Unit I
Ordinary Differential Equations
Ordinary Differential Equations
1.1 Definition and Examples of Differential Equations
Definition 1.1 An ordinary differential equation is an equation that contains one
or several derivatives of an unknown function (of single variable) y(x)
that may contain the function itself and given functions of x.
Example 1 a) y = cos x
c)
 4u
4x

b) x 2 y  + x + (x 2  4) y = 0
 4u

 x2  y 2
 4u
 y4
 2u
d)
= 2x  y.
x y
0
If the derivatives that appear in the differential equation are not partial derivatives, then the equation
is called ordinary differential equation. If partial derivatives occur, then the equation is called partial
differential equation.
Thus (a) and (b) are ordinary differential equations while (c) and (d) are partial differential
equations.
Definition 1.2 Order of a D.E.
The order of a differential equation is the order of the highest
derivative which appears in the equation.
Example 2 a) y  + 4y = e x is of order 2.
b) x 2 y y + 2 e x y  = (x 2 + 2) y 2 is of order 3.
c)
4u
2
x y
2
+
 2u
x
2
+
u
= 0 is of order 4.
x
1.1.1 First Order Differential Equations
First order differential equation contains only y and may contain y and given functions of x, hence can
be written as:
F(x, y, y ) = 0
Or
(1)
y = f(x, y)
Example 3 a) y  cos x = 0
b) x y  2y = 0
c) y y  2x = 0
are all first order differential equations.
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Ordinary Differential Equations
Solution of a Differential Equation
A solution of a given first order differential equation (1) on some open interval a < x < b is a function
y = h(x) that has a derivative
y = h (x) and satisfies (1) for a < x < b.
Example 4 Show that y = x 2 is a solution of x y  2y = 0.
Solution
y = 2x . Hence x y = 2 x 2 = 2 y.
Thus y = x 2 is a solution of x y  2y = 0.
Example 5 Show that y = e
Solution
x
is a solution of y + y = 0.
y =  e  x . Hence y + y = e  x  e  x = 0.
Thus y = e
x
is a solution of y + y = 0.
Remark: The solution of a differential equation may appear as an implicit function in the
Form:
H(x, y) = 0
Example 6 x 2  y 2  1is an implicit solution of the differential equation y y + x = 0.
Since
d 2
( x  y 2  1)  0
dx
 2 x  2 y y  0
 x  y y  0.
Hence H(x, y) = x 2  y 2  1.
Note that: The graph of any solution of a d. e. is called an integral curve.
Example 7 y = sin x + c is a solution of y = cos x. For different c we get the following curves.
y
x
for c = 0
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Ordinary Differential Equations
General and Particular Solutions
Solutions of d. e. that involves an arbitrary constant c are called general solutions and if we choose
specific constant c*, we obtain what is called a particular solution of the d. e.
Example 8 y = cos x.
y (x) = sin x + c is a general solution while y (x) = sin x +  is a particular solution.
1.1.2 Initial Value Problems
Given the first order differential equation
y = f (x, y) with additional condition y ( x0 ) = y0 .
The additional condition y ( x0 ) = y0 is called an initial condition.
A differential equation together with an initial condition is called an initial value problem. The initial
condition is used to determine the value of the arbitrary constant c.
Example 9 y = 2x, y (2) = 1.
The general solution is y = x 2 + c and from the initial condition y (2) = 4 + c = 1  c =  3.
Therefore, the solution of the initial value problem is y = x 2  3.
Example 10 y = 2xy, y (1) = e.
The general solution is y = c e
x2
and from the initial condition y (1) = e  c e = e  c = 1.
Therefore, the solution of the initial value problem is y = e
Example 11 y = y, y (0) = 2.
x2
.
The general solution is y = c e x and from the initial condition y (0) = 2  c = 2.
Therefore, y = 2 e x is the solution of the given initial value problem.
Example 12 y = x, y (0) = 3.
The general solution is y =
Therefore, y =
1 2
x + c and from the initial condition y (0) = 3  c = 3.
2
1 2
x + 3 is the solution of the given initial value problem.
2
1.2 Finding Solutions of Differential Equations
1.2.1 Separable Differential Equations
Many first order differential equations can be reduced to the form:
g(y) y = f(x)
(1)
This can also be written as:
g(y) dy = f(x) dx.
(2)
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Since the variables x and y in the above equations are separated, we call such equations separable
differential equations.
An equation of the form (1) can be solved by integration.
 g ( y) y dx
 g ( y) dy



 f ( x) dx
 f ( x) dx
 c
 c
(3)
(3) is the general solution of (1).
Example 13 Solve the d. e.
a) 9y y + 4x = 0
b) (1 y 2) dx = y (1 x) dy
Solutions a) 9y y + 4x = 0  9y dy =  4x dx

 9 y dy
   4 x dx
 c
9 2
y =2x2+c
2
x2
y2

+
=c
9
4

x2
y2
Thus
+
= c where c  0, is the general (implicit) solution of the d. e.
9
4
b) (1 y 2) dx = y (1 x) dy
dx
y
=
dy
1 x
1 y 2
dx
y
 c = 
 
dy
1 x
1 y 2

 

1
2
n 1  y   n 1  x
2
n 1  y
 n
2
1 y2
1  x 
2
 2 n 1  x
c
 c0 where c0 =  2 c.
c
c
2
= c0  1  y 2  e 0 1  x  1  y 2 =  e 0 (1  x) 2
c
Now let c0 =  e 0 .
Therefore, c0 (1  x) 2 + y 2 =1 for some c0 , is the general solution of the d. e.
Example 14 Solve the following initial value problems.
a) y = ay; y (0) = 3 and a is a constant.
b) y + 5 x 4 y 2 = 0 and y (0) = 1.
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Solutions a) y = ay; y (0) = 3.
y = ay 
dy
 a dx 
y

dy

y
 a dx
 c0
c
 n y  ax  c0   y  = e 0 e ax
c
 y = c e ax , where c =  e 0
Now y (0) = 3  c = 3.
Therefore, y = 3 e ax is the solution of the initial value problem.
b) y + 5 x 4 y 2 = 0
dy
y + 5x4y2 = 0 
2
=  5 x 4 dx 

dy
2

  5x
4
y
y
1
1
  x5  c  y = 5
 
0
y
x  c0
dx  c 0
Now y (0) = 1  c0 = 1.
Therefore, y =
1
x5  1
is the solution of the initial value problem.
Example 15 Find the general solution for the differential equation
y = r(y  a) (y  b) where r, a and b are constants.
Solution y = r(y  a) (y  b)
dy
dy
 r dx  

( y  a ) ( y  b)
( y  a ) ( y  b)
1
1
1
(

) dy  r x  c0

(a  b)
ya
y b


 r dx
 c0
ya
ya
 1 
 rx  c0  n
 a  b  rx  c0 
 n
yb
yb
 a b 
 


ya
yb
= e
(a  b) rx
e
(a  b) c0
ya
a  bc e ( a b) rx
( a  b) rx
 ce
 y =
yb
1  c e ( a b) rx
Therefore, y =
a  bc e ( a b) rx
1  c e ( a b) rx
is the general solution.
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Ordinary Differential Equations
Exponential Growth and Exponential Decay
In exponential growth and decay the rate of change of the substance at time t is proportional to the
amount y (t) of the substance present. i.e.
dy
is proportional to y.
dt
dy
 k y for some k  
dt
(1)
The general solution of (1) is of the form y = c e kt.
Exponential growth where k  0 and Exponential decay where k  0
Example 16 Assume that a radioactive substance is under consideration for each time t, let
N (t) denotes the amount of this substance present at time t. N´ (t) is proportional to N (t).
Find the solution to the initial value problem.
N´ (t) = k N (t) and No (t) = No.
Solution The solution is given by
N (t) = c e kt for some k  0.
From No (t) = No we get c = No e  k t .
Therefore, the solution of the initial value problem is N (t) = No e (t  to) kt.
Example 17 A radioactive substance decomposes at a rate proportional to the amount present.
Starting from a given amount of substance say m grams, at a certain time t = to what can be
said about the amount available at a later time?
Solution N’(t) = k N (t) and N 0 (t) = m where k  0.
The general solution is given by N (t) = c e kt.
From N 0 (t) = m we get c = m e  k to.
Therefore, the solution is N (t) = m e (t  to) k.
Example 18 Newton’s Law of Cooling.
A copper ball is heated to a temperature of 100 o c. Then at time t = 0 it is placed in water that is
maintained at a temperature of 30 o c. At the end of 3 min. the temperature of the ball reduced to
70 o c. Find the time at which the temperature of the ball is reduced to 31 o c.
Solution Let T denotes the temperature of the ball at any time t. Then
dT
is proportional to
dt
(T  T0 ) where T0 is the temperature of the surrounding medium.
dT
dT
 k dt  ℓn  T  To  = k t + c1  T = To + c e kt.
= k (T  To) 
dt
T  To
But at t = 0, T (0) = 100 and T0 = 30. Thus 100 = c + 30  c = 70.
Thus
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Ordinary Differential Equations
Now let us determine k.
T(t) = 30 + 70 e kt and at t = 3 min. T(3) = 70 and hence 70 = 30 + 70 e 3k.
 e 3k =
4
1
4
 k=
ℓn ( ).
3
7
7
Thus, k   0.186538596.
Therefore, T(t) = 70 e  0.186538596 t + 30.
Now T(t) = 31  1 = 70 e  0.186538596 t  e  0.186538596 t =
1
70
t
4
1
t
7
ℓn ( ) = ℓn (
)
ℓn ( ) = ℓn (70)
3
70
3
4
7
3 ( n 7  n 10 )
 t 
 t  22.775422
n 7   n 4

Therefore, t  22.78 min.
1.3 Optional Reduction to Separable Form
Certain first order differential equations can be made separable by simple change of variables. This
holds for equation of the form:
y = g (
y
y
) where g is any function of
.
x
x
du
y
. Then y = u x.  y = u + x u´ where u´ =
.
x
dx
u
1
du
dx

and hence

.
Thus g (u) = x u´ + u 
g (u )  u
x
g (u )  u
x
Set u =
Exercises Solve:
1. 2xy y  y2 + x2 = 0.
2. y =
2 x3 cos x 2
y
+
x
y
,y (  )  0
3. (2x  4y + 5) y + x + 2y + 3 = 0.
1.4 Exact Differential Equations
In this section we consider differential equation of the form:
M (x, y) + N (x, y) y = 0
(1)
which is also written as:
M (x, y) dx + N (x, y) dy = 0
(2)
An equation of the form (1) or (2) is called exact if there is a function U (x, y), with continuous
partial derivatives, such that
u
u
= M and
= N.
x
y
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If there exists such a function U (x, y) , then (1) can be written as
u
u
dx +
dy = 0
x
y
(3)
But (3) is called the total differential, du of U (x, y) = c, constant.
i.e. du =
u
u
dx +
dy = 0
x
y
The general solution (implicit solution) of the differential equation (1), if it exists, is given by the
form:
U (x, y) = c0
Example 19 Show that every separable differential equation is exact.
Solution A separable differential equation has the form:
dy
+ f (x) = 0  g (y) dy =  f (x) dx
dx
u
u
Now let U (x, y) = g ( y) dy  f ( x) dx . Then
= f (x) and
= g (y).
x
y
g (y)

(4)

Therefore, (4) is exact.
Theorem 4.1 If M, N and their first partial derivatives are continuous in a
rectangular region in a plane, then equation (1) is exact if and only if
M
y
=
N
x
If (1) is exact, the function U (x, y) can be found in the following ways.
i) U (x, y) =
 M ( x, y) dx
 k ( y) , where k (y) is a constant of integration.
To determine k (y) we use
where G (x, y) =
ii) U (x, y) =
=
dk
G
+
= N (x, y)
dy
y
 M (x, y) dx
 N ( x, y) dy
 l ( x) , where l (x) is a constant of integration.
To determine l (x) we use
where F (x, y) =
u
y
u
x
=
dl
F
+
= M (x, y)
dx
x
 N ( x, y) dy
dy
 0 is exact and find the
Example 20 Show that the differential equation 2 x 3 y 2  x 4 y
dx
general solution.
Solution M (x, y) = 2 x 3 y 2 and N (x, y) = x 4 y .
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Both M and N are continuous and have continuous partial derivatives on R.
Hence, by theorem 4.1, the given differential equation is exact if and only if
M
y
But
=
N
.
x
M
N
= 4x3y =
.
y
x
Thus the given differential equation is exact.
To find the general solution
U (x, y) =
and N (x, y) =
 M ( x, y) dx
u
y

3 y 2 dx  k ( y) = 1 x 4 y 2 + k (y)
2
d
k
x4 y = x4 y +
dy
 2x
 k ( y) =
dk
= 0  k (y) = c1 , constant.
dy

Hence the general solution is given by:
U (x, y) = c0 
1 4 2
x y + c 1 = c0 
2
x 4 y 2 = 2 ( c0 c1) = c
Therefore, the general solution is given by:
x 4 y 2 = c.
Example 21 Initial Value Problem
Solve the initial value problem
(sin x cosh y) dx  (cos x sinh y) dy = 0; y (0) = 0.
Solution M (x, y) = sin x cosh y and N (x, y) =  cos x sinh y.
Now
Hence
M
y
M
y
=
= sin x sinh y and
N
= sin x sinh y.
x
N
.
x
Therefore, the differential equation is exact.
Now U (x, y) =
 N ( x, y) dy
 l ( x) =
  cos x sinh y dy
 l ( x)
=  cos x cosh y + l (x).
Since sin x cosh y =
u
= sin x cosh y + l  (x), l  (x) = 0.
x
Thus l (x) = c1, constant.
Hence the general solution of the differential equation is:
U (x, y) = c0.
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  cos x cosh y + c1 = c0.  cos x cosh y = c1  c0 = c.
From the initial condition y (0) = 0 we get:
c = 1, since cos 0 = cosh 0 = 1.
Therefore, the solution of the initial value problem is:
cos x cosh y = 1.
Exercises 2 Given U (x, y) = c, find the exact differential equation du = 0.
a) x2 + y2 = U (x, y)
b) ℓn (x2y2) = U (x,y)
c) cos ( x2  y2) = U (x, y)
d) cosh (x3  y) = U (x, y)
Exercises 3 Solve the following differential equations.
a) e  x dy  y e  x dx  0
b) (2 x  e y ) dx  x e x dy  0
c) e x cos y dx  e x sin y dy  0
d) sin y  (y  x cos y) y = 0
e)
1
n x
n y  (
 sin y ) y = 0
x
y
g) ℓn (1 + y2 ) dx +
f) 4y2 + 8xy
1
dy
= 0: y (3) =
dx
2
2 xy
dy  0 ; y (2)  e  1
1  y2
h) 2 sin y   cos  y

dy
= 0 ; y (0) =
dx
2
i) 2 xy y = x2 + y2; y (1) = 2.
j) (2xy3  ye  x ) dx + (3x2y2 + e  x  4) dy = 0
Integrating Factor
Given the differential equation
P (x, y) dx + Q (x, y) dy = 0
(1)
that is not exact.
If there is a function F (x, y) such that
F P dx + F Q dy = 0
(2)
is exact, then we call the function F (x, y) an integrating factor of (1).
Example 22 The differential equation y dx  x dy = 0 is not exact.
But if we multiply by F (x, y) =
1
, then the differential equation
x2
y
1
dx  dy  0 is exact. (Verify!).
2
x
x
Hence F (x, y) =
1
is an integrating factor.
x2
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How to find integrating factors
Integrating factors can be found by:
i) inspection
or ii) using the fact that equation (2) is exact.
Now (2) is exact implies that


( F P) 
( F Q) .
y
x
 Fy P + F Py = Fx Q + F Qx.
(*)
In general (*) would be difficult to solve.
Hence we need to look for an integrating factor depending on one variable.
i) If F (x, y) = F (x), then Fx = F  and Fy = 0.
Thus (*) becomes F

P
F
Q

Q F
.
y
x
x
1 dF
1 P
Q

(

).
F dx
Q y
x
(3)
ii) Similarly if F (x, y) = F (y), then (*) becomes
1 dF
1 Q
P

(

)
F dx
P x
y
(4)
Hence we have the following theorems.
Theorem 4.2 (Integrating factor F (x)).
~
If (1) is such that the right hand side of (3), call it R , depends only on x, then
(1) has an integrating factor F = F (x) given by:
F (x) = exp
 R ( x) dx .
~
Theorem 4.3 (Integrating factor F (y)).
~
If (1) is such that the right hand side of (3), call it R , depends only on y, then
(1) has an integrating factor F = F (y) given by:
F (y) = exp
 R ( y) dy .
~
Example 23 Find an integrating factor for the differential equation
2 sin y2 dx + xy cos y2 dy = 0 and solve.
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Solution Let P = 2 sin y2 and Q = xy cos y2.
Now
P
y
Q
= y cos y2 .
x
= 4y cos y2 and
i) Integrating factor F = F (x).
1 dF
1
3

(4 y cos y 2  y cos y 2 ) 
2
F dx
x
x y cos y

dF 3
 dx  n F ( x)  3 n x
F
x
Thus we can take F (x) = x3.
Hence 2 sin y2 dx + xy cos y2 dy = 0  2 x3 sin y2 dx + x4y cos y2 dy = 0.
M
y
Thus, M = 2 x3 sin y2 and N = x4y cos y2 and
= 4 x3 y cos y2 =
N
.
x
1
Now U (x, y) = 2 x 3 sin y 2 dx = x 4 sin y 2 + k (y).

But N =
2
u
dk
= x 4 y cos y 2 

dx
y
dk
dx
= 0  k (y) = c1 .
Thus, the implicit solution for the differential equation is:
U (x, y) = c0 
1 4
x sin y 2 + c1 = c0
2
 x 4 y cos y 2 = 2( c0 – c1 ) = c
 x 4 y cos y 2 = c.
ii) Integrating Factor F (y)
1 dF
1
3 y cos y 2
3
2
2
2

(
y
cos
y
 4 y cos y )  
  y cot y
2
2
F d y 2 sin y
2 sin y 2
3
3
dF
3

  y cot y 2 dy  n F ( x)    y cot y 2 dy   n sin y 2
2
2
2
F

Therefore, F (y) = sin y
Now U (x, y) =
 xy cos y sin 

d
dx

3
4.
2
u
But N =
= sin y 2
x

2

1
4
3
2 4
y


dy   ( x) = 2 x sin y
d

 2 x sin y 2
dx

1
4
2

1
4
  ( x)
  ( x)
= 0  ℓ(x) = c~1 .
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Unit I
Ordinary Differential Equations
Thus, the implicit solution for the differential equation is:

1
2 4
y


U (x, y) = c0  2 x sin

+ c~1 = c0
1
1
1
(c0  c~1 ) 4 = c.
 2 x sin y 2 4 = ( c0 – c~1 )  x 4 sin y 2 =
2
16
Therefore, x 4 sin y 2 = c is the general solution.
Example 24 Solve the initial value problem
2 xy dx  (4 y  3 x 2 ) dy  0 , y (1) = 2
Solution Let P = 2 xy and Q = 4 y  3x 2 .
Now
P
Q
= 2x and
= 6x . Thus, the differential equation is not exact.
x
y
Let us find integrating factor F = F (y).
1 dF
1 Q P
dF
2
 dy  F ( y )  y 2 .
 (

) 

y
F
y
F dy
P x
Thus, M = 2xy3 and N = 4 y 3  3 x 2 y 2

and hence, U (x, y) = 2 xy3 dx  k ( y ) = x 2 y 3  k ( y )
but N =
U
dk
dk
= 3x 2 y 2 
= 4 y 3  3 x 2 y 2 . Thus,
= 4 y 3  k (y) = y 4 + c~1 .
y
dy
dy
The general solution of the differential equation is:
U (x, y) = c0  x 2 y 3 + y 4 + c~1 = c0  x 2 y 3 + y 4 = c.
From the initial condition y (1) = 2 we get:
C = 24.
Therefore, x 2 y 3 + y 4 = 24 is the solution of the initial value problem.
Ordinary Linear Differential Equations
Linear Differential Equations
A first order differential equation is said to be linear if it can be written as
y   P ( x) y  r ( x)
(1)
Note that: Equation (1) is linear in y and y where as P and r can be any functions of x.
Examples 25 y  x 2 y  e x , y  2 y  6 and xy  3 y  x 2  4  0 are examples of linear
differential equations while y  xy2  x and  y  
2
3xy  e x are not linear
differential equations.
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Unit I
Ordinary Differential Equations
In (1) if r (x)  0, i.e. y  P ( x) y  0 , then it is called a homogeneous differential equation;
otherwise it is said to be non-homogeneous.
Now we need to find a formula for the general solution of (1). To do so we proceed by finding a
formula for the general solution of the associated homogeneous differential equation
y  P ( x) y  0
This can be done as follows:
y  P ( x) y  0 
dy
dy
  P ( x) y 
  P ( x) dx
dx
y
c

 n y ( x)   P ( x) dx  c1  y ( x)  e 1 e
  P ( x) dx
  P ( x ) dx
c
 y ( x)  c e
, where c   e 1 .
Note that: y (0)  0 is the trivial solution of such differential equations.
Example 26 Solve the following:
b) y  ( x 2  1) y  0 , y (0) = 2.
a) y  x 2 y  0
dy
1
dy
  x2 y 
Solutions a) y  x 2 y  0 
  x 2 dx  n y ( x)   x 3  c1
dx
3
y
1 3
 x
 y ( x)  c e 3 .
Therefore, y ( x) 
 1 x3
c e 3 is
b) y  ( x 2  1) y  0 
the general solution of the given differential equation.
dy
dy
  ( x 2  1) y 
  ( x 2  1) dx
dx
y
1
 x3  x
1
 n y ( x)  x 3  x  c1  y ( x)  c e 3
.
3
From the initial condition we get:
C = 2.
 1 x3  x
Therefore, y ( x)  2 e 3
is the solution of the given initial value problem.
To find the solution of the non-homogeneous differential equation we proceed as follows:
y   P ( x) y  r ( x)  dy  P ( x) y  r ( x) dx  0
Now put P = P (x) y – r (x) and Q (x) = 1. Thus,
(*)
Q
P
 0 . Consequently, this
 P (x) and
x
y
differential equation is not exact.
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Unit I
Ordinary Differential Equations
We need to find an integrating factor.
1 P Q
1 dF
(

)  P ( x) 
.

x
Q y
F dx
Hence, F (x) = e 
P ( x) dx
is an integrating factor.
Multiplying both sides of the second equation in (*) we get:
P ( x) y  r ( x)dx  0
P ( x) dx
P ( x) y  r ( x) and N = e  P ( x) dx . Hence,
e
e
Now M
P ( x) dx
dy  e 
P ( x) dx
(**)
N
M
P ( x) dx
= P ( x) e 
=
.
x
y
Thus, (**) is exact.
Consequently the general solution of (**) and hence (*) is of the form
U (x, y) = c0 , where
U (x, y) =
 M ( x, y) dx  k ( y)
Thus, U (x, y) =
Now M =
e 
P ( x) dx
or U (x, y) =
dy   ( x)  y e 
U
P ( x ) dx

= y P ( x) e 
x
 y P ( x) e 
d

=
dx
P ( x ) dx
 r ( x) e

d
dx
 P ( x) dx
 N ( x, y) dy   ( x)
P ( x) dx
  ( x)
d
dx
= yP ( x) e 
P ( x) dx
 r ( x) e
 P ( x) dx
 ℓ (x) =   r ( x) e  P ( x) dx  c
1

Thus, yP ( x) e 
= c0  c1 = c.
 r ( x) e
Therefore, the general solution of the differential equation is
P ( x) dx
y ( x)  e
P ( x) dx
  P ( x ) dx 
 c   r ( x) e 

P ( x ) dx
dx 
.

Example 27 Solve the following linear differential equations.
a) y  y  e x
b) t y  2 y  sin t , t  0.
Solutions a) y  y  e x .
 P ( x) dx
 dx
x
Now put P (x) = – 1 and r (x) = e x . Then e 
 e  e
and  r ( x ) e  P ( x ) dx dx   dx  x .
Therefore, y (x) = ( x  c) e x is the general solution.
b) t y  2 y  sin t , t  0  y  
2
sin t
y
.
t
t
2

2
sin t
t
 P ( x) dx
Now put P (t) =
and r (t) =
. Then e 
=e
t
t
dt
e
 2 n t
 t 2
and  r (t ) e  P (t ) dt dt   e t t 2 dt .
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