Differential Equations
Summary 1
Definition: an equation containing derivatives of one or more unknown functions with
respect to one or more independent variables, is said to be a differential equation (DE)
Independent variables: one or more variables with respect to which the derivatives are
taken.
Dependent variables: the variables of which the derivatives occur.
Classification of differential equations: by type, by order, and by linearity
Classification of DE by type:
 Ordinary differential equation (ODE): contains only ordinary derivatives of one
or more functions with respect to a single variable.
Examples :
d2y
dx
2
x
dy
 5y  sin x,
dx
dx dy

 xy  3
dt dt
 Partial Differential Equations (PDE): involves partial derivatives
Example :
u  v

0
y  x
Mathematical Notation:
Notation
Independent
Variable
Leibniz Notation
x
Dependent
Variable
y
Leibniz Notation
t
x
Leibniz Notation
t
y
Prime notation
Dot notation
x
t
y
s
Subscript
notation
x
w
Expression
dy d 2 y d  dy  d 3 y d  d 2 y 
,


 ,
dx dx 2 dx  dx  dx 3 dx  dx 2 
dx d 2 x d  dx  d 3 x d  d 2 x 
,
  ,

dt dt 2 dt  dt  dt 3 dt  dt 2 
dy d 2 y d  dy  d 3 y d  d 2 y 
,
  ,

dt dt 2 dt  dt  dt 3 dt  dt 2 
y , y , y 
s 
ds
d 2s
d 3s
, s 
, s 
dt
dt 2
dt 3
w
 2 w   w 
wx 
, w xx 



x
x 2 x  x 
  w 
w xy 


y  x 
-1-
Classification of DE by order: the order of a differential equation is the order of the
highest derivative in the equation.
Order: the order of a differential equation is the highest ordered derivative appearing in
the equation. The symbol y
(n )

dn y
represents the nth derivative.
dx n
An nth order differential equation can be express by the general form F (t , y, y ,... y ( n) )  0
Example: t 2 y   ty   5 y  sin t is an ordinary differential equation of order 3.
Classification of DE by Linearity:
 Linear: a DE equation is said to be linear if it can be express in the form
a n ( x)
dn y
dx
n
 a n 1 ( x)
d n 1y
dx
n 1
 . . .  a1 x 
dy
 a 0 ( x)y  g( x) , neither y nor any of its
dx
derivatives is raised to an exponent other than 1. The coefficients of y and its derivatives
are functions of x.
 Nonlinear: at least y or any of its derivatives is raised to a power other than 1 or at
least one of the coefficient functions is a function of y.
Example: yy’ –3y =2 is nonlinear.
1  y   2 y  3 is nonlinear
Homogeneous and non-homogeneous equations:
In the equation a n ( x)
dn y
dx n
 a n 1 ( x)
d n 1y
dx n 1
 . . .  a1 x 
dy
 a 0 ( x)y  g( x) , if g(x)=0, the
dx
DE is said to be linear and homogenous. If g(x)  0 it is said to be non-homogenous.
Notation:
dy
d 2 y d  dy  dy 
d 3 y d  dy  


 y ,



y
,

 

  y   y (3)
2
3
dx
dx  dx  dx
dx  dx 
dx
dx
y (n ) 
y 
dn y
dx n
dy
d2y
d3y
, y 
, y 
dt
dt 2
dt 3
Solution of a differential equation: any function y defined on some interval I, which
when substituted into a differential equation reduces the equation to an identity, is said
to be a solution of the equation on the interval I. The interval I is called the interval of
definition, or the interval of existence, or the interval of validity, or the domain of
the solution of the DF. If can be an open interval (a, b), a closed interval [a, b], an
infinite interval, and so on.
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Example: the function y 
1
is a solution of the DE ty  y  0 defined on an interval not
t
containing 0.
( , 0)  (0,  ) is the interval of definition of the solution. Note:
1
1
y   t 1  y  1t  2  
t
t2
 1
ty  y  t 
 t2
 1
1 1
      0
t t
 t
Family of solutions: a solution of a DE containing an arbitrary constant c is called a
one-parameter family of solutions. c is called a parameter.
When solving an nth-order differential equation we need an n-parameter family of
solutions.
A particular solution of a DE, is a solution of the equation that is free of parameter.
1. Show that ln 2  X  t a solution dX  (2  X )(1  X )
1 X
dt
2 X
dt
1
1
 t  t  ln( 2  x)  ln(1  x ) 

( 1) 
( 1)
1 X
dx 2  x
1 x
 (1  x)  ( 2  x)
dt
1
1
1




dx 2  x 1  x
( 2  x)(1  x
( 2  x)(1  x)
dx

 ( 2  x )(1  x )
dt
ln
2. Verify that y  c1e2t  c2te2t is a solution of y   4y   4y  0
y  c1e 2t  c 2 te 2t  y   2c1e 2t  2c 2 te 2t  c 2 e 2t
y   4c1e 2t  4c 2 te 2t  2c 2 e 2t  2c 2 e 2t  4c1e 2t  4c 2 te 2t  4c 2 e 2t
 y   4y   4y

 
 4c1e 2t  4c 2 te 2t  4c 2 e 2t  4 2c1e 2t  2c 2 te 2t  c 2 e 2t  4 c1e 2t  c 2 te 2t

 4c1e 2t  4c 2 te 2t  4c 2 e 2t  8c1e 2t  8c 2 te 2t  4c 2 e 2t  4c1e 2t  4c 2 te 2t



 4c 2 te 2t  4c 2 te 2t  8c 2 te 2t  4c1e 2t  4c1e 2t  8c1e 2t  4c 2 e 2t  4c 2 e 2t
 0 0 0
0
-3-

3. Verify that the system
 d 2x
1 t

 2  4y  e t
 x  cos 2t  sin 2t  5 e

is a solution of  dt2

1 t
 y   cos 2t  sin 2t  e
 d y  4x  e t
5

 dt 2
Solution:
1 t
1 t

 dx
 x  cos 2t  sin 2t  5 e
 dt   2 sin 2t  2 cos 2t  5 e


1
dy
1
y   cos 2t  sin 2t  et
  2 sin 2t  2 cos 2t  et
5
5

 dt
d2x
1
 4 cos 2t  4 sin 2t  e t

2
5
 dt
 2
 d y  4 cos 2t  4 sin 2t  1 e t
 2
5
 dt
d2x
1 

 4  cos 2t  sin 2t  e t   e t  4y  e t

5 
 2

  dt 2
 d y  4 cos 2t  sin 2t  1 e t   e t  4x  e t



5 

 dt 2
 t 2 , t  0
4. Verify that the piecewise defined function y  
2
t ,
t0
is a solution of the
differential equation ty   2 y  0 on the int erval (, )
Solution:
  t 2 , t  0
y 2
 t , t  0
  2t 2 , t  0
  2t , t  0
 y  
 ty   
 2t 2 , t  0
 2t , t  0
  2t 2 , t  0   t 2 , t  0   2t 2  2t 2 , t  0
 ty   2y  
 2 2

 0, for ( ,  )
 2t 2 , t  0
 t , t  0  2t 2  2t 2 , t  0
-4-
5.. The function y  (1  sin t )

1
2
is a solution of the differential 2 y   y 3 cos t on an
interval I. Determine at least one such interval.
y  (1  sin t )

1
2

1
1  sin t
 1  sin t  0  sin t  1  t 
1
  2n where n is an int eger.
2


Domain of y   t : t  Re a, s, t  ( 4n  1) 
2

Note:
y  (1  sin t )
 y 

1
2
3
3
1


 
1
1 
2
 y    (1  sin t ) (  cos t )   (1  sin t ) 2  (cos t )

2
2 


1 3
 y (cos t )  2y   y 3 cos t
2
One such interval for I is
  
 , 
 2 2
6. Find values of m such that y  e mx is a solution of the DE y   5 y   6 y  0
Such values are called characteristic values or eigenvalues of the DE.
y  e mx  y   me mx  y   m 2 e mx
y  e mx is a solution of the DE y  - 5y   6y  0  m 2 e mx  5me mx  6e mx  0


 e mx m 2  5m  6  0  m 2  5m  6  0 because e mx is never zero
 m  2(m  3)  0  m  2 and m  3
 y  e 2x and y  e 3x are solutions of the DE
7. Find values of m such that y  t m is a solution of the DE t 2 y   7ty   15 y  0
Such values are called characteristic values or eigenvalues of the DE.
y  t m  y   mt m 1
y   m(m  1)t m  2
y  t m is a solution of t 2 y   7ty   15y  0  t 2 m(m  1)t m  2  7tmt m 1  15t m  0
 m(m  1)t m  7mt m  15t m  0  t m m(m  1)  7m  15  0


t m m 2  8m  15  0  t m m  3(m  5)  0  t  0 or m  3 or m  5
Therefore , y  0, y  t 3 and y  t 5 are solutions of the DE
y  0 is called the tribial solution .
-5-
Initial value problems (IVP): a differential equation has many solutions. A differential
equation with an initial condition is called an initial value problem. We are interested in
a problem in which we seek a solution y(x) of a differential equation so that y(x)
satisfies prescribed side conditions.
y(0)  y 0 : initial value of the function
x(0)  x 0 : initial value of x
Solutions of an IVP problem. An IVP problem can have no solution, one solution or
several solutions.
Problem 1.

 

Solve the IVP differenti al equation 3x 2 y  3y 2 dx  x 3  6xy dy  0, y (1)  1
given that x 3 y  3xy 2  c  0 is the general solution .
Solution :
x 3 y  3xy 2  c  0
y (1)  1  1  3  c  0  c  2
Therefore the IVP is given by x 3 y  3xy 2  2  0
Pr oblem 2. Solve the IVP , xy   y  2x 2 , y(1)  5 given that y  2x 2  cx is the general solution /
y  2x 2  cx
y (1)  5  5  2  c  c  3
 y  2x 2  3x is the solution
Boundary value problem (BVP): A linear DE of order two or greater in which the
dependent variable y or its derivatives are specified at different points.
Problem 7.
Solve the boundary value problem y   y  x 2  1, y(0)  5, y (1)  0 given
that Y  c1 cos x  c 2 sin x  x 2  1 is the general solution .
 Y  c1 cos x  c 2 sin x  x 2  1
c1  6

5  c1  1


y (0)  5, y (1)  0  

6 cos 1
0  6 cos 1  c 2 sin 1  1  1 c 2   sin 1  6 cot 1
 Y  6 cos x  6cot 1 sin x  x 2  1
-6-