)  4 x  3 y
f y( x, y )  3 x  3 y
(6)
2
(7)
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To determine f we first integrate 6 with respect to x :
f ( x, y)  2 x  3xy  g ( y)
2
(8)
Now we differentiate 8 with respect to y:
f y( x, y)  0  3x  g ( y)  3x  g ( y)
(9)
Comparing 7 and 9,we see that
g ( y )  3 y 2 and so
g ( y)  y3
We do not need the arbitrary constant here
Thus
f ( x, y)  2 x  3xy  y
So
2 x2  3xy  y3  C
2
3
is the solution.
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Example 2
Solution
Solve the differential equation
sin y  (1  x cos y) y  0
Here
P( x, y)  sin y
and
Q( x, y)  1  x cos y
have continuous partial derivatives on R2
Also
P
Q
 cos y 
y
x
So the differential equation is exact by theorem .
thus there exists a function f such that
f x( x, y)  sin y
f y( x, y )  1  x cos y
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To determine f we first integrate
f x( x, y)  sin y
f ( x, y)  x sin y  g ( y)
with respect to x :
Now we differentiate it with respect to y:
f y( x, y)  x cos y  g ( y)
Comparing them,we see
1  x cos y  x cos y  g ( y)
that
g ( y )  1
then
g ( y)  y
We do not need the arbitrary constant here
Thus
So
f ( x, y)  x sin y  y
x sin y  y  C
is the solution .
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Integrating factors
The differential equation
2 x  y 2  xyy  0
is not exact.
If the differential equation
P ( x, y )  Q ( x, y )
dy
0
dx
is not exact,
we look for an integrating factor I(x,y)
Such that ,after multiplication by I(x,y),
the resulting equation
dy
I ( x, y ) P( x, y )  I ( x, y )Q( x, y )  0
(10 )
dx
is exact.
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Equation (10) is exact if


( IP) 
( IQ)
Y
x
That is
I y P  IPy  I xQ  IQx
or
PI y  QI x  I (Qx  Py)
(11)
In general,it is harder to solve this partial differential
equation than the original differential equation.
But it is sometimes possible to find I that is a function of x
or y alone.
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Suppose I is a function of x alone.
Then
I y  0
So equation 11 becomes
if
Py  Qx
Q
dI Py  Qx

I
dx
Q
(12)
is a function of x alone ,
Then Equation 12 is a first –order linear (and separable )
ordinary differential equation.
It can be solved for I(x)
Then Equation 10 is exact and can be solved In Example 1 or 2.
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Example 3
Solve the differential equation
2 x  y  xyy  0
2
Solution :
Here P( x, y)  2 x  y 2
P
 2y
y
since
But
P
y
Py  Qx
Q
 Q

x
Q ( x, y )  xy
Q
 y
x
the given equation is not exact .
2y  y 1

xy
x
is a function of x alone.
So by Equation 12 there is an integrating factor I that satisfies
dI 1
 I
dx x
we get
I ( x )  x.
dI I

dx x
( just a particcular one )
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Multiplying the original equation by x ,we get
x(2 x  y 2  xyy)  0
(13)
2 x 2  xy 2  x 2 yy  0
If we let
p( x, y)  2 x 2  xy 2 , q( x, y)  x 2 y
p
q
Then
 2 xy 
y
x
So Equation 13 is now exact.thus there is a function f such that
f x ( x, y )  2 x 2  xy 2 ,
f y ( x, y)  x2 y
Integrating the first of these equations,we get
2 3 1 2 2
f ( x, y )  x  x y  q ( y )
3
2
so
f y ( x, y)  x2 y  g ( y)
Comparison then gives g ( y )  0
,so g is a constant
(which we can take to be 0)
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therefore
2 3 1 2 2
f ( x, y )  x  x y
3
2
2 3 1 2 2
The solution is x  x y  C
3
2
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15.3homework
15.4 Strategy for solving firstorder equations
In solving first-order differential equations we
used the technique for separable equations in
Sections 8.1 and the method for linear equations in
Section 15.2.(on the text).
We also develops methods for solving
homogeneous equations in Section 15.1 (on the text)
and exact equations in Section 15.3 (on the text)。
In this section we present a miscellaneous
collection of first-order differential equations and
part of the problem is to recognize which technique
should be used on each equation.
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As with the strategy of integration (Section7.6)
and the strategy of testing series (Section
10.7),the main idea is to classify the equation
according to its form.
Here , however , the important thing is not so
much the form of the functions involved e as it
is the form of the equation itself .
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Recall that a separable equation can be written in
the form
(1)
dy
 g ( x) f ( y )
dx
that is ,the expression for dy/dx can be factored
and a product of a function of x and a function
of y.A linear equation can be put into the form
(2)
dy
 P( x) y  Q(x)
dx
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A homogeneous equation can be expressed in the
form
dy
y
(3)
 g( )
dx
x
An exact equation has the form
(4)
dy
P ( x, y )  Q ( x, y )  0
dx
P
where
y
 Q
x
If an equation has none of these forms ,we can ,as
a last resort , attempt to find and integrating
factor and thus equation exact .
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In each of these cases ,some preliminary
algebra may be required in order to put a
given equation into one of the preceding
forms .
(This step is analogous to Step l of the
strategy for integration: algebraic
implification. )
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It may happen that a given equation is of more than one
type .
For instance, the equation
xy  y
is separable because dy/dx can be written as
dy
1
 ( )y
dx
x
(compare with Equation 1)
It is also linear ,since we can write the equation as
dy
1
 ( ) y  0
dx
x
dy y

dx x
(compare with Equation 2)
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Furthermore ,it is also homogeneous because we
can write it as
dy y

dx x
(compare with Equation 2).
In such a case we could solve the equation using
any one of the corresponding methods ,although
one of the methods might be easier than the others .
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In the following examples we identify the
type of each equation without working our
the details of the solutions .
Example1
y  1  x  y  xy
Initially ,this equation may not appear to be in
any of the forms of Equations 1,2,3,or 4,but
observe that we can factor the right side and
therefore write the equation as
y  (1  x)(1  y )
We now recognize the equation ad being
separable and we can solve it using the methods
of Section 15.1.
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Example 2
x 2  y 2  2 xyy  0
The equation is clearly mot separable , nor
separable, nor is it linear .
p y  2 y
Since
it is not exact .
qy  2 y
But if we solve for y’, we get


y2  x2 1  y
1 
y 


2 xy
2x y 

x 
Which shows that y’ is a function of y/x and the
equation is homogeneous (see Equation 3).
(We could have anticipated this because the
expressions x2,y2,and 2xy are all of degree 2.)
The change of variable v=y\x converts the
equation into a separable equation .
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3x 2  2 xy 3
y  
3x 2 y 2
Example3
This equation is not separable
homogeneous .
,linear ,or
We suspect is might be exact, so we write it in the
form
(3x 2  2 xy 3 )  (3x 2 y 2 ) y  0
If
then
p( x, y)  3x2  2 xy3 , q( x, y)  3x 2 y 2
py  6 xy 2  qx
Therefore ,the equation is indeed exact and can be
dolled by the methods of Section 15.3.
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15.4homework
Example4
( x  1) y  xy  x3  x 2
If we put the equation in the
form
x 1
y 
x
y  x  x2
We recognize it as having the form of Equation 2.
It is therefore linear and can be solved using
the integrating factor
e 
 (1 1 )dx
x
15.6 Nonhomogeneous Linear Equations
In this section we deal with second-order
nonhomogeneous linear differential equations with constant
coefficient :
(1)
ay  by  cy  G (x)
where a, b, c are constants and G is a continuous function.
The related homogeneous equation
(2)
ay  by  cy  0
is called the complementary equation and plays an important
role in the solution of the original nonhomogeneous equation
(1).
(3)Theorem The general solution of the
nonhomogeneous differential equation (1) can be written
as
) x ( cy  ) x ( p y  ) x ( y
where y p is a particular solution of Equation 1 and y c
is the general solution of the complementary Equation 2.
Proof We want to verify that if y is any solution of
Equation 1, then y  y p is a solution of the
complementary Equation 2.
Indeed
) p y  y (c  ) p y  y (b  ) p y  y ( a
p
yc  yc  pyb  yb  pya  ya 
) p yc  pyb  pya(  ) yc  yb  ya( 
0  )x ( G  )x ( G 
We know from Section 15.5 that the solution of the
complementary equation is 2y 2c  1y 1c  cy , where 1y
and 2y are linearly independent solutions of Equation 2.
Therefore, Theorem 3 says that we know the general
solution of the nonhomogeneous equation as soon as we
know a particular solution.
There are two methods for finding a particular
solution. The method of undetermined coefficients is
straightforward but works only for a restricted class of
functions G. The method of variation of parameters
works for every function G but is usually more difficult
to apply in practice.
•The method of undetermined coefficients
We first illustrate the method of undetermined
coefficients for the equation
ay  by  cy  G (x)
Case 1 G(x) is a polynomial
It is reasonable to guess that the particular solution y p
is a polynomial of the same degree as G because if y is a
polynomial, then ay  by  cy is also a polynomial.
We therefore substitute
y p (x)  a polynomial (of the
same degree as G ) into the differential equation and
determine the coefficients.
Example 1 Solve the equation
2



y  y  2y  x .
Case 2 G(x) is of the form Ce kx , where C and k are
constants.
We take as a trial solution a function of the same form
kx
y

Ae
` p
, because the derivatives of e kx are constant
kx
e
multiples of
.
Example 2
Case 3
3x


Solve the equation y  4 y  e .
G(x) is either Ccoskx or Csinkx
Because of the rules for differentiating the sine and
cosine functions, we take as a trial particular solution a
function of the form
y p ( x)  A cos kx  B sin kx
Example 3
Solve the equation y  y  2 y  sin x.
Case 4 G(x) is a product of functions of the preceding
types
We take the trial solution to be a product of functions
of the same type. For instance, in solving the differential
equation
y  2 y  4 y  x cos 3 x
we would try
y p ( x)  ( Ax  B) cos 3x  (Cx  D) sin 3x
Case 5 G(x) is a sum of functions of these types
We use the easily verified principle of superposition,
which says if y p1 and y p 2 are solutions of
ay  by  cy  G1 ( x)
ay  by  cy  G2 ( x)
respectively, then y p1  y p 2 is a solution of
ay  by  cy  G1 ( x)  G2 ( x)
Example 4 Solve
y  4 y  xex  cos 2 x.
Finally we note that the recommended trial solution y p
sometimes turns out to be a solution of the complementary
equation and therefore cannot be a solution of the
nonhomogeneous equation. In such cases we multiply the
recommended trial solution by x (or x 2 if necessary) so
that no term in y p (x) is a solution of the complementary
equation.
Example 5
Solve
y  y  sin x.
Summary of undetermined coefficient
G(x) =
First try
yp 
Cn x n    C1 x  C0
An x n   A1 x  A0
Ce kx
Aekx
C cos kx  D sin kx
A cos kx  B sin kx
Modification: If any term of y p is a solution of the
complementary equation, multiply y p by x (or x 2 if
necessary) .
•The method of variation of parameters
Suppose we have already solved the homogeneous
equation ay  by  cy  0 and written the solution as
(4)
)x ( 2y 2c  )x ( 1y 1c  )x (y
where
y1 and
y 2 are linearly independent solutions.
Let us replace the constants c1 and c2 in Equation
4 by arbitrary functions u1 ( x) and u2 ( x) .
We look for a particular solution of the nonhomogeneous
equation
of the form
ay  by  cy  G (x)
(5)
)x ( 2y )x ( 2u  )x ( 1y )x ( 1u  )x ( p y
Differentiating Equation 5, we get
(6)
) 2y 2u  1y 1u(  ) 2y 2u  1y 1u(  py
Since u1 and u2 are arbitrary functions, we can impose
two conditions on them. One condition is that y p is a
solution of the differential equation; we can choose the
other condition so as to simplify our calculations. In view
of the expression in Equation 6, let us impose the
condition that
(7)
0  2y 2u  1y 1u
Then
2y 2u  1y 1u  2y 2u  1y 1u  py
Substituting in the differential equation, we get
G  ) 2y 2u  1y 1u(c  ) 2y 2u  1y 1u(b  )2y 2u  1y 1u  2y 2u  1y 1u(a
or
(8) G  ) 2y 2u  1y 1u(a  ) 2yc  2yb  2ya( 2u  ) 1yc  1yb  1ya( 1u
But y1 and
equation, so
y 2 are solutions of the complementary
ay1  by1  cy1  0 and
and Equation 8 simplifies to
(9)
G  ) 2y 2u  1y 1u(a
ay2  by2  cy2  0
Equation 7 and 9 form a system of two equations in the
unknown functions u1 and u2 . After solving this
system we may be able to integrate to find u1 and u2
then the particular solution is given by Equation 5.
Example 6 Solve the equation
y  y  tan x,
0  x   2.
15.8 Series Solutions
So far, the only second-order differential equations
that we have been able to solve are linear equations with
constant coefficients.
Even a simple-looking equation like
(1)
y  2 xy  y  0
is not easy to solve. But it is important to be able to solve
equations such as Equation 1 since they arise from
physical problems and, in particular, in connection with
the Schrödinger equation(薛定谔方程) in quantum
mechanics.
In such a case, we use the method of power series; that is,
we look for a solution of the form

y  f ( x)   cn x n  c0  c1 x  c2 x 2  c3 x 3  
n 0
The method is to substitute this expression into the
differential equation and determine the values of the
coefficients c0 , c1 , c2 
Example 1 Use power series to solve the equation
y   y  0.
Solution We assume there is a solution of the form
(2)

y  f ( x)  c0  c1 x  c2 x 2  c3 x 3     cn x n
n 0
By Theorem 10.9.2 we can differentiate power series
term by term. So
(3)

y  c1  2c2 x  3c3 x 2     ncn x n 1
n 1

y  2c2  2  3c3 x     n(n  1)cn x n  2
n2
Rewrite y  as

y   (n  2)( n  1)cn  2 x n
n 0
In order to compare the expression for y and y  more
easily, we substituting these expressions into the
differential equation, we obtain


n
(
n

2
)(
n

1
)
c
x

c
x

 n 0
n2
n
n 0
n 0
or
(5)

n
[(
n

2
)(
n

1
)
c

c
]
x
0

n2
n
n 0
If two power series are equal, then the corresponding
coefficients must be equal (from Section 10.10). Therefore,
the coefficients of x n in Equation 5 must be 0:
(6)
(n  2)(n  1)cn 2  cn  0
cn
cn  2  
(n  1)( n  2)
Equation 6 is called a recursion relation. If c0 and c1
are known, this equation allows us to determine the
remaining coefficients recursively by putting n = 0, 1, 2,
3, …in succession.
Put
n  0:
Put
n  1:
Put
n  2:
Put
n  3:
Put
n  4:
Put
n  5:
c0
c2  
1 2
c1
c3  
23
c0
c0
c2
c4  


3  4 1  2  3  4 4!
c3
c1
c1
c5  


4  5 2  3  4  5 5!
c0
c0
c4
c6  


56
4! 5  6
6!
c5
c1
c1
c7  


67
5! 6  7
7!
By now we see the pattern:
For the even coefficients,
For the odd coefficients,
c0
(2n)!
c1
n
c2 n 1  (1)
(2n  1)!
c2 n  (1) n
Putting these values back into Equation 2, we write
the solution as
y  c0  c1 x  c2 x 2  c3 x 3  c4 x 4  c5 x 5  
2n
x2 x4 x6
x
 c0 (1 


   (1) n
 )
2! 4! 6!
(2n)!
2 n 1
x3 x5 x7
x
 c1 ( x  

   (1) n
 )
3! 5! 7!
(2n  1)!
2n
2 n 1

x
x
 c0  (1) n
 c1  (1) n
(2n)!
(2n  1)!
n 0
n 0

Notice that there are two arbitrary constants c0 and c1 ,
which is to be expected.
Note 1: We recognize the series obtained in Example 1
as being the Maclaurin series for cosx and sinx. (See
Equation 10.10.16 and 10.10.15.) Therefore, we could
write the solution as
y( x)  c0 cos x  c1 sin x
But we are not usually able to express power series
solutions of differential equations in terms of known
functions.
Example 2
y  2 xy  y  0.
Solve
Solution We assume there is a solution of the form
y  f ( x )   cn x n
n 0

Then
y   ncn x n 1
n 1

and
y   n(n  1)cn x
n2
n2

  (n  2)( n  1)cn  2 x n
n 0
Substituting in the differential equation, we get


 (n  2)(n  1)cn2 x  2 x ncn x
n
n 0
n 1
n 1


  cn x n  0
n 0


n
(
n

2
)(
n

1
)
c
x

2
nc
x

c
x

 n  n 0
n2
n
n 0
n
n 1

n 0
n
[(
n

2
)(
n

1
)
c

(
2
n

1
)
c
]
x
0

n2
n
n 0
This is true if the coefficient of xn
is 0:
(n  2)( n  1)cn  2  (2n  1)cn  0
(7)
cn  2
2n  1

cn
(n  1)( n  2)
n  0, 1, 2, 3, 
We solve this recursion relation by putting n = 0, 1, 2, 3, …
successively in Equation 7:
Put
n  0:
Put
n  1:
Put
n  2:
Put
n  3:
1
c2 
c0
1 2
1
c3 
c1
23
3
3
3
c4 
c2  
c0   c0
3 4
1 2  3  4
4!
5
1 5
1 5
c5 
c3 
c1 
c1
45
2 3 4 5
5!
Put
Put
Put
Put
7
3 7
3 7
n  4 : c6 
c4  
c0  
c0
56
4! 5  6
6!
9
1 5  9
1 5  9
n  5 : c7 
c5 
c1 
c1
67
5! 6  7
7!
11
3  7 11
n  6 : c8 
c6  
c0
7 8
8!
13
1 5  9 13
n  7 : c9 
c7 
c1
89
9!
In general, the even coefficients are given by
c2 n
3  7 11  (4n  5)

c0
(2n)!
and the odd coefficients are given by
1 5  9  (4n  3)
c2 n 1 
c1
(2n  1)!
The solution is
y  c0  c1 x  c2 x 2  c3 x 3  c4 x 4  
1 2 3 4 3  7 6 3  7 11 8
 c0 (1  x  x 
x 
x  )
2!
4!
6!
8!
1 3 1 5 5 1 5  9 7 1 5  9 13 9
 c1 ( x  x 
x 
x 
x  )
3!
5!
7!
9!
Or
(8)
1 2  3  7    (4n  5) 2 n
y  c0 (1  x  
x )
2!
(2n)!
n2
1  5  9    (4n  3) 2 n 1
 c1 ( x  
x )
(2n  1)!
n 1

Note 2: In Example 2 we had to assume that the
differential equation had a series solution. But now we
could verify directly that the function given by Equation
8 is indeed a solution.
Note 3: Unlike the situation of Example 1, the power
series that arise in the solution of Example 2 do not
define elementary functions. The functions
and
1 2  3  7    (4n  5) 2 n
y1 ( x)  1  x  
x
2!
(2n)!
n2
1 5  9    (4n  3) 2 n 1
y2 ( x)  x  
x
(2n  1)!
n 1

are perfectly good functions but they cannot be
expressed in terms of familiar functions. We can use
these power series expressions for y1 and
y2 to
compute approximate values of the functions and even
to graph them.
Note 4: If we are asked to solve the initial-value
problem
y  2 xy  y  0
y (0)  0
y(0)  1
we would observe from Theorem 10.10.5 that
c0  y(0)  0
c1  y(0)  1
This would simplify the calculations in Example 2,
since all of the even coefficients would be 0. The
solution to the initial-value problem is
1 5  9    (4n  3) 2 n 1
y ( x)  x  
x
(2n  1)!
n 1
