‫المحاضرة األولى‬
Differential Equation
Def :An equation relating an unknown function and one or more of its
derivatives is called a differential equation .
d2y
1) 2  x 2  y 2
dx
2) 4 x 2 y   y  0
* Differential Equation are classified by type , order , degree and linearity
1) The order of differential equation :
Is the order of the highest derivatives that appears in it
Example:
d3y
dy
1) 3  5( ) 2  y  e x
dx
dx
d2y 3
dy
2)( 2 )  5  y  sin x
dx
dx
2)the degree of differential equation :
The degree of differential equation is the power of the highest
derivatives in the equation
1
Example :
d2y 3
dy
( 2 )  6( ) 5  1
dx
dx
This of degree (3) and the order (2) .
* classification a linear or Non – linear differential equation .
The ordinary differential equation f ( x , y , y / , …… , y n-1 , y n ) = 0
is said to be linear if f is a linear in the variables y , y / , …… , y n-1 , y n .
thus a linear second order equation can be written in the form
A ( x ) y// +B ( x ) y / + C ( x ) y = f ( x )
* Properties of linear differential equation :
1 – The dependent variable ( y ) and all its derivatives are of the first
degree that is the power of each term involving ( y ) is 1 .
2 – Each coefficient depends only on the independent variable x function
of ( y )such as ( sing ) or function of the derivatives of ( y ) such as e-y can
not appear in appear in linear equation .
* A differential equation that is not linear is said to be non linear
equation .
Example :
2
y  6 y  xy  e x
are linear
( y – x ) dx + 4x dy = 0
(1  y ) y   2 y  e x
d2y
 sin y  0
dx 2
d3y
 5y2  0
3
dx
Non – linear
* Solution of a differential equation :
Def : Any function of defined on same interval I, which when substituted
into a differential equation reduce the equation to an identity , is said to
be a solution of the equation on the interval .
Example:
verify that y ( x ) = x2 is a solution of the differential equation
x 2 y   4 xy   6 y  0
Home work :
3
Q1 verify that y ( x ) = x lin x is solution of the diff. eq.
x3y(3) – x2y // + 2xy / - 2y = 0
Sol :
y  x ln x
1
y   x  ln x  1  ln x
x
1
y  
x
1
y   2
x
 x  x  2 x  2 x ln x  2 ln x  0
Q2 Determine the order and degree of the following diff. eq.
1. x 2  y   4 x( y ) 2  6 y  0
2. ( y (3) ) 2  3 y   4( y ) 3  12 y  0
Q3 Spiffy each of the linear or nonlinear :
2
1. x y   2 xy   3 y  cos x
3
2. y   xy   2 y ( y )  xy  0
x
1
3. e y   (cos x) y   (1  x ) y  tan x
4. y   yy 
2
3
5. y   3( y )  4 y  0
Q4 Determim the functions
4
1. y1 = e2x
and
y2 = e-2x are linear independent or linear
dependent .
2. y1=cosh2x
,
y2 =sinh2x
Second – Order Linear Equations :
Consider the general second – order linear equation :
y   p( x) y   q( x) y  f ( x) …………. (1)
where the coefficient function p , q and f are continuous on the open
interval I .
we will discuss first the associated homogeneous equation
y  p ( x) y  q ( x) y  0 ………… (2)
Theorem (1) :Existence and Uniqueness for linear Equation
Suppose that the functions p , q and f are continuous on the open
interval I containing the point a . Then, given any two numbers b0 and
b1, the equation
y   p( x) y   q( x) y  f ( x) ………… (1)
has a unique ( that is, one and only one ) solution on the enter interval I
that satisfies the initial conditions
y (a) = b0 ,
y / (a) = b1
Theorem (2) Principle of superposition for Homogeneous Equations :
Let y1 and y2 be two solutions of the homogeneous linear equation in
(2) on the interval I . If c1 and c2 are constants, then the linear
combination
5
y = c1y1 + c2y2
is also a solution of eq.(2) on I .
Example :
Verify that the functions y1(x)=ex
and
y2(x)=x ex
Are solution of the differential equation y   2 y   y  0
and then find a solution satisfying the initial conditions y(0)=3 , y /=1
Sol :
The general solution
y(x) = y = c1y1 + c2y2
y(x) = c1ex + c2x ex
for which y /(x) = c1ex + c2 ex+ c2x ex
y /(x)= (c1 + c2) ex + c2x ex
3=c1 + 0  c1 = 3
1=c1 +c2  c2 = -2
Hence the solution of the original initial value problem is
y(x) =3ex -2xex
Definition Linear independence of two functions
6
Two functions defined on an open interval I are said to be linear
independent on I provided that neither is a constant multiple of the
other.
* We can always determine whether two given functions f and g are
linearly dependent on an interval I by noting at a glance whether either
of the two quotients f / g and g / f is a constant – valued function on I
.
Example :
Thus it is clear that the following pairs of functions are linearly
independent on the entire real line
sinx
and
cosx
ex
and
e-2x
ex
and
x ex
x + 1 and x2
That is, neither sinx / cosx = tanx
nor cosx / sinx = cotx
is a constant – valued function .
neither
ex / e-2x = e3x
nor
e-2x / ex is a constant – valued function .
But the identically zero function f(x) 0 and any other function g are
linear dependent on every interval , because 0.g(x) =0=f(x) . Also the
functions f(x)= sin2x and g(x) = sin cosx are linearly dependent on any
interval because f(x) = 2g(x) and sin2x = 2sinx cosx .
7
Theorem (3) : Wronskians of solutions :
Suppose that y1 and y2 are two solutions of the homogeneous second –
order linear equation :
y  p ( x) y  q ( x) y  0
On an open interval I on which p and q are continuous
a) If y1 and y2 are linearly dependent, then w(y1 , y2)  0 on I .
b) If y1 and y2 are linearly independent, then w(y1 , y2)  0 at each point
of I .
* Given two function f and g, the wronskian of f and g is the
determinant
f g
 fg   f g
f  g
w
Example:
w(cosx , sinx) =
cos x sin x
 cos 2 x  sin 2 x  1
 sin x cos x
linearly independent .
and
w(e , xe ) 
x
x
ex
e
x
xe x
e  xe
x
x
 e 2 x linearly independent
on the other hand, if the functions f and g are linearly independent,
with f = kg, then
8
w( f , g ) 
kg g
 kgg   kgg   0
kg  g 
: ‫المحاضرة الثالثة‬
Homogeneous second – order linear differential equation with constant
coefficients :
Lets begin with second order equation with constant coefficients we shall
find solution of the second order :
y   ay   by  0.......(1) ( a ) and ( b ) are constants .
As first order :
y   ay  0 ( a constant ) for which all solution
are constant multiple of e-ax thus for ( 1 ) also some form of exponential
function would be a choose and the properly that the differential of an
exponential function erx always yield a constant multiplied by erx .
Thus we try y = erx and find the values of r, we have :
r 2 e rx  are rx  be rx  0
(r 2  ar  b)e rx  0  (r 2  ar  b)  0...............(2)
Hence erx is a solution of 1 if r is a solution of equation (2) is called
characteristic equation of (1) for the roots of (2) we have the following
three cases :
1. Distinct Real Roots :
9
If the roots r1 and r2 of the characteristic eq . in (2) are real and distinct
then y ( x ) = c1 er1x + c2 er2x is a general solution of eq . ( 1 )
Example :
y   6 y   xy  e x
( y  x)dx  4 xdy  x 2
are linear
Example : find the general solution of :
2 y   7 y   3 y  0
Sol : we can solve characteristic equation
2r 2  7r  3  0  (2r  1)( r  3)  0 the roots r1 = 1/2 and r2 =3
 the general solution
x
2
y( x)  c1e  c2 e 3 x
2. Repeated Roots :
If the characteristic equation in ( 2 ) has equal ( necessarily real ) roots
r1 = r2 , then
y ( x ) = ( c1 + c2x )er1x is a general solution of eq. (1)
Example : To solve the initial value problem
y   2 y   y  0, y (0)  5, y (0)  3
10
sol: the characteristic equation
r 2  2r  1  (r  1) 2  0
Has equal roots r1 +r2 = -1 , hence the general solution
y ( x ) = c1e-x +c2xe-x
y ( x)  c 1 e  x  c 2 xe x  c 2 e  x
5  c1  0  c1  5
 3  c1  0  c 2  c 2  0
The particular solution of the initial value problem is
3. Complex Roots
If the characteristic equation has an unrepeated pair of complex
conjugate roots a  bi ( with b  0 ) then the corresponding part of a
general solution of eq . ( 1 ) has the form :
y  e ax (c1 cos bx  c 2 sin bx)
r1  a  ib
 r2  a  ib
y  c1e rx1  c 2 e r2 x
 y  c1e ( a ib) x  c 2 e ( a ib) x
 the solution is  c e ax .e ibx  c e ax .e bix
1
2
 e ax (c1e ibx  c 2 e ibx )
But Euler, s formula
eibx  cos bx  i sin bx
e ibx  cos bx  i sin bx
11
y  e ax [c1 (cos bx  i sin bx)  c 2 (cos bx  i sin bx)]
ax
then  e [c1 (cos bx  c1i sin bx  (c 2 cos bx  c 2 i sin bx)]
 e ax [(c1  c 2 cos bx  (c1  c 2 )i sin bx]
where A=c1+c2 , B=i(c1 – c2 ) .
2
Example : solve the equation y   b y  0.....(b 0)
Sol :
r2+b2=0 with roots r= ib with a=0
give the general solution
y (x)=c1 cosbx +c2 sinbx
Example: Find the particular solution of
y   4 y   5 y  0
for which y(0) =1 and y (0)  5
Sol:
12
r 2  4r  5  0
r
 b  b 24 ac 4  16  20

 2i
2a
2
Hence a=2 and b=1 gives the the general solution
y( x)  e 2 x (c1 cos x  c2 sin x)
then, y ( x)  2e 2 x (c1 cos x  c2 sin x)  e 2 x (c1 sin x  c2 cos x)
So the initial conditions give
y(0)  c1  1
and
y(0)  2c1  c2  5  c2  3
So the desired particular solution is
y( x)  e 2 x (cos x  3 sin x)
solution :
r2 - 4r +5 =0
 b  b 2  4ac 4  16  20
r

 2i
2a
2
Hence a=2 and b=1 give the general solution
y( x)  e 2 x (c1 cos x  c2 sin x)
then
y ( x)  2e 2 x (c1 cos x  c2 sin x)  e 2 x (c1 sin x  c2 cos x)
So the initial conditions give
y(0)=c1=1
and
y(0)  2c1  c2  5  c2  3
So the desired particular solution is :
y(x)=e2x (cosx +3sinx )
13
Repeated Complex Roots :
If the conjugate pair a  ib has multiplicity k, then the corresponding
part of the general solution has the form
(A1+A2x + ……… + Akxk-1 ) e(a+ib) + ( B1 +B2x + ………. + Bkxk-1 ) e ( a – ib )
k 1
=  x p e ax (ci cos bx  di sin bx)............(7)
p 0
It can be shown that the 2k functions
xpeax cosbx
,
xpeax sinbx
that appear in eq.(7) are linearly independent .
Example : find the general solution of
y (3)  y   10 y  0
Solution :
The cubic equation r3+r-10 = 0
r 3  r  10  (r  2)( r 2  2r  5)  0
r 2  2r  5  0  r 
 2  4  20  2  4i

 1  2i
2
2
 the general solution y(x) = c1e2x +e-x(c2 cos2x + c3 sin2x )
H.W
14
The roots of the characteristic equation of a certain differential equation
are 3 , -5 , 0 , 0 , 0 , 0 , -5 , 2 3i , and 2 3i .
Write a general solution of this homogeneous differential equation .
Sol :
y(x) =c1 + c2 x +c3 x2 +c4 x3 +c5 e3x +c6 e-5x +c7 x e-5x
: ‫المحاضرة الرابعة‬
General Theory of the order linear Equation
An n th order linear differential equation of the form :
P0 ( x)
dny
d n 1 y
( x)

P
(
x
)
 .............  Pn y  G( x).........(1)
1
n
n 1
dx
dx
We assume that the function P0(x) , P1(x) , ………….., Pn(x) and G are
Continuous real – valued function on some interval I: < x < B, and that
P0 is nowhere zero in this interval . Then, dividing Eq.(1) by P0(x), we
obtain :
L[ y ] 
dny
d n 1 y

P
(
x
)
 ............  Pn ( x) y  g ( x).............(2)
1
dx n
dx n 1
The linear differential operator L of order n defined by Eq.(2) .
Theorem (1): If the function P1 , P2 , ……. , Pn and g are continuous on the
open interval I, then there exists exactly one solution y= (x) of the
differential equation (2) that also satisfied the initial conditions
15
y(x0)=y
, y /(x0) =y0/ , ……….. , y(n-1) (x0) =y0(n-1) …………(3)
This solution exists throughout the interval I .
The Homogeneous Equation :
As in the corresponding second problem, we first discuss the
homogeneous equation
L[x]=y(x) +P1(x)y(n-1) + ……. +Pn(x)y = 0 …………. (4)
the function y1 , y2 , …….. , yn are solution of eq. (4), then it follows by
direct computation that the linear combination
y = c1y1(x) + c2y2(x) + ……… + cnyn(x) ………… (5)
where c1 , c2 , ………. , cn are arbitrary constants is also a solution of Eq.(4).
* Hence a necessary and sufficient condition for the existence of
a solution of equations
c1y1(x0) + c2y2(x0) + ……….. +cnyn/(x0) = y0/



c1y1(n-1) (x0) + c2y2(n-1) (x0) + ………+ cnyn(n-1) (x0) = y(n-1) .
for arbitrary values of y0 , y0/ , ………. , y0(n-1) is that the wronskian
16
W ( y1 ,......, yn ) 
y1
y1
y2
y2






is not zero
( n1)
 yn
( n1)
y1
y2
( n1)
yn
yn
attttttttttttt
Theorem (2) Wronskians of solutions :
Suppose that y1 , y2 , ……. , yn are n solution of the homogeneous n th –
order linear equation
y(n) + P1(x)y(n-1) + ………… + Pn-1(x)y / + Pn(x)y = 0
on an open interval I, where each Pi is continuous .
Let W = W ( y1 , y2 , ………. , yn ) .
a) If y1 , y2 , ………. , yn are linear dependent, then W  0 on I .
b) If y1 , y2 , ………. , yn are linear independent, then W  0 at each point
of I .
Example : Determine whether the function f1(x) =1 , f2(x) = 2 + x ,
f3(x) = 3 - x2 and f4(x) = 4x + x2 are linearly independent or dependent on
any interval I .
solution :
W
1 2  x 3  x2
0
1
 2x
0
0
0
0
2
0
4x  x 2
4  2x
2
0
17
W = 0  f1 , f2 , f3 and f4 linearly dependent .
Theorem (3) :
If y1(x) , y2(x) , ………. , yn(x) is fundamental set of solution of Eq.(4)
L[y] = y(n) + P1(x)y(n-1) + ………… + Pn-1(x)y / + Pn(x)y = 0
on an interval I, then y1(x) , y2(x) , ………. , yn(x) linearly independent on I .
Conversely, if y1(x) , y2(x) , ………. , yn(x) are linearly independent solution
of Eq.(4) on I, Then they form a fundamental set of solutions on I
*Homogeneous Equation with constant coefficients :
Consider the n th order linear homogeneous differential equation
L[y] = a0 y(n) + a1 y(n-1) + ………… + an(x)y = 0……..(5)
where a0 , a1 , ……….. , an are real constants .
1)Real and unequal Roots :
If the roots of the characteristic equation are real and no two are equal,
then we have n distinct solution
er x , er x ,…….. , er x of Eq.(5) .
1
2
n
If these function are linearly independent, then the general solution of
eq.(5) is :
y = c1 er1x +c2 er2x +…….. +cn ernx
( 4)
Example : Find the general solution of y  y   7 y   y   6 y  0
sol : We must determine r by solving the polynomial equation
r4 + r3 -7r2 – r +6 = 0
18
r1 =1
(r – 1)(r – 2)(r2 +4r + 3) = 0
(r – 1)(r – 2)(r +1)(r +3) = 0
 r1 = 1 , r2 = 2 , r3 = -1 r4 = -3
 the general solution
y =c1ex +c2e2x +c3e-x +c4e-3x
Example: Find the general solution of y(4) –y =0 .
Also find the solution that satisfies the initial conditions
y(0) = 7/2 , y /(0)= -4 , y //(0) = 5/2
Sol: We find that the characteristic equation is :
r4 – 1 = (r2 – 1)(r2 + 1) = (r – 1)(r + 1)(r2 + 1) = 0
 the roots are r = 1 , - 1 , i , - i and the general solution is :
y = c1ex + c2e-x + c3cosx +c4sinx
y / = c1ex - c2e-x - c3sinx +c4cosx
y /// = c1ex + c2e-x - c3cosx - c4sinx
7/2 =c1 + c2 + c3……….(1)
-4= c1 – c2 + c4 ………..(2)
5/2 = c1 + c2 – c3…………(3)
)1( ‫ ) من معادلة‬3( ‫وبطرح معادلة‬
19
2c3 = 1  c3 =1/2
c1+ c2 = 7/2 – 1/2 = 6/2 = 3  c1 = 3 – c2
substitute in eq(2)
-4 = 3 – c2 – c2 + c4  c4 = -4 -3 + 2 c2 = -7 + 2c2
Let c2 = 3  c4 = -1 , c1 = 0
 y = 3e-x + 1/2 cosx – sinx
Repeated Roots : If the roots of the characteristic equation are not
distinct – that is if some of the roots are repeated, has multiplicity s
( where s n) , then :
er x , x er x , x2 er x …….. , xs-1 er x
1
1
1
1
Repeated Complex Roots : If the conjugate pair a ib has multiplicity k,
then the corresponding part of the general solution has the form
y  ( A1  A2 x  ..........  Ak x k 1 )e ( a ib) x  ( B1  B2 x  ........  Bk x k 1 )e ( a ib)
k 1
  x p e ax (ci cos bx  di sin bx)
p 0
It can be shown that the 2k functions xp eaxcosbx , xpeax sinbx are
linearly independent .
Example :
The roots of the characteristic equation of a certain - differential
equation are 3 , -5 , 0 , 0 , 0 , 0 , -5 , 2 ,  3i , and 2 3i and 2  3i write a
general solution of this h homogeneous diff.eq .
20
Sol :
y(x) = c1 + c2 x + c3 x2 + c4 x3 + c5 e3x + c6 e-5 + c7 x e-5
Theorem ( 4 ) : Solution of Non homogeneous Equation
Let Yp be a particular solution of the non homogeneous equation
y(n) + p1 (x)y(n-1) + p2 (x)y(n-2) + ……… + pn (x)y =g(x)………. (6)
On an open interval I, when the functions pi and g are continuous, let y1 ,
y2 , ………. , yn be linearly independent solution of the associated
homogeneous equation if Y is any solution what sorer of Eq. (6) on I .
then there exist numbers c1 , c2 , ………. , cn such that
Y(x) = c1y1 + c2y2 + ………. + cn yn (x) + yp(x) for all x in I .
Example :
Show that the function y1(x) = e-3x , y2(x) = cos2x and y3(x) = sin2x are
linearly independent .
Sol: Their wronskian is
e 3 x
W   3e 3 x
qe 3 x
 e 3 x
cos 2 x
sin 2 x
 2 sin 2 x 2 cos 2 x
 4 cos 2 x  4 sin 2 x
 2 sin 2 x 2 cos 2 x
cos 2 x
sin 2 x
cos 2 x
sin 2 x
 3e 3 x
 qe 3 x
 4 cos 2 x  4 sin 2 x
 4 cos 2 x  4 sin 2 x
 2 sin 2 x 2 cos 2 x
 26e 3 x  0
Because W  0 everywhere, it follows that y1 , y2 and y3 are linearly
independent on any open interval
Example :
21
Show first that the three solutions y1(x) = x , y2(x) = x lnx and y3(x)= x2
of the third – order equation x3y(3) – x2y //+2xy /-2y =0 ……(6)
are linearly independent on the open interval x > 0 .
then find a particular solution Eq.(6) that satisfied the initial jjjjjjjjjjjjjjj
Solution :
Note that for x > 0 , we could divide each term in (6) by x2 
y ( 3) 
1
2
2
y   2 y   3 y  0
x
x
x
when we compute the wronskian of the three given solution , we find
that
x
x ln
x2
w  1 1  ln x 2 x  x
1
0
2
x
thus w( ) 0 for x>0, so y1 , y2 and y3 are linearly independent
on the interval x>0, To find the desired particular solution
y(x) = c1x + c2x lnx + c3 x2
y / (x) = c1 + c2 ( 1 + lnx ) +2c3 x
y //(x) = 0 + c2 /x + 2c3
y(1) = c1 + + c3 = 3
y /(1) = c1 + c2 +2c3 = 2
22
y //(1) = c2 +2c3 =1
we solve to find c1=1, c2 =-3 and c3 =2 . thus the particular solution in
question is y (x) = x – 3x lnx + 2x2
Q
y1 = x , y2 = x lnx , y3 = x2
y = c1x + c2x lnx + c3 x2
y   c1  c2 ln x  c2  2c3 x
c2
 2c3
x
c
y   22
x
c
c
x 3 ( 22 )  x 2 ( 2  2c3 )  2 x(c1  c2 ln x  c2  2c3 x)  2(c1 x  c2 x ln x  c3 x 2 ) 
x
x
 c2 x  c2 x  2c3 x 2  2c1 x  2c2 x ln x  2c2 x  4c3 x 2  2c1 x  2c2 x ln x  2c3 x 2 
y  
‫المحاضرة الخامسة‬
Non homogeneous Equation and Undetermined coefficients
The general non homogeneous nth – order linear equation with constant
coefficients has form
an y ( n )  an1 y ( n1)  ........  a1 y   a0 y  f ( x) ………..(1)
By Theorem (4) a general solution of Eq.(1) has the form
y = yc + yp
23
Where the complementary function yc(x) is a general solution of the
associated homogeneous equation
an y ( n )  an1 y ( n1)  ........  a1 y   a0 y  0
and yp(x) is a particular solution of Eq.(1)
Example (1) :
find a particular solution of
solution:
Here f(x) =3x +2 is a polynomial of degree 1, so our guess is that
yp (x) = Ax + B
Then y p  A and y p  0 , so yp will satisfy the differential equation
provided that
(0)+ 3 (A)+4(A+B) = 3x + 2
 (4A)x + (3A +4B )=3x +2
4A =3

A=3/4
3A + 4B =2  9/4 + 4B =2  4B=2 - 9/4
4B = - 1/4
 B= - 1/16
24
Thus we have found the particular solution yp(x) = (3/4) x- 1/16
Example (2) :
Find a particular solution of
y   4 y  2e 3 x
solution :
yp(x)=Ae3x
then y p  qAe
3x
qAe3x – 4(Ae3x) =2e3x
 5A=2  A 
2 3x
2
e .
.Thus our particular solution is yp (x)=
5
5
Example :( 3 )
Find a particular solution of 3 y   y   2 y  2 cos x .
solution :
A first guess might be yp(x) =Acosx, but the presence of y  on the left –
hand side signals that we probably need a term involving sinx as well . So
we try
yp(x) = A cosx + B sinx
y p ( x)   A sin x  B cos x
y p ( x)  A cos x  B sin x
3(-Acosx – Bsinx )+ (-Asinx + Bcosx )-2(Acosx +Bsinx )=2cos
 (-5A+B)cosx +(-A-5B)sinx = 2 cosx
25
-5A +B = 2
-A – 5 B=0
with readily found solution A 
5
1
, B=
13
13
Hence a particular solution is
y p ( x) 
5
1
cos x 
.
13
13 sin x
Example(4):
Find a particular solution of y   4 y  2e 2 x
solution :
If we try yp(x) = Ae2x , we find that
yp  4 y p  4 Ae 2 x  4 Ae 2 x  0  2e 2 x
Therefore, we should begin with a trial function yp (x), wkose derivative
involves both
e2x and something else that can cancel upon substitution
into the differential Eq.to leave the e2x term that we need . Areasonable
guess is
yp(x)=Axe2x for which
y p ( x)  Ae 2 x  2 Axe2 x and y p ( x)  4 Ae 2 x  4 Axe2 x
substitution into the original differential equation yields
(4Ae2x + 4Axe2x ) – 4 ( Axe2x ) = 2e2x
26
The terms involving xe2x obligingly cancel, leaving only 4Ae2x = 2e2x, so
1
2
that A= ,  a particular solution is
yp(x) =
1 2x
xe
2
Rule (1) : Method of undetermined coefficients :
suppose that no term appearing either in f(x) or in any of its derivatives
satisfies the associated homogeneous equation Ly then take as a trial
solution for yp a linear combination of all linearly independent such
terms and their derivatives . Then determine the coefficients by
substitution of this trial solution into the non homogeneous eq. Ly = f (x)
Example (5) :
3
Find a particular solution of y   4 y  3x
solution:
The (familiar) complementary solution of eq.is
yc(x)= c1cos2x + c2sin2x
The function f(x) =3x3 and its derivatives are constant multiples of the
linearly independent function x3 , x2 , x and 1.
Because non of these appears in yc , we try
yp= Ax3 + Bx2 +Cx +D
27
y p  3 Ax 2  2 Bx  C
y p  6 Ax  2 B
y p  4 y p  (6 Ax  3B )  4( Ax 3  Bx 2  Cx  D)
 4 Ax 3  4 Bx 2 (6 A  4C ) x  (2 B  D)  3 x 2
4A  3
4B  0
6 A  4C  0
2B  D  0
with solution of is y p ( x) 
3 3 9
x  x
4
8
Example (6) : Solve the initial value problem
y   3 y   2 y  3e  x  10 cos 3x
y (0)  1
y (0)  2
solution : The characteristic equation r2 – 3r +2 =0
has roots r=1 and r = 2, so the complementary function is
yc(x) =c1ex + c2e2x
The terms involved in f(x) = 3 e-x – 10cos3x and its derivatives are
e-x ,
cos3x and sin3x . Because non of these appears in yc , we try
y p  Ae  x  B cos 3x  C sin 3x
y p   Ae  x  3B sin 3x  3C cos 3x
y p  Ae  x  9 B cos 3x  9C sin 3x
After we substitution the expressions into the differential equation and
collect coefficients, we get
28
y p  3 y p  2 y p  6 Ae  x  (7 B  9C ) cos 3 x  (9 B  7C ) sin 3 x  3e  x  10 cos 3 x
 6A  3
 7 B  9C  10
9 B  7C  0  A 
and C
1
7
,B 
2
13
9
13
This gives the particular solution
y p ( x) 
1 x 7
9
e  cos 3 x  sin 3x
2
13
13
To satisfy those initial conditions, we begin with the general solution
y(x)= yc(x) + yp(x)
 c1e x  c 2 e 2 x 
1 x 7
9
e  cos 3 x  sin 3 x
2
13
13
withderivative
1
21
27
y ( x)  c1e x  2c 2 e 2 x  e  x  sin x 
2
13
13 cos 3 x
The initial conditions lead to the equation
1 7
 1
2 13
1 27
y (0)  c1  2c 2  
2
2 13
y (0)  c1  c 2 
1
2
with solution
6
c2 
13
c1 
The desired particular solution is therefore
29
y ( x) 
 1 x 6 2x 1 x 7
9
e  e  e  cos 3x  sin 3x
2
13
2
13
13
Example (7):
find the general form of a particular solution of
y(3) +9y / = x sinx +x2e2x
solution:
The characteristic equation r3 +9r = 0 has roots r=0, r = -3i and
r = 3i , so the complementary function is
yc(x) = c1 +c2cos3x + c3sin3x
The derivatives of the right – hand side in eq.involve the term cosx, sinx,
xcosx, xsinx, e2x, xe2x and x2e2x because there is no duplication with the
terms of the complementary function trial solution takes the form
yp(x) = A cosx + B sinx +cx cosx + Dx sinx + E e2x +Fxe2x +Gx2e2x
Upon substitution yp in diff.eq. and equation coefficients of linear terms,
we get seven equations determining the seven coefficients A,B,C,D,E,F
and G .
Rule (2):Method of Undetermined Coefficients .
If the function f(x) is of either form pm(x)erxcoskx or pm(x)erxsinkx
where pm(x) is a polynomial in x of degree m, take as the trial solution
yp(x) = x2[(A0 +A1x + A2x2 + …… + Amxm ) erx coskx
30
+( B0 + B1x +B2x2 +……. + Bmxm ) erxsinkx ]
where s is the smallest nonnegative integer such that no term in yp
duplicates a term in the complementary function yc .
Then determine the coefficients by substituting yp into the
non homogeneous equation .
substituting in the method of undetermined coefficients .
f(x)
yp
om = b0 + b1x + b2x2 + …. +bmxm
xs (A0 +A1x + A2x2 + …… + Amxm )
a coskx + bsinkx
xs( Acoskx + Bsinkx )
erx(a coskx + bsinkx )
xs erx ( Acoskx + Bsinkx )
pm(x) erx
xs(A0 +A1x + A2x2 +…… +Amxm )erx
pm(x)( a coskx + bsinkx )
xs[ (A0 +A1x + A2x2 +…… + Amxm)
coskx+( B0 + B1x +B2x2 +……. + Bmxm
) erxsinkx]
31
Example (8) :
( 3)
x
2
Find a particular solution of y  y   3e  4 x
solution :
The characteristic equation r3+r2 = 0 has roots r1 = r2 = 0 and r3 = -1, so
the complementary function is yc(x) = c1 + c2x + c3 e-x .
As a first step toward our particular solution, we form the sum
(Aex) + ( B + Cx +Dx2 )
The part Aex corresponding to 3ex does not duplicate any part of the
complementary function, but the part B + Cx +Dx2 must be multiplied by
x2 to eliminate duplication . Hence we take
yp = Aex + B x2 +C x3 +D x4
y p  Ae x  2 Bx  3Cx 2  4 Dx 3
y p  Ae x  2 B  6Cx  12 Dx 2
and yp(3) = Aex +6C +24Dx
substitution of these derivatives in differential equation yield
2 Aex + (2B +6C) + (6C +24D)x + 12Dx2 = 3ex +4x2
The system of equations
2A = 3 , 2B + 6C = 0 , 6C + 24D = 0 , 12D = 4
32
has the solution
3
2
B4
4
C
3
and
1
D
3
A
3
2
4
3
1
3
x
2
3
4
Hence the desired particular solution is y p ( x)  e  4 x  x  x
Example (9):
Determine the appropriate form for a particular solution of
y   6 y   13 y  e 3 x cos 2 x
Solution :
The characteristic equation r2 + 6x +13 =0 has roots -3 2i, so the
complementary function is yc(x) = e-3x(c1 cos2x + c2 sin2x ) .
This is the same form as a first attempt e-3x(A cos2x + B sin2x ) at a
particular solution, so we must multiply by x to eliminate duplication .
Hence we would take yp(x) = e-3x(A cos2x + B sin2x )
Example (10):
Find a particular solution of y   3 y   4 y  3e 2 x  2 sin x  8e x cos 2 x
solution
33
r2 – 3r -4 = 0  ( r – 4 ) ( r + 1 ) = 0  r1 = 4 , -1
yc = c1 e4x + c2 e-x
yp = A e2x + B sinx + C cosx +D excos2x + E exsin2x
1
2
5
B
17
3
C
17
10
D
13
2
E
13
A
Q H.w
Find the general solution of the given diff.eq
1.
2 y   3 y   y  x 3  3 sin x
x
2. y   2 y   5 y  4e cos 2 x
2x
3. y   2 y   3 y  3xe
4.
y   4 y  x 2  3e x
5.
y   y  3 sin 2 x  x cos 2 x
‫المحاضرة السابعة والثامنة‬
Variation of Parameters
34
consider ay   by   cy  f (x) ………. (1) where a , b and c are real
constant; let (y1) and (y2) be two linearly independent solution of the
associated homogeneous equation of (1).with complementary function
yc(x)= c1y1(x) + c2y2(x) ……… (2)
we must to find function u1 and u2 such that
yp= u1(x) y1(x) +u2(x) y2(x) ……….. (3)
is a particular solution of eq. (1)first, to impose the condition
L[ yp ] = f(x),we must compute the derivatives
y p and y p . The product
value gives y p  (u1 ( x) y1 ( x)  u 2 y 2 )  (u1 y1  u 2 y 2 ) ………… (4)
To avoid the appearance of the second derivatives
u1 and u 2 , the
additional condition that we now impose is that the second sum here
must vanish
u1 y1  u 2 y2  0 ……… (5)
Then
y p  u1 y1  u 2 y 2 ………..(6)
and the product rule gives
y p  (u1 y1  u 2 y 2 )  (u1 y1  u 2 y 2 ) ……. (7)
But both y1 and y2 satisfy the homogeneous eq.
y   by   cy  0
associated with the non homogeneous eq. by substitute into equation(1)
we get :
35
(u1 y1  u2 y2 )  (u1 y1  u2 y2 )  b[u1 y1  u2 y2 ]  c[u1 y1  u2 y2 ]  f ( x)
u1 ( y1  by1  cy1 )  u2 ( y2  by2  cy2 )
The first two term are zero since both y1 and y2 are solution to the
homogeneous equation we get :
u1 y1  u2 y2  f ( x) and since u1 y1  u2 y2  0
u1 y1  u 2 y 2  0
u1 y1  u 2 y 2  f ( x)
by using gramers rule we can find u1 and u2
0
f ( x) y 2
y1 y 2
y1 y 2
 u1 
 u1 
y2
 y2 f
w
u1   
,
u 2 
,
y 2 f ( x)
dx
w( y1 , y 2 )
y1
0
y  f ( x)
u 2  1
y1 y 2
y1 y 2
,
y1
f
w
u2  
y1 f ( x)
dx
w( y1 , y 2 )
Then we are ready to write particular solution :
y p  u1 y1  u 2 y 2   y1 
y 2 f ( x)
y f ( x)
dx  y 2  1
dx
w( y1 , y 2 )
w( y1 , y 2 )
y = c1 y1 + c2 y2 + yp
where c1 and c2 are determined by initial condition .
Example :
36
y   y  sec x
Solution :
r2 +1 = 0  r =  i
yc = c1 cosx + c2 sinx
 y1 = cosx
 w( y1 , y 2 ) 
,
y2 = sinx
cos x sin x
 cos 2 x  sin 2 x  1
 sin x cos x
u1   
sin x. sec x
dx  ln cos x 
1
u2  
cos x. sec x
dx  x
1
 y p  cos x. ln cos x  x sin x
y = c1 cosx + c2 sinx + cosx ln cosx + x sinx
Example :
Find a particular solution of the equation
y   y  tan x
Solution :
The complementary function is yc(x) = c1 cosx + c2sinx
37
y1= cosx
;
y1   sin x
;
y2 = sinx
y2  cos x
Hence the equation in (14) are :
(u1 )(cos x)  (u 2 )(sin x)  0
(u1 )(  sin x)  (u 2 )(cos x)  tan x
w
cos x
sin x
 cos 2 x  sin 2 x  x
 shnx cos x
we easily solve these equations for
y p ( x)   cos x 
sin x. tan x
cos x. tan x
dx  sin x 
dx
1
1
sin 2 x
  cos x 
dx  sin x  sin xdx
cos x
(1  cos

  cos x
cos x
2
x)
dx  sin x. cos x
  cos x  (sec x  cos x)dx  sin x.c0 sx
  cos x(ln sec x  tan x  sin x)  sin x. cos x
 y p ( x)  (cos x) ln sex  tan x
Example:
38
x
y
y   y    4e 2
4
1
1
r 2  r   0  r1  r2 
4
2
Solve
y c  c1e
y1  e
1
2x
 c 2 xe
1
2x
1
2x
y 2  xe
1
2x
1
2x
e
w( y1 , y 2 )  1 1
e 2x
2
xe
1
2x
1
1
1 2x
xe  e 2 x
2
1
1
 e x ( x  1)  xe x  e x
2
2
u1   
u2  
1
2x
x
2
xe .4e
dx  2 x 2
x
e
1
2x
x
2
e .4e
 4x
ex
 y p  2 x e
2
1
2x
 4x e
2
1
x
2
 2x e
2
1
2x
the general solution is
y  c1e
1
2x
 c2 xe
1
2x
 2x e
2
1
2x
Not:
We can find solution of second order homogeneous linear equation when
one solution is give consider :
y  a1 ( x) y  a2 ( x) y  0 …………… (1)
39
if y1 solution of ( 1 )
then y 2  y1 
Ex : If y1 
y  
 a1( ( x ) dx
e 
y1
2
dx
sin x
is solution of D.E
x
2
y   y  0 find y2
x
Sol :
e 

 a ( x ) dx
2
 x dx
sin x e
dx
x  sin 2 x
y1
x2
1
 2 ln x
sin x e
sin x
x 2 dx

dx

x  sin 2 x
x  sin 2 x
x2
x2
y 2  y1 

2
dx 
sin x
sin x
sin x  cos x
 cos x
csc x 2 dx 
( cot x) 
(
)
 y2

x
x
x
sin x
x
Example:
If y1= x is a solution of y  
1
y
y   2  0 find y2 solution of D.E.
x
x
Sol :
40
a
1
x
1
  dx
e 
e x
y 2  y1 
dx  x  2 dx
y 21
x
1
ln x 1
e
 x  2 dx  x  x2 dx
x
x
x 2
1
1
x


 2  2x
2x
 a ( x ) dx
 the general solution is :
y = c1 x – c 2
1
2x
H.W: Solve the following diff . eq.
5x
1. y   4 y   3 y  e
y (0) = 3
, y (0)  9
4x
2. y   8 y   16 y  e
y (0) = 0
,
y (0)  1
3. y  y  x cos x
2
 The method of variation of parameters for determining a particular
solution of the no homogeneous nth order linear differential equation
L[ y ]  y ( n )  p1 ( x) y ( n 1)  ......  p n 1 ( x) y   p n ( x) y  g ( x)
u m ( x) 
g ( x).Wm ( x)
, m  1,2,......, n
W ( x)
Here W(x)=W( y1 , y2 , ……. , yn )(x) and Wmis the determinant obtained
from W by replacing the nth column by the column ( 0 , 0 , ….. , 1 ) . With
this notation a particular solution of Eq.(1)is given by
y ( x)  m 1 y m ( x) 
n
g ( x)Wm ( x)
dx
W ( x)
41
Ex: Given that y1(x)=ex , y2(x) = xex and y3(x) = e-x are solution of
the homogeneous equation corresponding to y  y  y  y  e
x
Sol:
ex
x
W(ex , xex , e-x )(x) = e
ex
xe x
xe x  e x
xe x  2e x
ex
 e x
e x
W( ex , xex , e-x )= ex[e-x (xex +ex )+ex ( xex +2ex )] – ex [ x – x – 2 ] + ex [ -x – x
–1]
= ex [ x + 1 + x + 2 ]+ 2ex -2xex -ex
= 2xex + 3ex +2ex-2xex – ex = 4ex
0
xe x
W1 ( x)  0 xe x  e x
1 xe x  2e x
ex
 ex
ex
W1(x)= -x –x -1 =-2x -1
ex
W2 ( x )  e x
ex
0 ex
0  e  x = ( - 1 )( - 1 – 1 ) = 2
1 e x
ex
W3 ( x)  e x
ex
xe x
xe x  e x
xe x  2e x
0
0 = ( x + 1 ) e2x –xe2x = e2x
1
42
x
x 2x
e x (2 x  1)
x e ( 2)
x e .e
dx

xe
dx

e
 4e x
 4e x dx
4e x
x2 x
1
1
 e x (
 )  xe x [ x]  e  x [e 2 x ]
4 4
2
8
y ( x)  e x 
Q Use the method of variation of particulars of determine the general
solution of the given differential eq.
1. y   y   x
4x
2. y   2 y   y   2 y  e
Q Use the method of variation of particularsto find a particular solution
of the given diff.eq. Then check your answer by using the method of
undetermined coefficients.
x
1. y   5 y   6 y  2e
x
2. y   y   2 y  2e
x
3. y   2 y   y  4e
x
y
Q Solve the D.E. y   y    4e 2
4
Sol :
43
1
0
4
1
1
1
(r  )( r  )  0  r1   r2
2
2
2
r2  r 
1
x
x
y c  c1e 2  c 2 xe 2
x
y p  Ax 2 e 2
x
x
1
y p  Ax 2 e 2  2 Axe 2
2
x
x
x
x
1
y p  Ax 2 e 2  Axe 2  Axe 2  2 Ae 2
4
A2
‫المحاضرة التاسعة‬
Operator " D "
If x is independent variable we can use the operator D in differential
equation as follows :
Dn 
dn
d2
d
2
..........
.,
D

,
D

dx
dx n
dx 2
For example
d 2 y dy
D y  2 , a  by  (aD  b) y
dx
dx
2
Dn y 
,
Dy 
dy
dx
dny
d3y
3
2
..........
,
D
y

D
(
D
y
)

dx n
dx 3
 We can write the non homogeneous linear differential eq.
44
dny
d n1 y
dy

a

......

a
 a n y  f ( x)...........(1)
1
n

1
dx
dx n
dx n 1
by formula the operator D
( Dn + a1Dn-1 +……..+an-1D + an ) y = f(x) ………(2)
 f (D) y = f(x)
where f(D) = Dn +a1Dn-1 + ……. +an-1D +an ……(3)
The eq.(3)is called polynomial operator
We can write f (D) in eq.(3) by Algebraic factors as follows
f ( D)  D n  a1 D n 1  ........  a n 1 D  a n
 ( D  m1 )( D  m2 )( D  m3 ).......( D  mn )
f ( D) y  ( D  m1 )( D  m2 )( D  m3 )..........( D  mn ) y
:
d 2x
dy
 a1
 a 2 y  f ( x)..............(4)
2
dx
dx
we can write the eq. (4) by (D) operator
( D2 +a1D +a2 )y = f(x)
f (D)y = ( D2 +a1D +a2 )y…………..(5)
we can write the eq. (5) by Algeria factors
f (D)y = (D-m1)(D-m2)y
where a1 = - (m1 + m2 )
a2 = m1.m2
45
 f (D)y = (D-m1) (D – m2)y
= (D – m2 ) (D – m1 )y
To prove that
(D – m1 ) (D – m2)y = D2y – D(m2 y) – m1 Dy +m1m2y
=D2y – (m2 + m1) Dy + m1m2y
= D2y + a1Dy +a2y
(D – m2) (D – m1)y = D2y – D(m1y) – m2Dy + m2m1y
= D2y – m1Dy –m2Dy +m1m2y
= D2y – (m1 + m2)Dy +m2m1y
= D2y +a1Dy +a2y
In the same way of order n
Example :
f (D)y = D2y+3Dy +2y
Sol:
f (D)y =(D + 2) (D + 1)y = (D + 1) (D + 2)y
because
(D + 2) (D + 1)y = D2y + 2Dy + Dy + 2y = D2y + 3Dy +2y
and
(D + 1) (D + 2)y = D2y + 3Dy +2y
Example:
46
(D2 – 1)y = (D + 1) (D – 1)y = (D – 1) (D + 1)y
(D2 +2D +1)y = (D + 1) (D + 1)y = (D + 1)2y
(D2 + 1) = (D + i) (D – i)y = (D – i) (D + i)y
:
Example :
(D +2x) (D + 1)y = D2y +2xDy + Dy + 2xy
y   2 xy   y   2 xy
but
(D + 1) (D + 2x)y = D2y + D(2xy) + Dy + 2xy
= D2y + 2xDy +2y + Dy +2xy
 y   2 xy  y   2 xy  2 y
 (D + 2x) (D + 1)y  (D + 1) (D + 2x)y
Properties of operator " D "
1. Dmf(x) + Dnf(x) = Dnf(x) + Dmf(x)
2. DmDn = Dn . Dm = Dm+n
wher n and m are non- negative
Dm [Dn f(x)] = Dn [Dm f(x)] = Dm+n f(x) integer number .
3. D [ f(x) + g (x) ] = Df(x) + Dg(x)
where f(x) and g(x) are differential function
4. If c is constant that D[cf(x)] = cDf(x)
5. If b is constant that f(D)ebx = f(b)ebx
proof:
47
 Dnebx = bnebn , D2ebx = b2ebx , Debx = bebx
and since
f(D)ebx = (Dn + a1Dn-1 + ……… +an-1D + an )ebx
= (bn + a1bn-1 + ………….+an-1b + an ) ebx
= f(b) ebx
Example :
Find (D2 + 3D + 2) e3x
Sol :
b=3 , f(D) = D2 + 3D + 2
f(3) = 32 +3 . 3 + 2 = 20
 (D2 + 3D + 2) e3x = f(3)e3x = 20e3x
6. If y is differentiable function and b is constant f(D)[ebxy] = ebx f(D +
b)y
Proof :
D{ebxy} = ebxDy + bebx = ebx(D +b)y
D2 { ebxy } = D { ebx(D + b)y} = ebx { (D + b )(D + b)y }
= ebx(D +b)2y
Dn[ebx] = ebx (D + b)ny
 f(D) = Dn + a1Dn-1 +………. +an-1D + an
f(D)[ebxy] = ebx f (D+b)
48
Example :
find (D2 -4D + 1)[e2x y]
Sol:
b = 2 , f(D) = D2 -4D + 1
 f(D + b) = f(D + 2) = (D + 2)2 – 4 (D + 2) + 1
= D2 + 4D + 4 – 4D – 8 + 1 = D2 – 3
i.e (D2 – 4D + i)[ e2xy] = e2xf(D + 2)y = e2x (D2 – 3)y
7. If b is constant then
f(D2)sinbx = f ( -b2)sinbx
f(D2)cosbx = f(-b2)cosbx
‫مع تبديل في‬
‫متعددة حدود‬
Proof :
D(sinbx) = bcosbx
D2(sinbx) = -b2sinbx
D4(sinx) = b4sinbx = (- b2)2 sinbx
 f(D) sinbx = (D2m + a1D2(n-1) + ……….. + an-1 D2 + an )sinbx
We substitute in operator " D" we obtain
f(D2)sinbx = [(-b2)n + a1(-b2)n-1 +……….. + an-1 (-b2) + an ] sinbx
In the same way we obtain
f(D2)cosbx = f(-b2)cosbx
49
‫حيث أن‬
Example :
find (D4 +3D2 – 1)sin2x
Sol:
b=2
,
f(D2) = D4 +3D2 – 1
f(-22)={ (-22)2 + 3(- 2) – 1}
=16 – 12 – 1 = 3
(D4 + 3D2 -1) sin2x = {(-22)2 + 3(-22) – 1}sin2x = 3sin2x
Example :
(D4 – 2D2)cos2x
Sol:
(D4 – 2D2)cos2x = { (-22)2 -2( -22) } cos2x
= (16 + 8 ) cos2x = 24cos2x
50
51