University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
Chapter One
Differential Equations
Definitions
1- Differential Equation (D.E.)
A Differential Equation is an equation that contains one or more derivatives of a
differentiable function. It is equation which contain x, y, y, y,
general as, F ( x, y, y , y ,
, y n . It can expressed in
, y n )  0 . A differential equation is linear if it can be put in
the form
dny
d n1 y
an  x  n  an1  x  n1 
dx
dx
 a1  x 
dy
 a0  x  y  F ( x)
dx
A differential equation is a relationship between an independent variable, x, a
dependent variable, y, and one or more differential coefficients of y with respect to x.
e.g.
dy
d2y
2
x  x y 2  xe x
dx
dx
2
x y  xyy  e x y  20
Differential equations represent dynamic relationships, i.e. quantities that change,
and are thus frequently occurring in scientific and engineering problems.
There are two types of differential equation:
An equation with ordinary derivatives that is, derivatives of a function of a single
variable, is called an Ordinary Differential Equation. While, an equation with partial
derivatives is called a Partial Differential Equation.
Page 1
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
e.g.
d2y
dy
5 2  4 x  3xy (ordinary )
dx
dx
2
 2u
u u
 4x

( partial )
2
x
y t
2- Order of D.E.
The order of a differential equation is the order of the equation’s highest order
derivative involved in the equation.
dy
 y2  0
is anequation of the1st order
dx
d2y
xy 2  y 2 sin x  e x is anequation of the 2nd order
dx
d3y
dy
 y  e 4 x  0 is anequation of the 33 d order
3
dx
dx
y  5 y  6 y  sin x is anequation of the 33 d order
x
3- Degree of D.E.
The degree of a differential equation is the power (exponent) of the equation’s
highest order derivative.
dy
dy
 5 y; 3  sin x  0
dx
dx
3
2
d y
d y
dy
( 3 ) 2  ( 2 )5 
 ex
dx
dx
dx
First order , first deg ree, linear
Third order ,sec ond deg ree, non  linear
Page 2
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
4- Solution of D.E.
To solve a differential equation, we have to find the function for which the equation
is true. This means that we have to manipulate the equation so as to eliminate all the
differential coefficients and leave a relationship between y and x (mean to find y=f(x), which
satisfies the given D.E.).
Solution of First Order Differential Equations
There are various methods of solving first-order differential equations, which we
shall study in this chapter. Second-order equations will be dealt with in a subsequent
scheme.
1) Separation of Variables
A first order differential equations is separable if it can be put in the form
f ( x)dx  g ( y )dy  0
Steps for Solving a Separable First Order D.E.
 Write the equation in the form
f ( x)dx  g ( y )dy , in which it is possible to separate
the variable x from y (the right-hand side is the product of a function of x and a
function of y).
 Integrate f with respect to x and g with respect to y to obtain an equation that relates
y and x (integrating both side).
 f ( x)dx   g ( y )dy  c
Page 3
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
Ex. (1)
x
Solve
dy
 cot( y )
dx
Sol:-
dy
dx

cot( y ) x
dx
x
1
tan(
y
)

dy


  dx
x
 ln  cos( y )   ln ( x)  c
 tan( y ) dy 
Ex. (2)
Solve dx  xydy  y dx  ydy
2
Sol:-
dx  y 2 dx  ydy  xydy
1  y  dx   y  xy  dy
1  y  dx  y 1  x  dy
2
2
1
y
 dx 
 dy
1 x
1  y2
1
y
 dx  
 dy

1 x
1  y2
1
 ln 1  x    ln 1  y 2   c
2
Page 4
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
H.W (1)
Solve the following differential equations of
3
x  2ln x  1
dy

;
dx sin y  y cos y
dy
2x
5

;
dx y  1
dy
 1  y 2  e x
dx
dy
4  e x  y  dx 
x
dy
6
 1  x 1  y 
dx
dy 1  y
7

;
dx 2  x
dy y 2  xy 2
8

;
dx x 2 y  x 2
1  ye x  y  dy  dx ;
dy y 2  1
9

;
dx
x
2
dy x 2  1
10  xy 
;
dx y  1
2) Homogeneous Equation
A first order D.E. in the form
dy
 f  x, y  is homogeneous if it does not depend
dx
y
x
or  . Homogeneous equation is in the form
y
x
on x and y separately, but only the ration 
 y
y  f  
x
Page 5
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
Steps for Solving a Homogeneous First Order D.E.
 Let
u
 And
y
which is also y  ux .
x
dy d  ux 
dx
du

u x
(by the Product Rule), which can be simplified to
dx
dx
dx
dx
dy
du
ux
 thenbecomes y  u  xu .
dx
dx


 Substitute into the original equation when  u  
du 
.
dx 
 du dx 
 .
x 
 u
 Now use separation of variables 
 With the variables now separated, the equation can now be solved by integrating
dx 
 du
  .
x 
 u
with respect to x and u  


 We can then return to x and y by substitution  u 
y

x
Page 6
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
Ex. (3)
Find the solution of the D.E. y  
y  y y
ln   
x x x
Sol:Let u 
y 
y
 y  xu  y  xu   u
x
y  y y
du 

ln    , where  u  

x x x
dx 

xu  u  u ln  u   u
 x
du
 u ln  u 
dx
du
dx

u ln  u  x
ln ln  u   ln  x   c
 y
ln ln    ln  x   c
x
Page 7
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
Ex. (4)
Solve the D.E.  x 2  3 y 2  dx  2 xy  dy  0
Sol:-
x 2  3 y 2  2 xy  y  0
2
 y
 y
1  3    2    y
x
 x
Let
y
 u  y  xu  u
x
2
 y
 y
1  3    2    y
x
 x
1  3u 2  2u   xu   u 
1  3u 2  2uxu   2u 2
du
dx
dx  2u 

du  separate var iable
x 1  u 2 
1  u 2  2ux
  y 
x  c 1    
 x 
2
Page 8
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
Ex. (5)
Solve the D.E.
y  2 y ln  x  ; and y  0   1
Sol:-
dy
 2 y  ln  x  
dx
dy
 ln  x    dx
2 y
1  12
 y  dy   ln  x   dx
2
y  x ln  x   x  c
y  0  1
 I  C 
1  0  0  c  c 1
y  x ln  x   x  1
y   x ln  x   x  1
2
Page 9
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
Ex. (6)

 x 
x

y
cot
Solve the D.E. 
 y    dy  y  dx
 

Sol:-

 x 
x

y
cot

 y    dy  y  dx
 

x
x
 cot    x
y
 y
Let
x


u
y
  x  yu


dx
du
 y u
dy
dy
du
u  cot  u   y  u
dy
dy
du

y cot  u 
x 
dy
  tan  u   du
y
ln  y  
Page 10
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
H.W (2)
Solve the following differential equations of
1  yy  2 y  x ;
2  x 2 y 2 y  xy 3  1
3  2 xyy  y  y  x ;
dy
x2  y 2
4  
dx
2 xy
dy
x  3y
5  
;
dx
2x
dy
2 xy  3 y 2
6   2
dx
x  2 xy
2
4
4
3) Linear Differential Equation
A first‐order differential equation is said to be linear if it can be expressed in the
form
y  P( x) y  Q( x)
where P and Q are functions of x.
Now, we are going to assume that there is some magical function somewhere out
there in the world, μ called an integrating factor, the integrating factor is defined by the
formula
  e
P  x dx
The solution is
  y  x      Q  x   dx
Page 11
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
Steps for Solving a Linear First Order D.E.
 Put it in standard form and identify the functions P and Q.
 Find an anti-derivative of P(x).
 Find the integrating factor 
 e
P  x dx
.
 Find y using the following equation   y  x      Q  x   dx .
Ex. (7)
Solve the equation
x
dy
 3 y  x2
dx
Sol:Step 1: Put the equation in standard form and identify the functions P and Q. To do so,
we divide both sides of the equation by the coefficient of dy/dx, in this case x, obtaining
dy 3
3
 y  x  P x   , Q x  x
dx x
x
Step 2: Find an anti-derivative of P(x).
3
1
P
x

dx



dx


3




  dx  3ln  x 
x
x
Step 3: Find the integrating factor μ.
 e
 P x dx
e
3ln  x 
e
ln x 3
e
ln
1
x3

1
x3
Page 12
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
Step 4: Find the solution.
  y  x      Q  x   dx
1
1

y

x  dx

x3
x3
1
1

y

 2  dx
x3
x
1
 y   x 2  dx
3
x
1
1
y  c
3
x
x
y   x 2  cx 3
Ex. (8)
Solve the equation xy  y  x
4
Sol:-
1
y  y  x 3
x
1
x
 P  x   dx    dx  ln  x 
  e
P  x dx
 eln x  x
x  y   x 4  dx
1
x  y  x5  c
5
Page 13
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
Ex. (9)
Solve the equation y  y  tan  x   sec  x 
Sol:-
 P  x   dx   tan  x   dx   ln  cos x 
  e
 e  lncos x   elncos x    cos x  
P  x dx
1
1
1
 sec  x 
 cos  x  
sec  x   y   sec  x   sec  x   dx   sec 2  x   dx
y  sec  x   tan  x   c
Ex. (10)
Solve the equation
 2 x  10 y 
3
dy
 y0
dx
Sol:-
dy
0
dx
3
2 x  10 y  y  x  0
y  x  2 x  10 y 3
 2 x  10 y   y 
3
x 
2
x  10 y 2
y
2
P
y

dy




  dy  2ln  y 
y
Page 14
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
  e
P  y dy
 e 2ln y   eln y   y 2
2
x  y 2   y 2   10 y 2   dy   10 y 4  dy
y5
2
x  y  10  c
5
x  y 2  2 y 5  c
H.W (3)
Solve the following differential equations of
dy
 y  ex
dx
2 y e x
2
1
3  1  x  dx   y  tan  x   dx  0; 4  y 

x
x
dy
5  1  x 2   dy  dx   2 xydx ;
6  x   3 y  x2
dx
x
2
7  cosh x  dy   y  sinh x  e  dx  0; 8   y  1 dx   2 xy  1 dy
1  e 2 y  dx  2  xe 2 y  y  dy  0;
2
2
Page 15
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
4) Bernoulli’s Equation
A first‐order differential equation is said to be Bernoulli if it can be expressed in
the form
y  P( x) y  Q( x) y n
where P and Q are functions of x, n is any real number but not 0 or 1.
When n = 0 the equation can be solved as a First Order Linear D.E., and when n = 1 the
equation can be solved using Separation of Variables D.E.
For other values of n we can solve it by substituting
  y1n
and turning it into a linear differential equation (and then solve that).
Ex. (11)
Solve the equation xy  y  xy
3
Sol:-
First Stage: Converting an equation from Bernoulli to linear D.E.
1) xy  y  xy 3
xy 3 y  y 2  x
2) Let   y 2
Page 16
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
3)  x  

 x
2
2
      2
x
 2 
        2
 x 
Second Stage: Solve the equation by linear D.E. with
1)   e
2)
 P x dx
e
2
  x dx
2
 e2ln x  eln x  x 2 
2
P  x    , Q  x   2
x
1
x2
1
1
    2   2   dx  c
2
x
x
1
2



c
x2
x
Return  to its origin    y 2
1 2 2
y  c
x2
x
1
2

c
2
2
x y
x
Ex. (12)
Solve the equation 3 xy  y  x y  0
2
4
Page 17
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
Sol:First Stage: Converting an equation from Bernoulli to linear D.E.
3xy  y  x 2 y 4  0
3xy 4 y  y 3  x 2  0
Let
  y 3   3 y 4 y    
3 x 

   x2  0
3
y 4 y 

3
 1
       x
 x
Second Stage: Solve the equation by linear D.E. with
1)   e
2)
 P x dx
e
1
  x dx
1
 e  ln x  eln x  x 1 
1
P x   , Q x  x
x
1
x
1
1
      x   dx  c
x
x
1
  xc
x
Return  to its origin    y 3
1
 xc
x  y3
Page 18
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
H.W (4)
Solve the following differential equations of
1  yy  y 2  x ;
2  cos  y   y  sin  y   x
dy 1
  y  xy 2 ;
dx x
dy
5  2 y  3  y 4  e3 x ;
dx
4  x2 y  x3 
3
dy
 y 4  cos  x 
dx
dy
6  y  2 x   x  x  1 y 3
dx
5) Exact Differential Equations
The equation M  x, y  dx  N  x, y  dy  0 is an exact differential equation if
there exists a function f  x, y  of two variables x and y having continuous partial
derivatives such that the exact differential equation definition is separated as follows
ux  x, y   M  x, y  and u y  x, y   N  x, y 
Therefore, the general solution of the equation is u  x, y   c , where “c” is an arbitrary
constant.
Testing for Exactness
Assume the functions M  x, y  and N  x, y  having the continuous partial
derivatives, and the differential equation is exact if and only if it satisfies the condition
M N

y
x
Page 19
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
Consider an example,
 The equation
derivatives
x
2
 y 2  dx   2 xy  cos  y   dy  0 is exact because the partial
M  2
  x  y2   2 y ;
y y
N 
  2 xy  cos  y    2 y
x x
are equal.
 The equation
derivatives
 x  3 y  dx   x
2
 cos  y   dy  0 is not exact because the partial
M 
  x  3y  3 ;
y y
N 
  2 xy  cos  y    2 x
x x
are not equal.
Steps for Solving an Exact First Order D.E.
 Match the equation to the form M  x, y  dx  N  x, y  dy  0 to identify M and N.
 Integrate M (of N) with respect to x (or y), writing the constant of integration as f(y)
(or g(x)).
 Differentiate with respect to y (or x) and set the result equal to N (or M) to find f′(y)
(or g′(x)).
 Integrate to find f (y) (or g(x)).
 Write the solution of the exact equation as f (x, y) = C.
Page 20
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
There are Two Types of Solutions
First Solution
Integrate the First Part  Integrate the Second Part 
Summation of the Two Parts Without Re petition
Second Solution
Integrate  Dervative  Compare withthe sec ond term  Bring f 
Page 21
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
Ex. (13)
Solve the differential equation
e y  cos  x  y   2 x  dx   xe y  cos  x  y   1 dy  0
Sol:Firs Stage: Check the equation is exact or not
M  y
  e  cos  x  y   2 x   e y  sin  x  y   0
y y
N 
  xe y  cos  x  y   1  e y  sin  x  y   0
x x
 M y  N x equation is Exact
1) ux  x, y    M  x, y  dx    e y  cos  x  y   2 x  dx
 e y  sin  x  y   x 2  f  y 
2) u y  x, y    N  x, y  dy    xe y  cos  x  y   1 dy
 xe y  sin  x  y   y  g  x 
3) uG . S  x, y   xe y  sin  x  y   x 2  y  c
Page 22
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
Another Solution by Using the Second Method
Step 1: Match the equation to the form M  x, y  dx  N  x, y  dy  0 to identify M.
e y  cos  x  y   2 x
Step 2: Integrate M with respect to x, writing the constant of integration as f  y  .
1) ux  x, y    M  x, y  dx    e y  cos  x  y   2 x  dx
 e y  sin  x  y   x 2  f  y 
Step 3: Differentiate with respect to y and set the result equal to N to find f   y  .
2)
Step 4:

 xe y  cos  x  y   f   y 
y
u
N
y
 xe y  cos  x  y   f   y   xe y  cos  x  y   1
Step 5: Integrate to find f  y  .
f  y    f   y  dy   1  dy
f  y   y
Step 6: Write the solution of the exact equation
 u  x, y   xe y  sin  x  y   x 2  y  c
Page 23
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
Ex. (14)
Solve the differential equation
y
2
 1 dx   2 xy  sin y  dy  0
Sol:Firs Stage: Check the equation is exact or not
M

  y 2  1  2 y
y y
N 
  2 xy  sin y   2 y
x x
 M y  N x The equation is Exact
1) u x  x, y    M  x, y  dx    y 2  1 dx
 y2 x  x  f  y 
2) u y  x, y    N  x, y  dy    2 xy  sin y  dy
 xy 2  cos  y   g  x 
3) uG . S  x, y   xy 2  x  cos  y   c
Page 24
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
Ex. (15)
Solve the differential equation
x
2
 y 2  dx   2 xy  cos  y   dy  0
Sol:Step 1: Match the equation to the form M  x, y  dx  N  x, y  dy  0 to identify M.
M  x, y   x 2  y 2
Step 2: Integrate M with respect to x, writing the constant of integration as f  y  .
x3
ux  x, y    M  x, y  dx    x  y  dx   xy 2  f  y 
3
2
2
Step 3: Differentiate with respect to y and set the result equal to N to find f   y  .
  x3

2
  xy  f  y    2 xy  f   y 
y  3

2 xy  f   y   2 xy  cos  y   f   y   cos  y 
Step 4: Integrate to find f  y  .
 f   y  dy   cos  y  dy  sin  y 
Step 5: Write the solution of the exact equation as f  x, y   c .
x3
 xy 2  sin  y   c
3
Page 25
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
Another Solution
Step 1: Match the equation to the form M  x, y  dx  N  x, y  dy  0 to identify N.
N  x, y   2 xy  cos  y 
Step 2: Integrate N with respect to y, writing the constant of integration as g  x  .
u y  x, y    N  x, y  dy    2 xy  cos  y   dy  xy 2  sin  y   g  x 
Step 3: Differentiate with respect to x and set the result equal to M to find g   x  .

xy 2  sin  y   g  x    y 2  g   x 

y
y2  g x   x2  y 2  g x   x2
Step 4: Integrate to find g  x  .
x3
 g   x  dx   x dx 
3
2
Step 5: Write the solution of the exact equation as f  x, y   c .
x3
 xy 2  sin  y   c
3
Page 26
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
H.W (5)
Solve the following differential equations of
1   3 x 2  6 xy  dx   3 x 2  2 y  dy  0
2   x 2  y 2  dx   2 xy  cos y  dy
3  2 xy dx   x 2  3 y 2  dy
4   6 x 2  y  3 dx   3 y 2  x  2  dy
6) Integration Factor Method
The equation in integrating factor is a similar form to the exact equation
M  x, y  dx  N  x, y  dy  0
In order to use the method of exact equations, we need to assess if the solution is a
linear equation that will result in a constant. To do so, we need to identify the terms of M
and the terms of y and take the partial derivative of M with respect to y, and the partial
derivative of N with respect to x, and then compare these results. If they are equal, it means
this first-order differential equation is exact, and we can follow through with this technique
to find the solution.
M N

y
x
But what if M y is not equal to N x
M y  Nx
Page 27
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
What do we do now? Well, this is how we end up with the integrating factor
technique    .
  M  x, y  dx    N  x, y  dy  0
Steps for Solving an Integrating Factor First Order D.E.
Below are the steps to solve the first-order D.E. using the integrating factor.
Step 1: Calculate the integrating factor    .
M y  Nx
f  x  dx
   e
N
M  Nx
g  y  dy
2) y
   e
M
1)
Step 2: Multiply the differential equation with integrating factor    in such a way;
  M  x, y  dx    N  x, y  dy  0
Set this new equation to be exact.
Step 3: Now, solve the new equation in step 2, and overlook the old equation (original
equation).
 M y  N x The equation is Exact
Page 28
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
1) ux  x, y    M  x, y  dx
2) u y  x, y    N  x, y  dy
 Write the solution of the exact equation as f (x, y) = C.
3) uG . S  x, y    x   y  c
Ex. (16)
Solve the differential equation  3xy 2  2 y  dx   2 x 2 y  x  dy  0
Sol:Firs Stage: Check the equation is exact or not
M 
  3 xy 2  2 y   6 xy  2
y y
N 
  2 x 2 y  x   4 xy  1
x x
 M y  N x The equation is not exact
M y  N x  6 xy  2    4 xy  1 6 xy  2  4 xy  1


2
N
2
x
y

x
x  2 xy  1



 2 xy  1  1
x  2 xy  1 x
Page 29
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
  e
f  x  dx
 eln x   x    x
   3xy 2  2 y  dx     2 x 2 y  x  dy  0
x   3 xy 2  2 y  dx  x   2 x 2 y  x  dy  0
 3x
2
y 2  2 xy  dx   2 x 3 y  x 2  dy  0
1) u x  x, y    M  x, y  dx    3x 2 y 2  2 xy  dx
 x3 y 2  x 2 y  f  y 
2) u y  x, y    N  x, y  dy    2 x 3 y  x 2  dy
 x3 y 2  x 2 y  g  x 
3) uG . S  x, y   x3 y 2  x 2 y  c
Page 30
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
Ex. (17)
Solve the differential equation  x 2  y 2  dx  2 xy  dx
Sol:-
x
2
 y 2  dy  2 xy  dx
2 xy  dx   x 2  y 2  dy  0
 2 xy  dx    x  y  dy  0
 2 xy  dx    x  y  dy  0
 2 xy  dx   y  x  dy  0
2
2
2
2
2
2
Firs Stage: Check the equation is exact or not
M

  2 xy   2 x
y y
N  2
  y  x 2   2 x
x x
 M y  N x The equation is not exact
M y  N x 2 x   2 x  2 x  2 x


M
  2 xy 
2 xy

4x
2

2 xy y
Page 31
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
 e
 g  y  dy
e

2
dy
y
2
 e 2ln y   eln  y   y 2 
1
y2
 
1
y2
   2 xy  dx     y 2  x 2  dy  0
1
1

2
xy
dx

  y 2  x 2  dy  0


2
2
y
y
 2x 
 x2 
 y   dx  1  y 2   dy  0
 


1) u x  x, y    M  x, y  dx  
2x
 dx
y
x2
  f  y
y
 x2 
2) u y  x, y    N  x, y   dy   1  2   dy
y 

x2
 y   g  x
y
x2
3) uG . S  x, y    y  c
y
Page 32
University of Anbar
College of Engineering
Department of Electrical Engineering (Stage: 2)
Engineering Mathematics
Dr. Zeyid Tariq Ibraheem
H.W (6)
Solve the following differential equations of
dy
dy
 y  x;
2   y  e x
dx
dx
dy
dy
3  x   2 y  10 x 2 ;
4  x   y  x2
dx
dx
dy
dy
5  x   2 y  x 4  sin  x  ;
6  x   2 y  x2
dx
dx
dy
dy
7   y  cot  x   cos ex  x  ; 8   y  cot  x   cos  x 
dx
dx
dy
dy
9   x 2  1   2 xy  x ;
10 
 y  tan  x   sec  x 
dx
dx
1
Page 33