Chapter 1
First-Order Differential Equations
Shurong Sun
University of Jinan
Semester 1, 2010-2011
Analogous to a course in algebra and trigonometry,
where a good amount of time is spent solving
equations such as
x 2  5 x  4  0 for the unknown variable x.
In this course one of our tasks will be to solve
differential equations such as y  2 y  y  0
for the unknown function y=y(t).
1.1
1.1 Separable Differential
Equations
• Many first order differential equations can
be written as:
g ( y ) y  f ( x)
'
or
dy
g ( y )  f ( x)
dx
 g ( y)dy  f ( x)dx
•We can solve these equations by integrating:
 g ( y)dy   f ( x)dx  c
1.2
Separable Differential Equation
A first order differential equation of the form
dy
 f ( x ) g( y ) is called a separable differential equation.
dx
I.
g ( y )  0.
Set
1
h( y ) 
and multiply by dx to obtain
g( y )
h( y )dy  f ( x )dx .
Integrating, we have
 h( y)dy   f ( x)dx  c.
1.3
II.
g(0 )  0, then y( x )  0
dy
 f ( x ) g( y ).
is one solution to the equation
dx
1.4
Separable Equation Example
dy
x
 .
Example Solve the equation
dx
y
Solution: This is a separable equation.
Separate variables and rewrite the equation in the form
ydy   xdx .
Integrating, we have
 ydy    xdx
or
y2
x2

 C ,(C  0).
2
2
Solving for y , we obtain the solution in explicit form as
y   C1  x 2 .
1.5
Separable Equation Example
• Solve the differential equation:
y'  1  y 2
Solution:
We first rewrite the DE in fractional form
dy
 1  y2
dx
Separate variables and rewrite the equation in the form
dy
 dx
2
1 y
Integrating, we get arctan y  x  c or y  tan( x  c )
1.6
Example Solve the IVP
dy y  1

, y(1)  0.
dx x  3
Solution: Separate variables and rewrite the equation
dy
dx

.
in the form
y 1
Integrating, we have
x3
1
1
dy

 y  1  x  3 dx
or
ln | y  1 | ln | x  3 |  C .
Solving for y , we obtain the solution in explicit form as
y  1  e C ( x  3) or y  1  C1 ( x  3), where C1  e C  0.
1.7
Note that y=1 is also a solution to the equation.
So the general solution to the equation is
y  1  K ( x  3), where K is an arbitrary constant.
Applying the initial condition directly, we have
1
or
0  1  2K
K  .
2
1
1
Thus y  1  ( x  3)   ( x  1).
2
2
1.8
Example Solve the equation
dy 6 x 5  2 x  1

.
y
dx
cos y  e
Solution: Separate variables and rewrite the equation
in the form (cos y  e y )dy  (6 x 5  2 x  1)dx .
Integrating, we have
y
5
(cos
y

e
)
dy

(6
x

  2 x  1)dx,
or
sin y  e y  x 6  x 2  C .
1.9
Example- Newton ’ s Law of Cooling
A copper sphere is heated to 100o c . At time t=0 it is
then placed in water that is maintained at 30o c .
After 3 minutes the sphere ’ s temperature is 70o c .
Derive an equation for the temperature T of the ball
as a function of time t.
•Newton ’ s law of cooling implies:
dT
 k (T  30)
dt
1.10
Example- Newton ’ s Law of Cooling
• Step 1 - Separate variables
dT
 k dt
T  30
• Step 2 - Integrate
ln | T  30 | kt  c
 T (t )  ce  30
kt
1.11
Example- Newton ’ s Law of Cooling
• Step 3 - Determine a particular solution
T (0)  ce  30  100
kt
 T (t )  70e  30
kt
•Step 4 - Determine k
T (3)  70e  30  70
3k
1 70  30
 k  ln
 0.1865
3
70
1.12
1.2 Reduction to Separable Form
• Sometimes first order differential equations
can be made separable by a simple change of
variable. Consider equations of the form:
(1) Homogeneous
Equation
dy
 y
 g 
dx
 x
Remark:
dy
 f ( x, y ) is homogeneous iff f ( tx , ty )  f ( x , y ).
dx
1.14
y
•This suggests we set: u 
x
• Writing the differential equation in terms of
u gives:
dy
du
y  xu 
 u x
dx du dx
dy
 g ( u)  u  x
dx
du
x
 g ( u)  u
dx
dx
 g ( u)
1.15
du
x
 g ( u)  u
dx
• Separating variables
gives:
du
dx

g (u )  u x
1.16
Example: Solve ( xy  y 2  x 2 )dx  x 2dy  0. (6)
Solution: We express (6) in the derivative form
dy xy  y 2  x 2 y
y 2

  ( )  1,
2
dx
x
x
x
then we see that the equation (6) is homogeneous.
dy
du
y
 u x .
Now let u 
and recall that
dx
dx
x
With these substitution, equation (6) becomes
du
x
 u2  1.
dx
1.17
The above equation is separable, and , on separating the
variables and integrating, we obtain
1
1
 u2  1 dv   xdx, arctan u  ln | x | C .
Hence,
u  tan(ln | x |  C ).
y
Finally, we substitute for u and solve for y
x
to get y  x tan(ln | x |  C ).
Also note that x=0 is a solution.
1.18
dy
(2) Equations of the Form
 G(ax  by )
dx
Set z  ax  by . Then the equation is transformed
into a separable one.
Example: Solve
dy
 y  x  1  ( x  y  2)1 . (8)
dx
z  x  y.
Solution:
Set
Then
dz
dy
 1 ,
dx
dx
and so
dy
dz
 1 .
dx
dx
1.19
Substituting into (8) yields
dz
  z  1  ( z  2)1 , or
dx
2
dz
(
z

2)
1
1
 z  2  ( z  2) 
.
dx
z2
1
Solving this separable equation, we obtain
z2
1
dz

dx
,
 ( z  2)2  1 
ln | ( z  2)2  1 | x  C ,
2
2
2x
from which it follows that ( z  2)  Ce
 1.
Finally, replacing z by x-y yields
( x  y  2)  Ce
2
2x
 1.
1.20
 a1 x  b1 y  c1 
dy
(3)
 f

dx
 a2 x  b2 y  c2 
where the ai s , bi s and ci s are constants.
I.
a1
b1
a2
b2
 0.
In this case, the equations can be reduced to
dy
the form
 G(ax  by )
dx
1.21
II. a1 b1
a2
b2
 0.
Then the system of equations
 a1 x  b1 y  c1  0

 a2 x  b2 y  c2  0
has a unique solution ( x0 , y0 ).
The above DE can be written in the form
 a1 ( x  x0 )  b1 ( y  y0 ) 
dy
 f

dx
 a2 ( x  x0 )  b2 ( x  x0 ) 
1.22
which yields the DE
 a1u  b1v 
dy
f

dx
 a2u  b2v 
Homogenous
Equation
after the translation of axes of the form
u  x  x0 , v  y  y0 .
1.23
Example: Solve
( 3 x  y  6)dx  ( x  y  2)dy  0.
Solution:
From
3 1
 4  0.
1 1
 3 x  y  6  0

 x y20
Hence, we let
we obtain x0  1, y0  3.
u  x  1, v  y  3.
1.24
The differential equation for v is
( 3u  v )du  ( u  v )dv  0,
or
v
3
dv 3u  v
u.


v
du u  v
1
u
v
The above equation is homogenous, so we let z  .
u
2
dz 3  z
dz
3

2
z

z
Then z  u du  1  z , or u

.
du
1 z
1.25
Separating variables gives
z 1
1
 z 2  2z  3dz    udu,
1
ln | z 2  2 z  3 |  ln | u | C ,
2
from which it follows that z 2  2 z  3  Cu 2 .
When we substitute back in for z, u, and v, we find
v 2
v
( )  2  3  cu2 , v 2  2uv  3u2  C ,
u
u
( y  3)2  2( x  1)( y  3)  3( x  1)2  C
1.26
1.2 Linear First-Order DE
(1) Definition Linear Equation
A first-order differential equation of the form
a1 ( x )
dy
 a0 ( x ) y  h( x )
dx
(1)
is said to be linear equation.
When h( x )  0, the linear equation is said
to be homogeneous; otherwise, it is non-homogenous
or inhomogenous.
1.27
a1 ( x )
dy
 a0 ( x ) y  h( x )
dx
(2) Standard Form
By dividing both sides of (1) by the lead coefficient
a1 ( x ), we obtain a more useful form, the stand
form, of a linear equation: dy  P ( x ) y  q( x ).
(2)
dx
We seek a solution of (2) on an interval I for which
both function P and f are continuous.
(3) Variation of parameters
1.28
I. The Homogeneous Equation dy  P ( x ) y  0.
(3)
dx
This is a separable equation.
Writing (3) as
dy
 P ( x )dx  0
y
and integrating .
  P ( x ) dx
Solving for y gives y  Ce
.
II. The Nonhomogeneous Equation
dy
 P ( x ) y  q( x ).
dx
(2)
Method of variation of constants
1.29
II. The Nonhomogeneous Equation
dy
 P ( x ) y  q( x ).
dx
(2)
Method of variation of constants
The basic idea is the constant C in the general solution of
the homogeneous equation (3) is replaced by a function
C(x). The calculation of an appropriate choice of C(x)
gives a solution of the nonhomogeneous equation (2) .
Substituting y  C ( x )e 
 P ( x ) dx
(C ( x )  C ( x ) P( x ))e 
 P ( x ) dx
into (2) gives
 P( x )C ( x )e 
 P ( x ) dx
 q( x )
1.30
so
Hence
and
C ( x )e 
 P ( x ) dx
P ( x ) dx

 q( x ) or C ( x )  e
q ( x)
P ( x ) dx

C ( x)   e
q ( x)  C ,
y  Ce 
 P ( x ) dx
e 
 P ( x ) dx
e 
 P ( x ) dx

P ( x ) dx

e
q( x )dx
P ( x ) dx

( e
q( x )dx  C ).
(4)
Thus if (2) has a solution, it must be of form (4).
Conversely, it is easy to verify that (4) constitutes
one-parameter family of solutions of equation (2),
that is a general solution of equation (2).
1.31
Example
Finding the general solution to
dy
( x  1)  ny  e x ( x  1)n1 .
dx
Solution:
We write the differential equation in standard form
dy
n

y  e x ( x  1)n .
dx x  1
The associated homogeneous equation is
dy
n

y  0.
dx x  1
Separating variable, we find
1.32
n
dy
dx  0,

y x 1
So the general solution to the homogeneous equation is
y  C ( x  1)n .
By the method of variation of parameter, substituting
into the equation gives
y  C ( x )( x  1)n
dc( x )
n
n
n 1
( x  1)  n( x  1) c( x ) 
c( x )( x  1)n  e x ( x  1)n .
dx
x 1
dc( x )
x
So
 ex ,
1
dx
Thus the general solution to ( x  1) dy  ny  e x ( x  1)n1 .
dx
c( x )  e  c .
1.33
is
y  ( x  1)n (e x  c1 ).
1.34
dy
n
 P  x  y  Q( x) y
Bernoulli Equations
dx
(9)
Remark: when n=0,1, equation (9) is also a linear equation.
n
For n  1, dividing equation (9) by y
dy

n
yields
y
 P ( x ) y1 n  Q( x ). (10)
dx
Taking v  y1 n , we find via the chain rule that
dv
 n dy
 (1  n) y
,
dx
dx
1.40
dv
 n dy
 (1  n) y
,
dx
dx
and so equation (10) becomes
1 dv
Linear
 P ( x )v  Q( x ).
equation
1  n dx
 n dy
y
 P ( x ) y1 n  Q( x ).
dy
5 3
dx
Example : Solve
(11)
 5 y   xy .
dx
2
Solution: This is a Bernoulli equation with n=3,
5x
P ( x )  5, and Q( x )  
.
2
1.41
We make the substitution
dv
3 dy
 2 y
,
dx
dx
1 dv
5

 5v   x,
2 dx
2
Since
2
v y .
the transformed equation is
dv
 10v  5 x .
dx
This is a linear equation, its solution is
ve 
 10 dx
10 dx

(  5 xe
 C )  e 10 x (  5 xe10 x dx  C )
x
1
 
 Ce 10 x .
2 20
1.42
ve 
 10 dx

10 dx

(  5 xe
 C )  e 10 x (  5 xe10 x dx  C )
x
1

 Ce 10 x .
2 20
Substituting
v y
2
give the solution
x 1
y    Ce 10 x .
2 20
2
Not included in the last equation is the solution
y  0 that was lost in the process of dividing
3
y
.
(11) by
1.43
1.3 Exact Differential Equations
and Integrating factors
Exact Differential Equations
• Recall that the total or exact differential
of a function u(x,y) is:
u
u
du 
dx  dy
x
y
• We will use the concept of exactness to
study differential equations of the form:
M ( x, y )dx  N ( x, y )dy  0
1.45
• Compare the following
u
u
dx 
dy  du (exact differential)
x
y
M ( x , y )dx  N ( x , y )dy  0
(ODE)
We say an ODE is exact if there exists a function
such that
u( x , y )  C 1 ( ),   R2
u
M ( x, y) 
x
u
and N ( x , y ) 
.
y
That is
du( x , y )  Mdx  Ndy
Note this means that we can now write our ODE as:
1.46
Note this means that we can now write our ODE as:
du( x , y )  0
In this case, its solution (in implicit form) is given by
u( x , y )  C
Remember we need:
u
M
x
u
and
N
y
1.47
Testing for Exactness
• In practice we often don ’ t know about u,
only about M and N and it is hard to check
that
u
M
x
u
and
N
y
• We want a technique of testing for
exactness based on knowing M and N, not u
1.48
• Let ’ s check the derivatives of M and N:
M
  u   2u
N   u   2u
  
and
   
y y  x  yx
x x  y  xy
If M(x,y) and N(x,y) are continuous functions and have
continuous first partial derivatives on some simply
connected region of xy-plane, then
• A necessary and sufficient condition for
exactness is:
M N

y
x
1.49
Solving Exact Equations
• Once we have checked the equation is exact
it can be readily solved by evaluating either:
u   Mdx  g( y )
u
from
M
x
u 

Mdx  g ( y )=N

y y
u
from
N
y
1.50
or u   Ndy  h( x )
u 

Ndy  h( x )=M

x x
u
from
N
y
u
from
M
x
1.51
Exact Equations: Example
2
2
(2
xy

sec
x
)
dx

(
x
 2 y )dy  0.
Example Solve
Solution
Here M ( x, y )  2 xy  sec2 x and N ( x, y )  x 2  2 y.
Since
M
N
 2x 
,
y
x
the equation given is exact.
u
2

2
xy

sec
x,
From
x
we have that
u( x, y )   (2 xy  sec2 x )dx  g( y )  x 2 y  tan x  g( y ).
1.52
Next we take the partial derivative of u with respect
2
x
 2 y for N:
to y and substitute
x 2  g( y )  x 2  2 y.
Thus g( y )  2 y, and g( y )  y 2 .
Here we drop the constant of integration that technically
should be present in g(y) since it will just get absorbed
into the constant we pick up in the next step.
Hence u( x, y )  x 2 y  tan x  y 2
So, the implicit solution to the differential equation is
x 2 y  tan x  y 2  C .
1.53
Exact Equations: Example
( x  3 xy )dx  (3 x y  y )dy  0
3
2
2
 M  x 3  3 xy 2 and
M

 6 xy
y
3
N  3x2 y  y3
N
and
 6 xy
x
(exact )
u   Ndy  h( x )   (3 x 2 y  y 3 )dy  h( x )
3 2 2 1 4
 x y  y  h( x )
2
4
1.54
Exact Equations: Example
• Solve for k(x) and then u:
u
dk
2
3
2
 M  3 xy 
 x  3 xy
x
dx
4
x
 h
c
4
1 4
2 2
4
 u  ( x  6x y  y )  c
4
1.55
Example Find the solution for the following IVP.
3 y 3e 3 xy  1  (2 ye 3 xy  3 xy 2e 3 xy ) y  0.
y(0)  1.
Solution
Let ’ s identify M and N and check that it ’ s exact.
3 3 xy
2 3 xy
3 3 xy
M  3 y e  1, M y  9 y e  9 xy e
N  2 ye 3 xy  3 xy 2 e 3 xy , N x  9 y 2e 3 xy  9 xy 3e 3 xy
So, it ’ s exact.
1.56
With the proper simplification integrating the second
one isn ’ t too bad. However, the first is already set
up for easy integration so let ’ s do that one.
u( x, y )   (3 y 3e 3 xy  1)dx  h( y )
ye
2 3 xy
 x  h( y )
Differentiate with respect to y and compare to N.
uy ( x, y )  2 ye  3 xy e  h( y )  N
So, we get h( y )  0  h( y )  0
3 xy
3 xy
2 3 xy
N  2 ye  3 xy 2e 3 xy
Recall that actually h(y) = k, but we drop the k
because it will get absorbed in the next step. That
gives us h(y) = 0.
1.57
Therefore, we get
u( x , y )  y e
2 3 xy
 x.
The implicit solution is then
2 3 xy
ye
xC
Applying the initial condition gives 1 = C
The implicit solution is then
y 2e 3 xy  x  1.
1.58
Remember Exact Equations
• We can check whether an equation in
this form is exact by checking if
M N

y
x
(Necessary and sufficient)
• Alternate method: we can often rearrange
linear first order ODEs into the form:
u
u
M ( x , y )dx  N ( x , y )dy 
dx 
dy
x
y
 du  0
1.59
Solve the following DE
( x 3  3 xy 2 )dx  (3 x 2 y  y 3 )dy  0
Solution:
Here
M  x  3 xy
3
2
and
N  3x y  y
2
3
We first check to see if we have an exact equation.
Since
M
 6 xy
y
N
and
 6 xy
x
it ’ s exact.
1.60
Note that ( x 3  3 xy 2 )dx  (3 x 2 y  y 3 )dy
 x dx  (3 xy dx  3 x y )dy  y dy
3 2 2 3 2 2
3
 x dx  ( y dx  x dy )  y 3dy
2
2
1 4 3 2 2 1 4
 d( x  x y  y )
4
2
4
3
2
2
3
So the general solution is
1 4
2 2
4
( x  6x y  y )  C
4
1.61
dx  dy  d ( x  y )
xdy  ydx  d ( xy )
1
xdx  ydy  d ( x 2  y 2 )
2
xdx  ydy
xdy  ydx
 d ln (xy)
xy
xdy  ydx
y
 d( )
2
x
x
xdy  ydx
y
 d [ln ( )]
xy
x
xdy  ydx
1 y
 d [ tan ( )]
2
2
x y
x
x2  y2
 d ( x2  y2 )
1.62
Example. Solve the homogeneous DE
dy x  y

dx x  y
Solution: This equation can be written in the form
( y  x )dx  ( x  y )dy  0.
which is an exact equation.
In this case, the solution in implicit form is
1 2 1 2
xy  x  y  C .
2
2
i.e. ,
2 xy  x 2  y 2  C .
1.65
Integrating Factors
• If the equation is not exact we can consider
multiplying it by a function  ( x , y )  0
such that the new equation is exact:
M ( x , y )dx  N ( x , y )dy  0
  ( x, y)
  Mdx   Ndy  0
1.66
Finding Integrating Factors
• For exactness we require:

M 
N
( M ) ( N )

M

N

y
y
x
x
y
x


N M
M
N
(

)
y
x
x
y
This equation can be simplified in special cases, two.
of which we treat next
1.67
• If we choose
  ( x)
d
M
N
N

dx
y
x
where
Let
M N

y
x
N

M N

d
y
x


dx
N
is just a function of x.
1  M N 
 ( x)  


N  y
x 


N M
M
N
(

)
y
x
x
y
1.68
Now solve for
1 d  1   M N 
1 d
 

  ( x)

 dx N  y x   dx

1

d    ( x )dx 
1
  d     ( x )dx
 ln     ( x )dx    e
  ( x ) dx
1.69
Conversely, if
Let
Then
M N

y
x
N
is just a function of x.
1  M N 
 ( x)  


N  y
x 
e
  ( x ) dx
is an integrating factor for Equation.
1.70
In a similar fashion, if equation has an integrating
factor that depends only y, then
M N

is just a function of y.
y
x
M N
M

y
x is just a function of y.
Conversely, if
M
M  N

Let  ( y )  y x .
Then   e  ( y ) dy
M
is an integrating factor for Equation.
1.71
Integrating Factors: Example
2
2
(2
x

y
)
dx

(
x
y  x )dy  0
Example: Solve
Solution: M  2 x 2  y , N  x 2 y  x ,
M
N
 1  2 xy  1 
.
y
x
The equation is not exact.
We compute
M N

2(1  xy )
2
y x 1  (2 xy  1)


 .
2
N
x y x
 x(1  xy )
x
1.72
So an integrating factor for the equation is given by
e
2

dx

x
2
x .
When we multiplying by   x
we get the exact equation
(2  yx 2 )dx  ( y  x 1 )dy  0.
2
,
Solving this equation, we obtain the implicit
solution
2
y
2 x  yx 1 
 C.
2
1.73
Note that the solution x=0 is lost in multiplying by
x .
2
2
y
Hence, 2 x  yx 1 
and
C
2
are solutions to the given equation.
x0
1.74
Linear Differential Equations
• Recall a differential equation is linear if it
can be written:
y ' p( x ) y  q( x )
• If q(x)=0 the equation is homogeneous,
otherwise the equation is nonhomogeneous.
1.75
Homogeneous Linear Case
• Separating variables and integrating:
y '  p( x) y  0
dy

  p ( x)dx
y
~
 ln | y |   p ( x)dx  c
 y ( x)  ce 
 p ( x ) dx
1.76
Nonhomogeneous Linear Case
• Nonhomogeneous linear equations can
be solved using the integrating factor
method. First rewrite the differential
equation as:
i.e.
( py  q )dx  dy  0
M  py  q,
N 1
1.77
Finding the integrating factor
• To find the integrating factor we first compute
1
N
 M N 


 ( M  py  q ,
x 
 y
N  1)
1   ( py  q )  (1) 
 


1
y
x 
 p, where p is a function of x
1.78
Nonhomogeneous Linear Case
• We can now find an integrating factor
pdx

( x)  e
• Now multiply the differential equation by
the integrating factor:
'
p dx
p dx
p dx





e
( y ' py )   e
y  e
q


1.79
Nonhomogeneous Linear
Case
pdx
pdx


(e
y )  e
q
(Integrate both sides w.r.t. x )
pdx
pdx


e
y   e q dx  c
 ye 
 pdx
pdx

(  e qdx  c )
or y   1 ( x )(   ( x )qdx  c ),
pdx

where  ( x )  e
.
1.80
Integrating Factors: Example
Example: Solve ydx  (2 x  ye y )dy  0
Solution:
Here
M  y, N  2 x  ye y ,
M
N
1 2
.
y
x
The equation is not exact.
We compute
M N

y x 1  2 1

 .
M
y
y
1.81
So an integrating factor for is given by
e
1
 dy
y
 y.
When we multiplying by   y ,
we get the exact equation
y 2dx  (2 xy  y 2e y )dy  0
Solving this equation, we obtain
2
2
y
xy  ( y  2 y  2)e  C
as the solution in implicit form.
1.82
In general, integrating factors are difficult to uncover.
If a differential equation does not have one of the forms
given above, then a search for an integrating factor likely
will not be successful, and other methods of solution are
recommended.
1.83
Example : Solve
dy
x
x 2
   1  ( ) ( y  0).
dx
y
y
Solution: Rewriting this equation in differential form,
we have ( x  x 2  y 2 )dx  ydy  0.
This differential equation is not exact, and no
integrating factor is immediately apparent.
Note however, that if terms are strategically
regrouped, the DE can be rewritten as
( xdx  ydy )  x 2  y 2 dx  0.
1.84
( xdx  ydy )  x 2  y 2 dx  0.
(1)
The first group of terms has many integrating
factors (see Table 2). One of these factors, namely
1
 ( x, y) 
,
2
2
x y
is an integrating factor for the entire equation.
1
Multiplying (1) by  ( x, y ) 
,
x2  y2
we find ( xdx  ydy )
 dx  0.
(2)
2
2
x y
1.85
Since (2) is exact, it can be solved using the steps
described previously.
Alternatively, we note from Table 1,
( xdx  ydy )
x2  y2
 d x2  y2
so that (2) can be rewrite as
d x 2  y 2  dx  0.
Integrating both sides of this last equation, we find
1.86
x 2  y 2  x  c.
or equivalently,
y 2  c(c  2 x ).
1.87
Solve
ydx  ( y  x )dy  0.
Solution: Here M  y , N  y  x
and, since
M
N
 1,
 1,
y
x
The differential equation is not exact.
M N

Since
2 is a function of y alone.
y x

M
y
we have an integrating factor
1.88
 ( x, y)  e


2
y
1
 2.
y
1
Multiplying the given DE by  ( x , y )  2 ,
y
we obtain the exact equation
1
( y  x)
dx 
dy  0,
2
y
y
or equivalently,
ydx  xdy 1
 dy  0,
2
y
y
1.89
Integrating both sides of this last equation, we find
x
 ln | y | c .
y
(ii) Note that the differential equation can be rewritten as
( ydx  xdy )  ydy  0.
The first group of terms has many integrating
factors (see Table 1). One of these factors, namely
1.90
1
 ( x, y)  2 ,
y
is an integrating factor for the entire equation.
1
Multiplying (1) by  ( x , y )  2 ,
y
we find ( ydx  xdy )  1 dy  0.
y2
y
Integrating both sides of this last equation, we find
x
 ln | y | c .
y
1.91
(iii) Rewriting this equation in the derivative form,
dy
y

.
dx x  y
(3)
then we see that the equation (3) is homogeneous.
dy
dv
y
v x .
Now let v 
and recall that
dx
dx
x
With these substitution, equation (3) becomes
dv
v
1 v
dx
x v 
or x 2 dv 
dx
1 v
v
x
1.92
The above equation is separable, and , by separating the
variables and integrating, we obtain
1
  ln | v | ln | x | C .
v
y
Finally , we substitute
for v to get
x
x
 ln | y |  C  0
y
1.93
(iv) We rewrite the differential equation in the form
dx x  y
dx x

or
  1,
dy
y
dy y
which is linear equation.
Its solution is
xe

1
dy
y
( e


1
dy
y
dy  c )  y(ln | y |  c ).
Also note that x=0 is a solution.
1.94
Table 1
1.95
Table 2
1.96
1.97
Exercise Solve
( x  y 2 )dx  y(1  x )dy  0.
1.98
1.99
1.100
1.101
1.102
1.103