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Introduction to Ordinary
Differential Equations
Chapter 1
Overview
II. Classification of Solutions
Chapter 1: Introduction to Differential Equations
I. Definitions
I. Definitions
Learning Objective
At the end of the section, you should be able to
define a differential equation and classify
differential equations by type, order and linearity.
I. Definitions
Basic Example
Consider
x
e
x
f 2
)
( 
x
e
x
f 2
'
2
)
( 
0
2
2
)
(
2
)
( 2
2
'



 x
x
e
e
x
f
x
f
satisfies the Differential Equation:
f
0
2
'

 y
y
I. Definitions
What is a Differential Equation
A differential equation (DE) is an equation containing the
derivatives of one or more dependent variables with
respect to one or more independent variables.
I. Definitions
Examples
0
2
3
)
1 '
'


 y
y
y
x
dt
dy
dt
dx
4
2
3
)
3 


1
)
2 3
2


 x
y
x
dx
dy
x
I. Definitions
Classification
Differential equations (DE) can be classified by:
• TYPE
• ORDER
• LINEARITY.
I. Definitions
Classification by Type
Two types of Differential equations (DE) exist:
• ORDINARY DIFFERENTIAL EQUATION (ODE).
An equation containing only ordinary derivatives of one
or more dependent variables with respect to a SINGLE
independent variable is said to be an Ordinary
Differential Equation (ODE).
I. Definitions
Examples of ODE
x
e
y
dx
dy

 5
)
1
0
6
)
2 2
2


 y
dx
dy
dx
y
d
y
x
dt
dy
dt
dx


 2
)
3
I. Definitions
• PARTIAL DIFFERENTIAL EQUATIONS (PDE).
An equation containing partial derivatives of one or more
dependent variables with respect to TWO or more
independent variables is said to be a Partial Differential
Equation (PDE).
I. Definitions
Examples of PDE
0
)
1 2
2
2
2






y
u
x
u
t
u
t
u
x
u








2
)
2 2
2
2
2
x
v
y
u






)
3
I. Definitions
Classification by Order
The order of a differential equation (ODE or PDE)
is the order of the highest derivative in the
equation.
I. Definitions
Examples of Orders
x
e
y
dx
dy

 3
5
0
6
2
2


 y
dx
dy
dx
y
d
x
e
y
dx
dy
dx
y
d








 4
5
3
2
2
is of order 1 (or first-order)
is of order 2
is of order 2
I. Definitions
First-order ODEs are occasionally written in differential
form :
Remark
0
)
,
(
)
,
( 
 dy
y
x
N
dx
y
x
M
I. Definitions
Classification by Linearity
   
   
     
x
g
y
x
a
y
x
a
...
y
x
a
y
x
a n
n
n
n 




 
 0
1
1
1
The general form for an nth-order ODE is:
       
x
g
y
x
c
y
x
b
y
x
a 





The general form for an 2nd-order ODE is:
I. Definitions
Examples for linear ODEs
  0
4
)
1 

 dy
x
dx
x
y
0
2
)
2 




 y
y
y
x
e
y
dx
dy
x
dx
y
d


 5
)
3 3
3
x
y
y
x 


 4
I. Definitions
Examples for non-linear ODEs
  x
e
y
y
y
- 

 2
1
)
1
0
sin
)
2 2
2

 y
dx
y
d
0
)
3 2
4
4

 y
dx
y
d
I. Definitions
Example:
For each of the following ODEs, determine the order and
state whether it is linear or non-linear:
I. Definitions
Solution:
ODE Order Linearity
  0
cos 

 dx
x
xy
dy
0
6
2
2



dt
dQ
dt
Q
d
  0
2
2









 xy
y
y
y
x
y
0




 y
y
x
ey
Linear
Linear
Non-linear
Non-linear
1
2
3
2
I. Definitions
Solution:
ODE Order Linearity


 sin



  0
1
2


 xdy
dx
y
2
2
2
1 







dx
dy
dx
y
d
Linear
Non-linear
Non-linear
1
1
2


 2
sin



I. Definitions
Exercise-I:
For each of the following ODEs, determine the order and
state whether it is linear or non-linear:
t
y
dt
dy
t
dt
y
d
t sin
2
2
2
2



t
e
y
dt
dy
t
dt
y
d
y 


 2
2
2
)
1
(
1
2
2
3
3
4
4




 y
dt
dy
dt
y
d
dt
y
d
dt
y
d
0
2

 ty
dt
dy
t
y
t
dt
y
d
sin
)
sin(
2
2



3
2
3
3
)
(cos t
y
t
dt
dy
t
dt
y
d



t
ty
dt
dy
t tan
1
2



II. Classification of Solutions
Learning Objective
At the end of this section, you should be able to
• verify the solutions to a given ODE
• identify the different types of solutions of an
ODE.
• Define IVP, BVP
Definition:
A solution of a DE is a function that satisfies the DE
identically for all in an interval , where is the
independent variable.
y
x I x
II. Classification of Solutions
Example
x
y ln

)
,
0
( 

I
is a solution of the DE: 0
'
" 
y
xy
x
y ln

x
y
1
'
2
1
"
x
y 

x
x
x
y
xy
1
)
1
(
'
'
' 2




Indeed,
II. Classification of Solutions
0
1
1




x
x
Definition:
A solution in which the dependent variable is expressed
solely in terms of the independent variable and constants
is said to be an explicit solution.
II. Classification of Solutions
Definition:
A solution in which the dependent and the independent
variables are mixed in an equation is said to be an implicit
solution.
II. Classification of Solutions
Examples:
x
y ln
 is an explicit solution of the DE: 0
'
" 
y
xy
9
2
2

 y
x
9
2
2

 y
x
is an implicit solution of the DE: 0
' 
x
yy
Indeed:
Implicit differentiation: 0
'
2
2 
 yy
x
0
'
 yy
x
1)
2)
II. Classification of Solutions
General or Particular solution
Example:
Consider the ODE: 0
' 
y
y
x
e
y  is a solution (particular)
x
ce
y  (where c is a constant) is a solution (general)
II. Classification of Solutions
x
e
y 2
 is also a solution (particular)
General or Particular solution
• A solution of a DE that is free of arbitrary parameters is
called a particular solution.
II. Classification of Solutions
Definitions:
• A solution of a DE representing all possible solutions is
called a general solution.
Example
II. Classification of Solutions
x
ce
y  is a 1-parameter family of solutions of the DE
0
' 
y
y
x
x
de
ce
y 

 is a 2-parameter family of solutions of the DE
0
" 
y
y
Example:
Verify that the indicated function is an explicit solution of
the given DE :
II. Classification of Solutions
Example:
1)
II. Classification of Solutions
2
;
0
2
x
e
y
y
y





2
2
1
'
x
e
y



2
2
)
2
1
(
2
'
2
x
x
e
e
y
y






0
2
2






x
x
e
e
Example:
2)
II. Classification of Solutions
t
e
y
;
y
dt
dy 20
5
6
5
6
24
20 




t
e
dt
dy
y 20
)
5
6
(
20
' 



 t
e 20
24 










 
 t
t
e
e
y
dt
dy 20
20
5
6
5
6
20
24
20
t
t
e
e 20
20
24
24
24 



 24

3)
II. Classification of Solutions
x
cos
e
y
;
y
y
y x
2
0
13
6 3







 
x
e
x
e
y x
x
2
sin
2
2
cos
3
' 3
3


 x
e
y x
2
sin
2
3 3


x
e
x
e
y
y x
x
2
cos
4
2
sin
6
'
3
" 3
3


 y
x
e
x
e
y x
x
4
2
sin
6
)
2
sin
2
3
(
3 3
3









 y
y
y 13
6 y
x
e
y
x
e
y x
x
13
)
2
sin
2
3
(
6
2
sin
12
5 3
3




Example:
x
e
y x
2
sin
12
5 3


0
13
2
sin
12
18
2
sin
12
5 3
3





 y
x
e
y
x
e
y x
x
4)
II. Classification of Solutions
   
x
tan
x
sec
ln
x
cos
y
;
x
tan
y
y 






      x
x
x
x
x
y sec
cos
tan
sec
ln
sin
' 



   1
tan
sec
ln
sin 

 x
x
x
  x
x
x
x
x
y sec
sin
tan
sec
ln
cos
" 

   x
x
x
x tan
tan
sec
ln
cos 


   
x
x
x
x
x
x
x
y
y tan
sec
ln
cos
tan
tan
sec
ln
cos 






 x
tan

Example:
5)
II. Classification of Solutions
  t
t
e
c
e
c
P
P
P
P
1
1
1
;
1





t
t
e
c
e
c
P
1
1
1

 
 2
1
1
1
1
1
1
1
'
t
t
t
t
t
e
c
e
c
e
c
e
c
e
c
P




   2
1
1
2
1
2
2
1
2
2
1
1
1
1 t
t
t
t
t
t
e
c
e
c
e
c
e
c
e
c
e
c






  











 t
t
t
t
e
c
e
c
e
c
e
c
P
P
1
1
1
1
1
1
1
1 











 t
t
t
t
t
e
c
e
c
e
c
e
c
e
c
1
1
1
1
1
1
1
1
 
'
1
2
1
1
P
e
c
e
c
t
t



Example:
6)
II. Classification of Solutions
x
x
xe
c
e
c
y
;
y
dx
dy
dx
y
d 2
2
2
1
2
2
0
4
4 




 
x
x
x
xe
e
c
e
c
y 2
2
2
2
1 2
2
' 

 x
x
x
xe
c
e
c
e
c 2
2
2
2
2
1 2
2 


  x
x
x
x
e
c
y
e
c
xe
c
e
c 2
2
2
2
2
2
2
1 2
2 




x
e
c
y
y 2
2
2
'
2
" 



 y
y
y 4
'
4
" 


 y
y
e
c
y x
4
'
4
2
'
2 2
2 y
y
e
c x
4
'
2
2 2
2 

  0
4
2
2
2 2
2
2
2 



 y
e
c
y
e
c x
x
Example:
Exercise-II:
Verify if the indicated functions are explicit solutions of the
given DE :
II. Classification of Solutions
t
t
t
t
t
y
t
t
y
y
t
t
t
y
t
t
y
t
y
ty
y
t
t
t
y
t
t
y
t
y
ty
y
t
t
e
t
y
t
t
y
t
y
y
y
t
t
y
t
y
y
t
t
sin
cos
ln
)
(cos
)
(
;
2
0
,
sec
'
'
)
5
ln
)
(
,
)
(
;
0
,
0
4
'
5
'
'
)
4
)
(
,
)
(
;
0
,
0
'
3
'
'
2
)
3
3
)
(
,
3
)
(
,
3
4
)
2
3
,
)
1
2
2
2
1
2
1
2
2
1
1
2
2
1
)
3
(
)
4
(
2
2


































Definition
A DE with initial conditions on the unknown function and its
derivatives, all given at the same value of the independent
variable, is called an initial-value problem, IVP.
0
x
II. Classification of Solutions
Examples
3
)
0
(
,
0
)
1 


 y
y
y
25
)
1
(
'
,
0
)
2 


 y
y
y
5
)
2
(
,
0
'
2
'
'
)
3 


 y
y
y
y
II. Classification of Solutions
Definition
A DE with initial conditions on the unknown function and its
derivatives, all given at different values (e.g. at and )
of the independent variable, is called a boundary-value
problem, BVP.
0
x 1
x
II. Classification of Solutions
Examples
  2
2
,
1
;
2
)
1 














 y
y
e
y
y x
    1
1
,
1
0
;
2
)
2 





 y
y
e
y
y x
II. Classification of Solutions
Examples
Find the solution of the IVP or BVP if the general solution is the
given one:
  ,
2
3
;
0
)
1 


 y
y
y   x
e
c
x
y 
 1
  3
1
3 
 e
c
y   2
3 
y
2
3
1 

e
c
3
1 2e
c 
  x
x
e
e
e
x
y 


 3
3
2
2
solution of the IVP:
II. Classification of Solutions
1
6
,
0
8
;
0
4
)
2 



















y
y
y
y   x
c
x
c
x
y 2
cos
2
sin 2
1 

8
2
cos
8
2
sin
8
2
1



c
c
y 







2
2
2
1 c
c


0
8







y  1
2 c
c 

Examples
II. Classification of Solutions
solution of the BVP:
6
2
cos
6
2
sin
6
2
1



c
c
y 







2
2
3 2
1 c
c


1
6







y  1
2
3 2
1

 c
c
2
3 1
1 
 c
c
1
3
2
1


c

 
1
3
2
2



c
 
x
x
y 2
cos
2
sin
1
3
2



Examples
II. Classification of Solutions
  x
c
x
c
x
y 2
cos
2
sin 2
1 

  0
cos
0
sin
0 2
1 c
c
y 
 2
c

  1
0 
y  1
2 
c
  ,
2
2
,
1
0
;
0
4
)
3 












y
y
y
y



cos
sin
2
2
1 c
c
y 







2
c


2
2







y  2
2 

c
2
2 

c
1
2 
c
IMPOSSIBLE NO SOLUTION
Examples
II. Classification of Solutions
Exercise-III
1) Determine and so that
will satisfy the conditions :
0
8







y 2
8









y
1
c 2
c   1
2
cos
2
sin 2
1 

 x
c
x
c
x
y
2) Determine and so that
will satisfy the conditions :
  x
e
c
e
c
x
y x
x
sin
2
2
2
1 


1
c 2
c
  0
0 
y   1
0 

y
II. Classification of Solutions
End Chapter 1

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first order differential equation for grad

  • 2. Overview II. Classification of Solutions Chapter 1: Introduction to Differential Equations I. Definitions
  • 3. I. Definitions Learning Objective At the end of the section, you should be able to define a differential equation and classify differential equations by type, order and linearity.
  • 4. I. Definitions Basic Example Consider x e x f 2 ) (  x e x f 2 ' 2 ) (  0 2 2 ) ( 2 ) ( 2 2 '     x x e e x f x f satisfies the Differential Equation: f 0 2 '   y y
  • 5. I. Definitions What is a Differential Equation A differential equation (DE) is an equation containing the derivatives of one or more dependent variables with respect to one or more independent variables.
  • 6. I. Definitions Examples 0 2 3 ) 1 ' '    y y y x dt dy dt dx 4 2 3 ) 3    1 ) 2 3 2    x y x dx dy x
  • 7. I. Definitions Classification Differential equations (DE) can be classified by: • TYPE • ORDER • LINEARITY.
  • 8. I. Definitions Classification by Type Two types of Differential equations (DE) exist: • ORDINARY DIFFERENTIAL EQUATION (ODE). An equation containing only ordinary derivatives of one or more dependent variables with respect to a SINGLE independent variable is said to be an Ordinary Differential Equation (ODE).
  • 9. I. Definitions Examples of ODE x e y dx dy   5 ) 1 0 6 ) 2 2 2    y dx dy dx y d y x dt dy dt dx    2 ) 3
  • 10. I. Definitions • PARTIAL DIFFERENTIAL EQUATIONS (PDE). An equation containing partial derivatives of one or more dependent variables with respect to TWO or more independent variables is said to be a Partial Differential Equation (PDE).
  • 11. I. Definitions Examples of PDE 0 ) 1 2 2 2 2       y u x u t u t u x u         2 ) 2 2 2 2 2 x v y u       ) 3
  • 12. I. Definitions Classification by Order The order of a differential equation (ODE or PDE) is the order of the highest derivative in the equation.
  • 13. I. Definitions Examples of Orders x e y dx dy   3 5 0 6 2 2    y dx dy dx y d x e y dx dy dx y d          4 5 3 2 2 is of order 1 (or first-order) is of order 2 is of order 2
  • 14. I. Definitions First-order ODEs are occasionally written in differential form : Remark 0 ) , ( ) , (   dy y x N dx y x M
  • 15. I. Definitions Classification by Linearity               x g y x a y x a ... y x a y x a n n n n         0 1 1 1 The general form for an nth-order ODE is:         x g y x c y x b y x a       The general form for an 2nd-order ODE is:
  • 16. I. Definitions Examples for linear ODEs   0 4 ) 1    dy x dx x y 0 2 ) 2       y y y x e y dx dy x dx y d    5 ) 3 3 3 x y y x     4
  • 17. I. Definitions Examples for non-linear ODEs   x e y y y -    2 1 ) 1 0 sin ) 2 2 2   y dx y d 0 ) 3 2 4 4   y dx y d
  • 18. I. Definitions Example: For each of the following ODEs, determine the order and state whether it is linear or non-linear:
  • 19. I. Definitions Solution: ODE Order Linearity   0 cos    dx x xy dy 0 6 2 2    dt dQ dt Q d   0 2 2           xy y y y x y 0      y y x ey Linear Linear Non-linear Non-linear 1 2 3 2
  • 20. I. Definitions Solution: ODE Order Linearity    sin      0 1 2    xdy dx y 2 2 2 1         dx dy dx y d Linear Non-linear Non-linear 1 1 2    2 sin   
  • 21. I. Definitions Exercise-I: For each of the following ODEs, determine the order and state whether it is linear or non-linear: t y dt dy t dt y d t sin 2 2 2 2    t e y dt dy t dt y d y     2 2 2 ) 1 ( 1 2 2 3 3 4 4      y dt dy dt y d dt y d dt y d 0 2   ty dt dy t y t dt y d sin ) sin( 2 2    3 2 3 3 ) (cos t y t dt dy t dt y d    t ty dt dy t tan 1 2   
  • 22. II. Classification of Solutions Learning Objective At the end of this section, you should be able to • verify the solutions to a given ODE • identify the different types of solutions of an ODE. • Define IVP, BVP
  • 23. Definition: A solution of a DE is a function that satisfies the DE identically for all in an interval , where is the independent variable. y x I x II. Classification of Solutions
  • 24. Example x y ln  ) , 0 (   I is a solution of the DE: 0 ' "  y xy x y ln  x y 1 ' 2 1 " x y   x x x y xy 1 ) 1 ( ' ' ' 2     Indeed, II. Classification of Solutions 0 1 1     x x
  • 25. Definition: A solution in which the dependent variable is expressed solely in terms of the independent variable and constants is said to be an explicit solution. II. Classification of Solutions
  • 26. Definition: A solution in which the dependent and the independent variables are mixed in an equation is said to be an implicit solution. II. Classification of Solutions
  • 27. Examples: x y ln  is an explicit solution of the DE: 0 ' "  y xy 9 2 2   y x 9 2 2   y x is an implicit solution of the DE: 0 '  x yy Indeed: Implicit differentiation: 0 ' 2 2   yy x 0 '  yy x 1) 2) II. Classification of Solutions
  • 28. General or Particular solution Example: Consider the ODE: 0 '  y y x e y  is a solution (particular) x ce y  (where c is a constant) is a solution (general) II. Classification of Solutions x e y 2  is also a solution (particular)
  • 29. General or Particular solution • A solution of a DE that is free of arbitrary parameters is called a particular solution. II. Classification of Solutions Definitions: • A solution of a DE representing all possible solutions is called a general solution.
  • 30. Example II. Classification of Solutions x ce y  is a 1-parameter family of solutions of the DE 0 '  y y x x de ce y    is a 2-parameter family of solutions of the DE 0 "  y y
  • 31. Example: Verify that the indicated function is an explicit solution of the given DE : II. Classification of Solutions
  • 32. Example: 1) II. Classification of Solutions 2 ; 0 2 x e y y y      2 2 1 ' x e y    2 2 ) 2 1 ( 2 ' 2 x x e e y y       0 2 2       x x e e
  • 33. Example: 2) II. Classification of Solutions t e y ; y dt dy 20 5 6 5 6 24 20      t e dt dy y 20 ) 5 6 ( 20 '      t e 20 24               t t e e y dt dy 20 20 5 6 5 6 20 24 20 t t e e 20 20 24 24 24      24 
  • 34. 3) II. Classification of Solutions x cos e y ; y y y x 2 0 13 6 3          x e x e y x x 2 sin 2 2 cos 3 ' 3 3    x e y x 2 sin 2 3 3   x e x e y y x x 2 cos 4 2 sin 6 ' 3 " 3 3    y x e x e y x x 4 2 sin 6 ) 2 sin 2 3 ( 3 3 3           y y y 13 6 y x e y x e y x x 13 ) 2 sin 2 3 ( 6 2 sin 12 5 3 3     Example: x e y x 2 sin 12 5 3   0 13 2 sin 12 18 2 sin 12 5 3 3       y x e y x e y x x
  • 35. 4) II. Classification of Solutions     x tan x sec ln x cos y ; x tan y y              x x x x x y sec cos tan sec ln sin '        1 tan sec ln sin    x x x   x x x x x y sec sin tan sec ln cos "      x x x x tan tan sec ln cos        x x x x x x x y y tan sec ln cos tan tan sec ln cos         x tan  Example:
  • 36. 5) II. Classification of Solutions   t t e c e c P P P P 1 1 1 ; 1      t t e c e c P 1 1 1     2 1 1 1 1 1 1 1 ' t t t t t e c e c e c e c e c P        2 1 1 2 1 2 2 1 2 2 1 1 1 1 t t t t t t e c e c e c e c e c e c                      t t t t e c e c e c e c P P 1 1 1 1 1 1 1 1              t t t t t e c e c e c e c e c 1 1 1 1 1 1 1 1   ' 1 2 1 1 P e c e c t t    Example:
  • 37. 6) II. Classification of Solutions x x xe c e c y ; y dx dy dx y d 2 2 2 1 2 2 0 4 4        x x x xe e c e c y 2 2 2 2 1 2 2 '    x x x xe c e c e c 2 2 2 2 2 1 2 2      x x x x e c y e c xe c e c 2 2 2 2 2 2 2 1 2 2      x e c y y 2 2 2 ' 2 "      y y y 4 ' 4 "     y y e c y x 4 ' 4 2 ' 2 2 2 y y e c x 4 ' 2 2 2 2     0 4 2 2 2 2 2 2 2      y e c y e c x x Example:
  • 38. Exercise-II: Verify if the indicated functions are explicit solutions of the given DE : II. Classification of Solutions t t t t t y t t y y t t t y t t y t y ty y t t t y t t y t y ty y t t e t y t t y t y y y t t y t y y t t sin cos ln ) (cos ) ( ; 2 0 , sec ' ' ) 5 ln ) ( , ) ( ; 0 , 0 4 ' 5 ' ' ) 4 ) ( , ) ( ; 0 , 0 ' 3 ' ' 2 ) 3 3 ) ( , 3 ) ( , 3 4 ) 2 3 , ) 1 2 2 2 1 2 1 2 2 1 1 2 2 1 ) 3 ( ) 4 ( 2 2                                  
  • 39. Definition A DE with initial conditions on the unknown function and its derivatives, all given at the same value of the independent variable, is called an initial-value problem, IVP. 0 x II. Classification of Solutions
  • 40. Examples 3 ) 0 ( , 0 ) 1     y y y 25 ) 1 ( ' , 0 ) 2     y y y 5 ) 2 ( , 0 ' 2 ' ' ) 3     y y y y II. Classification of Solutions
  • 41. Definition A DE with initial conditions on the unknown function and its derivatives, all given at different values (e.g. at and ) of the independent variable, is called a boundary-value problem, BVP. 0 x 1 x II. Classification of Solutions
  • 42. Examples   2 2 , 1 ; 2 ) 1                 y y e y y x     1 1 , 1 0 ; 2 ) 2        y y e y y x II. Classification of Solutions
  • 43. Examples Find the solution of the IVP or BVP if the general solution is the given one:   , 2 3 ; 0 ) 1     y y y   x e c x y   1   3 1 3   e c y   2 3  y 2 3 1   e c 3 1 2e c    x x e e e x y     3 3 2 2 solution of the IVP: II. Classification of Solutions
  • 44. 1 6 , 0 8 ; 0 4 ) 2                     y y y y   x c x c x y 2 cos 2 sin 2 1   8 2 cos 8 2 sin 8 2 1    c c y         2 2 2 1 c c   0 8        y  1 2 c c   Examples II. Classification of Solutions
  • 45. solution of the BVP: 6 2 cos 6 2 sin 6 2 1    c c y         2 2 3 2 1 c c   1 6        y  1 2 3 2 1   c c 2 3 1 1   c c 1 3 2 1   c    1 3 2 2    c   x x y 2 cos 2 sin 1 3 2    Examples II. Classification of Solutions
  • 46.   x c x c x y 2 cos 2 sin 2 1     0 cos 0 sin 0 2 1 c c y   2 c    1 0  y  1 2  c   , 2 2 , 1 0 ; 0 4 ) 3              y y y y    cos sin 2 2 1 c c y         2 c   2 2        y  2 2   c 2 2   c 1 2  c IMPOSSIBLE NO SOLUTION Examples II. Classification of Solutions
  • 47. Exercise-III 1) Determine and so that will satisfy the conditions : 0 8        y 2 8          y 1 c 2 c   1 2 cos 2 sin 2 1    x c x c x y 2) Determine and so that will satisfy the conditions :   x e c e c x y x x sin 2 2 2 1    1 c 2 c   0 0  y   1 0   y II. Classification of Solutions