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1 | 4818-4821 | Some examples of this type of
differential equation are
dx
dy +x
= cos y
2
dx
x
dy
+−y
= y2e – y
To solve the first order linear differential equation of the type
P
dy
y
dx
= Q (1)
Multiply both sides of the equation by a function of x say g (x) to get
g(x) dy
dx + P (g(x)) y = Q g (x) |
1 | 4819-4822 | (1)
Multiply both sides of the equation by a function of x say g (x) to get
g(x) dy
dx + P (g(x)) y = Q g (x) (2)
© NCERT
not to be republished
MATHEMATICS
408
Choose g (x) in such a way that R |
1 | 4820-4823 | (g(x)) y = Q g (x) (2)
© NCERT
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MATHEMATICS
408
Choose g (x) in such a way that R H |
1 | 4821-4824 | g (x) (2)
© NCERT
not to be republished
MATHEMATICS
408
Choose g (x) in such a way that R H S |
1 | 4822-4825 | (2)
© NCERT
not to be republished
MATHEMATICS
408
Choose g (x) in such a way that R H S becomes a derivative of y |
1 | 4823-4826 | H S becomes a derivative of y g (x) |
1 | 4824-4827 | S becomes a derivative of y g (x) i |
1 | 4825-4828 | becomes a derivative of y g (x) i e |
1 | 4826-4829 | g (x) i e g(x) dy
dx + P |
1 | 4827-4830 | i e g(x) dy
dx + P g(x) y = d
dx [y |
1 | 4828-4831 | e g(x) dy
dx + P g(x) y = d
dx [y g (x)]
or
g(x) dy
dx + P |
1 | 4829-4832 | g(x) dy
dx + P g(x) y = d
dx [y g (x)]
or
g(x) dy
dx + P g(x) y = g(x) dy
dx + y g′ (x)
⇒
P |
1 | 4830-4833 | g(x) y = d
dx [y g (x)]
or
g(x) dy
dx + P g(x) y = g(x) dy
dx + y g′ (x)
⇒
P g(x) = g′ (x)
or
P =
( )
( )
g
x
g x
′
Integrating both sides with respect to x, we get
∫Pdx
=
( )
( )
g x dx
′g x
∫
or
P dx
∫⋅
= log(g (x))
or
g(x) =
P dx
e∫
On multiplying the equation (1) by g(x) =
e∫P dx
, the L |
1 | 4831-4834 | g (x)]
or
g(x) dy
dx + P g(x) y = g(x) dy
dx + y g′ (x)
⇒
P g(x) = g′ (x)
or
P =
( )
( )
g
x
g x
′
Integrating both sides with respect to x, we get
∫Pdx
=
( )
( )
g x dx
′g x
∫
or
P dx
∫⋅
= log(g (x))
or
g(x) =
P dx
e∫
On multiplying the equation (1) by g(x) =
e∫P dx
, the L H |
1 | 4832-4835 | g(x) y = g(x) dy
dx + y g′ (x)
⇒
P g(x) = g′ (x)
or
P =
( )
( )
g
x
g x
′
Integrating both sides with respect to x, we get
∫Pdx
=
( )
( )
g x dx
′g x
∫
or
P dx
∫⋅
= log(g (x))
or
g(x) =
P dx
e∫
On multiplying the equation (1) by g(x) =
e∫P dx
, the L H S |
1 | 4833-4836 | g(x) = g′ (x)
or
P =
( )
( )
g
x
g x
′
Integrating both sides with respect to x, we get
∫Pdx
=
( )
( )
g x dx
′g x
∫
or
P dx
∫⋅
= log(g (x))
or
g(x) =
P dx
e∫
On multiplying the equation (1) by g(x) =
e∫P dx
, the L H S becomes the derivative
of some function of x and y |
1 | 4834-4837 | H S becomes the derivative
of some function of x and y This function g(x) =
e∫P dx
is called Integrating Factor
(I |
1 | 4835-4838 | S becomes the derivative
of some function of x and y This function g(x) =
e∫P dx
is called Integrating Factor
(I F |
1 | 4836-4839 | becomes the derivative
of some function of x and y This function g(x) =
e∫P dx
is called Integrating Factor
(I F ) of the given differential equation |
1 | 4837-4840 | This function g(x) =
e∫P dx
is called Integrating Factor
(I F ) of the given differential equation Substituting the value of g (x) in equation (2), we get
P
P
P
dx
dx
dy
e
e
y
dx
=
P
Q
e dx
or
Pdx
d
dxy e
=
P
Q
dx
e
Integrating both sides with respect to x, we get
y e Pdx
=
P
Q
dx
e
dx
or
y =
P
P
Q
C
dx
dx
e
e
dx
which is the general solution of the differential equation |
1 | 4838-4841 | F ) of the given differential equation Substituting the value of g (x) in equation (2), we get
P
P
P
dx
dx
dy
e
e
y
dx
=
P
Q
e dx
or
Pdx
d
dxy e
=
P
Q
dx
e
Integrating both sides with respect to x, we get
y e Pdx
=
P
Q
dx
e
dx
or
y =
P
P
Q
C
dx
dx
e
e
dx
which is the general solution of the differential equation © NCERT
not to be republished
DIFFERENTIAL EQUATIONS
409
Steps involved to solve first order linear differential equation:
(i)
Write the given differential equation in the form
P
Q
dy
dx +y
=
where P, Q are
constants or functions of x only |
1 | 4839-4842 | ) of the given differential equation Substituting the value of g (x) in equation (2), we get
P
P
P
dx
dx
dy
e
e
y
dx
=
P
Q
e dx
or
Pdx
d
dxy e
=
P
Q
dx
e
Integrating both sides with respect to x, we get
y e Pdx
=
P
Q
dx
e
dx
or
y =
P
P
Q
C
dx
dx
e
e
dx
which is the general solution of the differential equation © NCERT
not to be republished
DIFFERENTIAL EQUATIONS
409
Steps involved to solve first order linear differential equation:
(i)
Write the given differential equation in the form
P
Q
dy
dx +y
=
where P, Q are
constants or functions of x only (ii)
Find the Integrating Factor (I |
1 | 4840-4843 | Substituting the value of g (x) in equation (2), we get
P
P
P
dx
dx
dy
e
e
y
dx
=
P
Q
e dx
or
Pdx
d
dxy e
=
P
Q
dx
e
Integrating both sides with respect to x, we get
y e Pdx
=
P
Q
dx
e
dx
or
y =
P
P
Q
C
dx
dx
e
e
dx
which is the general solution of the differential equation © NCERT
not to be republished
DIFFERENTIAL EQUATIONS
409
Steps involved to solve first order linear differential equation:
(i)
Write the given differential equation in the form
P
Q
dy
dx +y
=
where P, Q are
constants or functions of x only (ii)
Find the Integrating Factor (I F) =
ePdx |
1 | 4841-4844 | © NCERT
not to be republished
DIFFERENTIAL EQUATIONS
409
Steps involved to solve first order linear differential equation:
(i)
Write the given differential equation in the form
P
Q
dy
dx +y
=
where P, Q are
constants or functions of x only (ii)
Find the Integrating Factor (I F) =
ePdx (iii)
Write the solution of the given differential equation as
y (I |
1 | 4842-4845 | (ii)
Find the Integrating Factor (I F) =
ePdx (iii)
Write the solution of the given differential equation as
y (I F) =
Q × I |
1 | 4843-4846 | F) =
ePdx (iii)
Write the solution of the given differential equation as
y (I F) =
Q × I F
C
dx
In case, the first order linear differential equation is in the form
1
1
P
Q
dx
x
dy +
=
,
where, P1 and Q1 are constants or functions of y only |
1 | 4844-4847 | (iii)
Write the solution of the given differential equation as
y (I F) =
Q × I F
C
dx
In case, the first order linear differential equation is in the form
1
1
P
Q
dx
x
dy +
=
,
where, P1 and Q1 are constants or functions of y only Then I |
1 | 4845-4848 | F) =
Q × I F
C
dx
In case, the first order linear differential equation is in the form
1
1
P
Q
dx
x
dy +
=
,
where, P1 and Q1 are constants or functions of y only Then I F =
e 1P dy
and the
solution of the differential equation is given by
x |
1 | 4846-4849 | F
C
dx
In case, the first order linear differential equation is in the form
1
1
P
Q
dx
x
dy +
=
,
where, P1 and Q1 are constants or functions of y only Then I F =
e 1P dy
and the
solution of the differential equation is given by
x (I |
1 | 4847-4850 | Then I F =
e 1P dy
and the
solution of the differential equation is given by
x (I F) =
(
)
1
Q × I |
1 | 4848-4851 | F =
e 1P dy
and the
solution of the differential equation is given by
x (I F) =
(
)
1
Q × I F
dy +C
∫
Example 19 Find the general solution of the differential equation
cos
dy
y
x
dx −
= |
1 | 4849-4852 | (I F) =
(
)
1
Q × I F
dy +C
∫
Example 19 Find the general solution of the differential equation
cos
dy
y
x
dx −
= Solution Given differential equation is of the form
P
Q
dy
y
dx +
=
, where P = –1 and Q = cos x
Therefore
I |
1 | 4850-4853 | F) =
(
)
1
Q × I F
dy +C
∫
Example 19 Find the general solution of the differential equation
cos
dy
y
x
dx −
= Solution Given differential equation is of the form
P
Q
dy
y
dx +
=
, where P = –1 and Q = cos x
Therefore
I F =
1 dx
x
e
e
Multiplying both sides of equation by I |
1 | 4851-4854 | F
dy +C
∫
Example 19 Find the general solution of the differential equation
cos
dy
y
x
dx −
= Solution Given differential equation is of the form
P
Q
dy
y
dx +
=
, where P = –1 and Q = cos x
Therefore
I F =
1 dx
x
e
e
Multiplying both sides of equation by I F, we get
x
x
dy
e
e
y
dx
−
−−
= e–x cos x
or
(
dy yex)
dx
−
= e–x cos x
On integrating both sides with respect to x, we get
ye– x =
cos
C
ex
x dx
−
+
∫ |
1 | 4852-4855 | Solution Given differential equation is of the form
P
Q
dy
y
dx +
=
, where P = –1 and Q = cos x
Therefore
I F =
1 dx
x
e
e
Multiplying both sides of equation by I F, we get
x
x
dy
e
e
y
dx
−
−−
= e–x cos x
or
(
dy yex)
dx
−
= e–x cos x
On integrating both sides with respect to x, we get
ye– x =
cos
C
ex
x dx
−
+
∫ (1)
Let
I =
excos
x dx
−
∫
= cos
( sin ) (
)
1
x
x
xe
x
e
dx
−
−
⎛
⎞ −
−
−
⎜
⎝−⎟
⎠ ∫
© NCERT
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MATHEMATICS
410
=
cos
sin
x
x
x e
x e
dx
−
−
−
−∫
=
cos
sin (–
)
cos
(
)
x
x
x
x e
x
e
x
e
dx
−
−
−
⎡
⎤
−
−
−
−
⎣
⎦
∫
=
cos
sin
cos
x
x
x
x e
x e
x e
dx
−
−
−
−
+
−∫
or
I = – e–x cos x + sin x e–x – I
or
2I = (sin x – cos x) e–x
or
I = (sin
cos )
2
x
x
x e−
−
Substituting the value of I in equation (1), we get
ye– x =
sin
cos
C
2
x
x
x e−
−
⎛
⎞
+
⎜
⎟
⎝
⎠
or
y =
sin
cos
C
2
x
x
x
e
−
⎛
⎞ +
⎜
⎟
⎝
⎠
which is the general solution of the given differential equation |
1 | 4853-4856 | F =
1 dx
x
e
e
Multiplying both sides of equation by I F, we get
x
x
dy
e
e
y
dx
−
−−
= e–x cos x
or
(
dy yex)
dx
−
= e–x cos x
On integrating both sides with respect to x, we get
ye– x =
cos
C
ex
x dx
−
+
∫ (1)
Let
I =
excos
x dx
−
∫
= cos
( sin ) (
)
1
x
x
xe
x
e
dx
−
−
⎛
⎞ −
−
−
⎜
⎝−⎟
⎠ ∫
© NCERT
not to be republished
MATHEMATICS
410
=
cos
sin
x
x
x e
x e
dx
−
−
−
−∫
=
cos
sin (–
)
cos
(
)
x
x
x
x e
x
e
x
e
dx
−
−
−
⎡
⎤
−
−
−
−
⎣
⎦
∫
=
cos
sin
cos
x
x
x
x e
x e
x e
dx
−
−
−
−
+
−∫
or
I = – e–x cos x + sin x e–x – I
or
2I = (sin x – cos x) e–x
or
I = (sin
cos )
2
x
x
x e−
−
Substituting the value of I in equation (1), we get
ye– x =
sin
cos
C
2
x
x
x e−
−
⎛
⎞
+
⎜
⎟
⎝
⎠
or
y =
sin
cos
C
2
x
x
x
e
−
⎛
⎞ +
⎜
⎟
⎝
⎠
which is the general solution of the given differential equation Example 20 Find the general solution of the differential equation
2
2
(
0)
xdy
y
x
x
dx +
=
≠ |
1 | 4854-4857 | F, we get
x
x
dy
e
e
y
dx
−
−−
= e–x cos x
or
(
dy yex)
dx
−
= e–x cos x
On integrating both sides with respect to x, we get
ye– x =
cos
C
ex
x dx
−
+
∫ (1)
Let
I =
excos
x dx
−
∫
= cos
( sin ) (
)
1
x
x
xe
x
e
dx
−
−
⎛
⎞ −
−
−
⎜
⎝−⎟
⎠ ∫
© NCERT
not to be republished
MATHEMATICS
410
=
cos
sin
x
x
x e
x e
dx
−
−
−
−∫
=
cos
sin (–
)
cos
(
)
x
x
x
x e
x
e
x
e
dx
−
−
−
⎡
⎤
−
−
−
−
⎣
⎦
∫
=
cos
sin
cos
x
x
x
x e
x e
x e
dx
−
−
−
−
+
−∫
or
I = – e–x cos x + sin x e–x – I
or
2I = (sin x – cos x) e–x
or
I = (sin
cos )
2
x
x
x e−
−
Substituting the value of I in equation (1), we get
ye– x =
sin
cos
C
2
x
x
x e−
−
⎛
⎞
+
⎜
⎟
⎝
⎠
or
y =
sin
cos
C
2
x
x
x
e
−
⎛
⎞ +
⎜
⎟
⎝
⎠
which is the general solution of the given differential equation Example 20 Find the general solution of the differential equation
2
2
(
0)
xdy
y
x
x
dx +
=
≠ Solution The given differential equation is
2
xdy
dx +y
= x2 |
1 | 4855-4858 | (1)
Let
I =
excos
x dx
−
∫
= cos
( sin ) (
)
1
x
x
xe
x
e
dx
−
−
⎛
⎞ −
−
−
⎜
⎝−⎟
⎠ ∫
© NCERT
not to be republished
MATHEMATICS
410
=
cos
sin
x
x
x e
x e
dx
−
−
−
−∫
=
cos
sin (–
)
cos
(
)
x
x
x
x e
x
e
x
e
dx
−
−
−
⎡
⎤
−
−
−
−
⎣
⎦
∫
=
cos
sin
cos
x
x
x
x e
x e
x e
dx
−
−
−
−
+
−∫
or
I = – e–x cos x + sin x e–x – I
or
2I = (sin x – cos x) e–x
or
I = (sin
cos )
2
x
x
x e−
−
Substituting the value of I in equation (1), we get
ye– x =
sin
cos
C
2
x
x
x e−
−
⎛
⎞
+
⎜
⎟
⎝
⎠
or
y =
sin
cos
C
2
x
x
x
e
−
⎛
⎞ +
⎜
⎟
⎝
⎠
which is the general solution of the given differential equation Example 20 Find the general solution of the differential equation
2
2
(
0)
xdy
y
x
x
dx +
=
≠ Solution The given differential equation is
2
xdy
dx +y
= x2 (1)
Dividing both sides of equation (1) by x, we get
2
dy
y
dx
+x
= x
which is a linear differential equation of the type
P
Q
dy
dx +y
=
, where
2
P
=x
and Q = x |
1 | 4856-4859 | Example 20 Find the general solution of the differential equation
2
2
(
0)
xdy
y
x
x
dx +
=
≠ Solution The given differential equation is
2
xdy
dx +y
= x2 (1)
Dividing both sides of equation (1) by x, we get
2
dy
y
dx
+x
= x
which is a linear differential equation of the type
P
Q
dy
dx +y
=
, where
2
P
=x
and Q = x So
I |
1 | 4857-4860 | Solution The given differential equation is
2
xdy
dx +y
= x2 (1)
Dividing both sides of equation (1) by x, we get
2
dy
y
dx
+x
= x
which is a linear differential equation of the type
P
Q
dy
dx +y
=
, where
2
P
=x
and Q = x So
I F =
e∫x2 dx
= e2 log x =
log2
2
x
e
=x
log
( )
[
( )]
f x
as e
f x
=
Therefore, solution of the given equation is given by
y |
1 | 4858-4861 | (1)
Dividing both sides of equation (1) by x, we get
2
dy
y
dx
+x
= x
which is a linear differential equation of the type
P
Q
dy
dx +y
=
, where
2
P
=x
and Q = x So
I F =
e∫x2 dx
= e2 log x =
log2
2
x
e
=x
log
( )
[
( )]
f x
as e
f x
=
Therefore, solution of the given equation is given by
y x2 =
( ) (2
)
C
x
x
dx +
∫
=
3
x dx +C
∫
or
y =
2
2
C
x4
x−
+
which is the general solution of the given differential equation |
1 | 4859-4862 | So
I F =
e∫x2 dx
= e2 log x =
log2
2
x
e
=x
log
( )
[
( )]
f x
as e
f x
=
Therefore, solution of the given equation is given by
y x2 =
( ) (2
)
C
x
x
dx +
∫
=
3
x dx +C
∫
or
y =
2
2
C
x4
x−
+
which is the general solution of the given differential equation © NCERT
not to be republished
DIFFERENTIAL EQUATIONS
411
Example 21 Find the general solution of the differential equation y dx – (x + 2y2) dy = 0 |
1 | 4860-4863 | F =
e∫x2 dx
= e2 log x =
log2
2
x
e
=x
log
( )
[
( )]
f x
as e
f x
=
Therefore, solution of the given equation is given by
y x2 =
( ) (2
)
C
x
x
dx +
∫
=
3
x dx +C
∫
or
y =
2
2
C
x4
x−
+
which is the general solution of the given differential equation © NCERT
not to be republished
DIFFERENTIAL EQUATIONS
411
Example 21 Find the general solution of the differential equation y dx – (x + 2y2) dy = 0 Solution The given differential equation can be written as
dx
x
dy
−y
= 2y
This is a linear differential equation of the type
1
1
P
Q
dx
x
dy +
=
, where 1
1
P
= −y
and
Q1 = 2y |
1 | 4861-4864 | x2 =
( ) (2
)
C
x
x
dx +
∫
=
3
x dx +C
∫
or
y =
2
2
C
x4
x−
+
which is the general solution of the given differential equation © NCERT
not to be republished
DIFFERENTIAL EQUATIONS
411
Example 21 Find the general solution of the differential equation y dx – (x + 2y2) dy = 0 Solution The given differential equation can be written as
dx
x
dy
−y
= 2y
This is a linear differential equation of the type
1
1
P
Q
dx
x
dy +
=
, where 1
1
P
= −y
and
Q1 = 2y Therefore
1
1
log
log( )
1
I |
1 | 4862-4865 | © NCERT
not to be republished
DIFFERENTIAL EQUATIONS
411
Example 21 Find the general solution of the differential equation y dx – (x + 2y2) dy = 0 Solution The given differential equation can be written as
dx
x
dy
−y
= 2y
This is a linear differential equation of the type
1
1
P
Q
dx
x
dy +
=
, where 1
1
P
= −y
and
Q1 = 2y Therefore
1
1
log
log( )
1
I F
dy
y
y
y
e
e
e
y
−
−
−
=∫
=
=
=
Hence, the solution of the given differential equation is
1
x y =
1
(2 )
C
y
⎛ydy
⎞
+
⎜
⎟
⎝
⎠
∫
or
x
y =
(2
)
dy +C
∫
or
x
y = 2y + C
or
x = 2y2 + Cy
which is a general solution of the given differential equation |
1 | 4863-4866 | Solution The given differential equation can be written as
dx
x
dy
−y
= 2y
This is a linear differential equation of the type
1
1
P
Q
dx
x
dy +
=
, where 1
1
P
= −y
and
Q1 = 2y Therefore
1
1
log
log( )
1
I F
dy
y
y
y
e
e
e
y
−
−
−
=∫
=
=
=
Hence, the solution of the given differential equation is
1
x y =
1
(2 )
C
y
⎛ydy
⎞
+
⎜
⎟
⎝
⎠
∫
or
x
y =
(2
)
dy +C
∫
or
x
y = 2y + C
or
x = 2y2 + Cy
which is a general solution of the given differential equation Example 22 Find the particular solution of the differential equation
cot
dy+
y
x
dx
= 2x + x2 cot x (x ≠ 0)
given that y = 0 when
2
x
=π |
1 | 4864-4867 | Therefore
1
1
log
log( )
1
I F
dy
y
y
y
e
e
e
y
−
−
−
=∫
=
=
=
Hence, the solution of the given differential equation is
1
x y =
1
(2 )
C
y
⎛ydy
⎞
+
⎜
⎟
⎝
⎠
∫
or
x
y =
(2
)
dy +C
∫
or
x
y = 2y + C
or
x = 2y2 + Cy
which is a general solution of the given differential equation Example 22 Find the particular solution of the differential equation
cot
dy+
y
x
dx
= 2x + x2 cot x (x ≠ 0)
given that y = 0 when
2
x
=π Solution The given equation is a linear differential equation of the type
P
Q
dy
dx +y
=
,
where P = cot x and Q = 2x + x2 cot x |
1 | 4865-4868 | F
dy
y
y
y
e
e
e
y
−
−
−
=∫
=
=
=
Hence, the solution of the given differential equation is
1
x y =
1
(2 )
C
y
⎛ydy
⎞
+
⎜
⎟
⎝
⎠
∫
or
x
y =
(2
)
dy +C
∫
or
x
y = 2y + C
or
x = 2y2 + Cy
which is a general solution of the given differential equation Example 22 Find the particular solution of the differential equation
cot
dy+
y
x
dx
= 2x + x2 cot x (x ≠ 0)
given that y = 0 when
2
x
=π Solution The given equation is a linear differential equation of the type
P
Q
dy
dx +y
=
,
where P = cot x and Q = 2x + x2 cot x Therefore
I |
1 | 4866-4869 | Example 22 Find the particular solution of the differential equation
cot
dy+
y
x
dx
= 2x + x2 cot x (x ≠ 0)
given that y = 0 when
2
x
=π Solution The given equation is a linear differential equation of the type
P
Q
dy
dx +y
=
,
where P = cot x and Q = 2x + x2 cot x Therefore
I F =
cot
log sin
sin
x dx
x
e
e
x
Hence, the solution of the differential equation is given by
y |
1 | 4867-4870 | Solution The given equation is a linear differential equation of the type
P
Q
dy
dx +y
=
,
where P = cot x and Q = 2x + x2 cot x Therefore
I F =
cot
log sin
sin
x dx
x
e
e
x
Hence, the solution of the differential equation is given by
y sin x = ∫(2x + x2 cot x) sin x dx + C
© NCERT
not to be republished
MATHEMATICS
412
or
y sin x = ∫ 2x sin x dx + ∫x2 cos x dx + C
or
y sin x =
2
2
2
2
2
sin
cos
cos
C
2
2
x
x
x
x
dx
x
x dx
⎛
⎞
⎛
⎞
−
+
+
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
∫
∫
or
y sin x =
2
2
2
sin
cos
cos
C
x
x
x
x dx
x
x dx
−
+
+
∫
∫
or
y sin x = x2 sin x + C |
1 | 4868-4871 | Therefore
I F =
cot
log sin
sin
x dx
x
e
e
x
Hence, the solution of the differential equation is given by
y sin x = ∫(2x + x2 cot x) sin x dx + C
© NCERT
not to be republished
MATHEMATICS
412
or
y sin x = ∫ 2x sin x dx + ∫x2 cos x dx + C
or
y sin x =
2
2
2
2
2
sin
cos
cos
C
2
2
x
x
x
x
dx
x
x dx
⎛
⎞
⎛
⎞
−
+
+
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
∫
∫
or
y sin x =
2
2
2
sin
cos
cos
C
x
x
x
x dx
x
x dx
−
+
+
∫
∫
or
y sin x = x2 sin x + C (1)
Substituting y = 0 and
2
x
=π
in equation (1), we get
0 =
2
sin
C
2
2
π
π
⎛
⎞
⎛
⎞ +
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
or
C =
2
4
−π
Substituting the value of C in equation (1), we get
y sin x =
2
2 sin
4
x
x
π
−
or
y =
2
2
(sin
0)
4 sin
x
x
πx
−
≠
which is the particular solution of the given differential equation |
1 | 4869-4872 | F =
cot
log sin
sin
x dx
x
e
e
x
Hence, the solution of the differential equation is given by
y sin x = ∫(2x + x2 cot x) sin x dx + C
© NCERT
not to be republished
MATHEMATICS
412
or
y sin x = ∫ 2x sin x dx + ∫x2 cos x dx + C
or
y sin x =
2
2
2
2
2
sin
cos
cos
C
2
2
x
x
x
x
dx
x
x dx
⎛
⎞
⎛
⎞
−
+
+
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
∫
∫
or
y sin x =
2
2
2
sin
cos
cos
C
x
x
x
x dx
x
x dx
−
+
+
∫
∫
or
y sin x = x2 sin x + C (1)
Substituting y = 0 and
2
x
=π
in equation (1), we get
0 =
2
sin
C
2
2
π
π
⎛
⎞
⎛
⎞ +
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
or
C =
2
4
−π
Substituting the value of C in equation (1), we get
y sin x =
2
2 sin
4
x
x
π
−
or
y =
2
2
(sin
0)
4 sin
x
x
πx
−
≠
which is the particular solution of the given differential equation Example 23 Find the equation of a curve passing through the point (0, 1) |
1 | 4870-4873 | sin x = ∫(2x + x2 cot x) sin x dx + C
© NCERT
not to be republished
MATHEMATICS
412
or
y sin x = ∫ 2x sin x dx + ∫x2 cos x dx + C
or
y sin x =
2
2
2
2
2
sin
cos
cos
C
2
2
x
x
x
x
dx
x
x dx
⎛
⎞
⎛
⎞
−
+
+
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
∫
∫
or
y sin x =
2
2
2
sin
cos
cos
C
x
x
x
x dx
x
x dx
−
+
+
∫
∫
or
y sin x = x2 sin x + C (1)
Substituting y = 0 and
2
x
=π
in equation (1), we get
0 =
2
sin
C
2
2
π
π
⎛
⎞
⎛
⎞ +
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
or
C =
2
4
−π
Substituting the value of C in equation (1), we get
y sin x =
2
2 sin
4
x
x
π
−
or
y =
2
2
(sin
0)
4 sin
x
x
πx
−
≠
which is the particular solution of the given differential equation Example 23 Find the equation of a curve passing through the point (0, 1) If the slope
of the tangent to the curve at any point (x, y) is equal to the sum of the x coordinate
(abscissa) and the product of the x coordinate and y coordinate (ordinate) of that point |
1 | 4871-4874 | (1)
Substituting y = 0 and
2
x
=π
in equation (1), we get
0 =
2
sin
C
2
2
π
π
⎛
⎞
⎛
⎞ +
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
or
C =
2
4
−π
Substituting the value of C in equation (1), we get
y sin x =
2
2 sin
4
x
x
π
−
or
y =
2
2
(sin
0)
4 sin
x
x
πx
−
≠
which is the particular solution of the given differential equation Example 23 Find the equation of a curve passing through the point (0, 1) If the slope
of the tangent to the curve at any point (x, y) is equal to the sum of the x coordinate
(abscissa) and the product of the x coordinate and y coordinate (ordinate) of that point Solution We know that the slope of the tangent to the curve is dy
dx |
1 | 4872-4875 | Example 23 Find the equation of a curve passing through the point (0, 1) If the slope
of the tangent to the curve at any point (x, y) is equal to the sum of the x coordinate
(abscissa) and the product of the x coordinate and y coordinate (ordinate) of that point Solution We know that the slope of the tangent to the curve is dy
dx Therefore,
dy
dx = x + xy
or
dy
dx −xy
= x |
1 | 4873-4876 | If the slope
of the tangent to the curve at any point (x, y) is equal to the sum of the x coordinate
(abscissa) and the product of the x coordinate and y coordinate (ordinate) of that point Solution We know that the slope of the tangent to the curve is dy
dx Therefore,
dy
dx = x + xy
or
dy
dx −xy
= x (1)
This is a linear differential equation of the type
P
Q
dy
dx +y
=
, where P = – x and Q = x |
1 | 4874-4877 | Solution We know that the slope of the tangent to the curve is dy
dx Therefore,
dy
dx = x + xy
or
dy
dx −xy
= x (1)
This is a linear differential equation of the type
P
Q
dy
dx +y
=
, where P = – x and Q = x Therefore,
I |
1 | 4875-4878 | Therefore,
dy
dx = x + xy
or
dy
dx −xy
= x (1)
This is a linear differential equation of the type
P
Q
dy
dx +y
=
, where P = – x and Q = x Therefore,
I F =
2
2
x
x dx
e
e
−
−∫
=
© NCERT
not to be republished
DIFFERENTIAL EQUATIONS
413
Hence, the solution of equation is given by
2
2
x
y e
⋅−
=
(
)
2
2
( )
C
x
x
dx
e
−
+
∫ |
1 | 4876-4879 | (1)
This is a linear differential equation of the type
P
Q
dy
dx +y
=
, where P = – x and Q = x Therefore,
I F =
2
2
x
x dx
e
e
−
−∫
=
© NCERT
not to be republished
DIFFERENTIAL EQUATIONS
413
Hence, the solution of equation is given by
2
2
x
y e
⋅−
=
(
)
2
2
( )
C
x
x
dx
e
−
+
∫ (2)
Let
I =
2
2
( )
x
x
dx
e
−
∫
Let
2
x2
t
−
= , then – x dx = dt or x dx = – dt |
1 | 4877-4880 | Therefore,
I F =
2
2
x
x dx
e
e
−
−∫
=
© NCERT
not to be republished
DIFFERENTIAL EQUATIONS
413
Hence, the solution of equation is given by
2
2
x
y e
⋅−
=
(
)
2
2
( )
C
x
x
dx
e
−
+
∫ (2)
Let
I =
2
2
( )
x
x
dx
e
−
∫
Let
2
x2
t
−
= , then – x dx = dt or x dx = – dt Therefore, I =
2
2
–
x
t
t
e dt
e
e
−
−
= −
=
∫
Substituting the value of I in equation (2), we get
2
2
x
y e
−
=
2
2 + C
x
e
or
y =
2
2
1
C
x
e
− + |
1 | 4878-4881 | F =
2
2
x
x dx
e
e
−
−∫
=
© NCERT
not to be republished
DIFFERENTIAL EQUATIONS
413
Hence, the solution of equation is given by
2
2
x
y e
⋅−
=
(
)
2
2
( )
C
x
x
dx
e
−
+
∫ (2)
Let
I =
2
2
( )
x
x
dx
e
−
∫
Let
2
x2
t
−
= , then – x dx = dt or x dx = – dt Therefore, I =
2
2
–
x
t
t
e dt
e
e
−
−
= −
=
∫
Substituting the value of I in equation (2), we get
2
2
x
y e
−
=
2
2 + C
x
e
or
y =
2
2
1
C
x
e
− + (3)
Now (3) represents the equation of family of curves |
1 | 4879-4882 | (2)
Let
I =
2
2
( )
x
x
dx
e
−
∫
Let
2
x2
t
−
= , then – x dx = dt or x dx = – dt Therefore, I =
2
2
–
x
t
t
e dt
e
e
−
−
= −
=
∫
Substituting the value of I in equation (2), we get
2
2
x
y e
−
=
2
2 + C
x
e
or
y =
2
2
1
C
x
e
− + (3)
Now (3) represents the equation of family of curves But we are interested in
finding a particular member of the family passing through (0, 1) |
1 | 4880-4883 | Therefore, I =
2
2
–
x
t
t
e dt
e
e
−
−
= −
=
∫
Substituting the value of I in equation (2), we get
2
2
x
y e
−
=
2
2 + C
x
e
or
y =
2
2
1
C
x
e
− + (3)
Now (3) represents the equation of family of curves But we are interested in
finding a particular member of the family passing through (0, 1) Substituting x = 0 and
y = 1 in equation (3) we get
1 = – 1 + C |
1 | 4881-4884 | (3)
Now (3) represents the equation of family of curves But we are interested in
finding a particular member of the family passing through (0, 1) Substituting x = 0 and
y = 1 in equation (3) we get
1 = – 1 + C e0 or C = 2
Substituting the value of C in equation (3), we get
y =
2
2
1
2
x
e
− +
which is the equation of the required curve |
1 | 4882-4885 | But we are interested in
finding a particular member of the family passing through (0, 1) Substituting x = 0 and
y = 1 in equation (3) we get
1 = – 1 + C e0 or C = 2
Substituting the value of C in equation (3), we get
y =
2
2
1
2
x
e
− +
which is the equation of the required curve EXERCISE 9 |
1 | 4883-4886 | Substituting x = 0 and
y = 1 in equation (3) we get
1 = – 1 + C e0 or C = 2
Substituting the value of C in equation (3), we get
y =
2
2
1
2
x
e
− +
which is the equation of the required curve EXERCISE 9 6
For each of the differential equations given in Exercises 1 to 12, find the general solution:
1 |
1 | 4884-4887 | e0 or C = 2
Substituting the value of C in equation (3), we get
y =
2
2
1
2
x
e
− +
which is the equation of the required curve EXERCISE 9 6
For each of the differential equations given in Exercises 1 to 12, find the general solution:
1 2
sin
dy
y
x
dx +
=
2 |
1 | 4885-4888 | EXERCISE 9 6
For each of the differential equations given in Exercises 1 to 12, find the general solution:
1 2
sin
dy
y
x
dx +
=
2 2
3
x
dy
y
e
dx
−
+
=
3 |
1 | 4886-4889 | 6
For each of the differential equations given in Exercises 1 to 12, find the general solution:
1 2
sin
dy
y
x
dx +
=
2 2
3
x
dy
y
e
dx
−
+
=
3 2
dy
y
x
dx
+x
=
4 |
1 | 4887-4890 | 2
sin
dy
y
x
dx +
=
2 2
3
x
dy
y
e
dx
−
+
=
3 2
dy
y
x
dx
+x
=
4 (sec )
tan
0
2
dy
x y
x
x
dx
π
⎛
⎞
+
=
≤
<
⎜
⎟
⎝
⎠
5 |
1 | 4888-4891 | 2
3
x
dy
y
e
dx
−
+
=
3 2
dy
y
x
dx
+x
=
4 (sec )
tan
0
2
dy
x y
x
x
dx
π
⎛
⎞
+
=
≤
<
⎜
⎟
⎝
⎠
5 cos2
tan
xdy
y
x
dx +
=
0
2
x
π
⎛
⎞
≤
<
⎜
⎟
⎝
⎠
6 |
1 | 4889-4892 | 2
dy
y
x
dx
+x
=
4 (sec )
tan
0
2
dy
x y
x
x
dx
π
⎛
⎞
+
=
≤
<
⎜
⎟
⎝
⎠
5 cos2
tan
xdy
y
x
dx +
=
0
2
x
π
⎛
⎞
≤
<
⎜
⎟
⎝
⎠
6 2
2
log
xdy
y
x
x
dx +
=
7 |
1 | 4890-4893 | (sec )
tan
0
2
dy
x y
x
x
dx
π
⎛
⎞
+
=
≤
<
⎜
⎟
⎝
⎠
5 cos2
tan
xdy
y
x
dx +
=
0
2
x
π
⎛
⎞
≤
<
⎜
⎟
⎝
⎠
6 2
2
log
xdy
y
x
x
dx +
=
7 2
log
log
dy
x
x
y
x
dx
x
+
=
8 |
1 | 4891-4894 | cos2
tan
xdy
y
x
dx +
=
0
2
x
π
⎛
⎞
≤
<
⎜
⎟
⎝
⎠
6 2
2
log
xdy
y
x
x
dx +
=
7 2
log
log
dy
x
x
y
x
dx
x
+
=
8 (1 + x2) dy + 2xy dx = cot x dx (x ≠ 0)
© NCERT
not to be republished
MATHEMATICS
414
9 |
1 | 4892-4895 | 2
2
log
xdy
y
x
x
dx +
=
7 2
log
log
dy
x
x
y
x
dx
x
+
=
8 (1 + x2) dy + 2xy dx = cot x dx (x ≠ 0)
© NCERT
not to be republished
MATHEMATICS
414
9 cot
0 (
0)
xdy
y
x
xy
x
x
dx +
−
+
=
≠
10 |
1 | 4893-4896 | 2
log
log
dy
x
x
y
x
dx
x
+
=
8 (1 + x2) dy + 2xy dx = cot x dx (x ≠ 0)
© NCERT
not to be republished
MATHEMATICS
414
9 cot
0 (
0)
xdy
y
x
xy
x
x
dx +
−
+
=
≠
10 (
)
1
dy
x
+y dx
=
11 |
1 | 4894-4897 | (1 + x2) dy + 2xy dx = cot x dx (x ≠ 0)
© NCERT
not to be republished
MATHEMATICS
414
9 cot
0 (
0)
xdy
y
x
xy
x
x
dx +
−
+
=
≠
10 (
)
1
dy
x
+y dx
=
11 y dx + (x – y2) dy = 0
12 |
1 | 4895-4898 | cot
0 (
0)
xdy
y
x
xy
x
x
dx +
−
+
=
≠
10 (
)
1
dy
x
+y dx
=
11 y dx + (x – y2) dy = 0
12 2
(
3
)
(
0)
dy
x
y
y
y
dx
+
=
> |
1 | 4896-4899 | (
)
1
dy
x
+y dx
=
11 y dx + (x – y2) dy = 0
12 2
(
3
)
(
0)
dy
x
y
y
y
dx
+
=
> For each of the differential equations given in Exercises 13 to 15, find a particular
solution satisfying the given condition:
13 |
1 | 4897-4900 | y dx + (x – y2) dy = 0
12 2
(
3
)
(
0)
dy
x
y
y
y
dx
+
=
> For each of the differential equations given in Exercises 13 to 15, find a particular
solution satisfying the given condition:
13 2 tan
sin ;
0 when
3
dy
y
x
x y
x
dx
π
+
=
=
=
14 |
1 | 4898-4901 | 2
(
3
)
(
0)
dy
x
y
y
y
dx
+
=
> For each of the differential equations given in Exercises 13 to 15, find a particular
solution satisfying the given condition:
13 2 tan
sin ;
0 when
3
dy
y
x
x y
x
dx
π
+
=
=
=
14 2
12
(1
)
2
;
0 when
1
1
dy
x
xy
y
x
dx
x
+
+
=
=
=
+
15 |
1 | 4899-4902 | For each of the differential equations given in Exercises 13 to 15, find a particular
solution satisfying the given condition:
13 2 tan
sin ;
0 when
3
dy
y
x
x y
x
dx
π
+
=
=
=
14 2
12
(1
)
2
;
0 when
1
1
dy
x
xy
y
x
dx
x
+
+
=
=
=
+
15 3 cot
sin2 ;
2 when
2
dy
y
x
x y
x
dx
π
−
=
=
=
16 |
1 | 4900-4903 | 2 tan
sin ;
0 when
3
dy
y
x
x y
x
dx
π
+
=
=
=
14 2
12
(1
)
2
;
0 when
1
1
dy
x
xy
y
x
dx
x
+
+
=
=
=
+
15 3 cot
sin2 ;
2 when
2
dy
y
x
x y
x
dx
π
−
=
=
=
16 Find the equation of a curve passing through the origin given that the slope of the
tangent to the curve at any point (x, y) is equal to the sum of the coordinates of
the point |
1 | 4901-4904 | 2
12
(1
)
2
;
0 when
1
1
dy
x
xy
y
x
dx
x
+
+
=
=
=
+
15 3 cot
sin2 ;
2 when
2
dy
y
x
x y
x
dx
π
−
=
=
=
16 Find the equation of a curve passing through the origin given that the slope of the
tangent to the curve at any point (x, y) is equal to the sum of the coordinates of
the point 17 |
1 | 4902-4905 | 3 cot
sin2 ;
2 when
2
dy
y
x
x y
x
dx
π
−
=
=
=
16 Find the equation of a curve passing through the origin given that the slope of the
tangent to the curve at any point (x, y) is equal to the sum of the coordinates of
the point 17 Find the equation of a curve passing through the point (0, 2) given that the sum of
the coordinates of any point on the curve exceeds the magnitude of the slope of
the tangent to the curve at that point by 5 |
1 | 4903-4906 | Find the equation of a curve passing through the origin given that the slope of the
tangent to the curve at any point (x, y) is equal to the sum of the coordinates of
the point 17 Find the equation of a curve passing through the point (0, 2) given that the sum of
the coordinates of any point on the curve exceeds the magnitude of the slope of
the tangent to the curve at that point by 5 18 |
1 | 4904-4907 | 17 Find the equation of a curve passing through the point (0, 2) given that the sum of
the coordinates of any point on the curve exceeds the magnitude of the slope of
the tangent to the curve at that point by 5 18 The Integrating Factor of the differential equation
22
xdy
y
x
dx −
=
is
(A) e–x
(B) e–y
(C) 1
x
(D) x
19 |
1 | 4905-4908 | Find the equation of a curve passing through the point (0, 2) given that the sum of
the coordinates of any point on the curve exceeds the magnitude of the slope of
the tangent to the curve at that point by 5 18 The Integrating Factor of the differential equation
22
xdy
y
x
dx −
=
is
(A) e–x
(B) e–y
(C) 1
x
(D) x
19 The Integrating Factor of the differential equation
2
(1
y) dx
yx
dy
−
+
=
( 1
1)
ay
y
is
(A)
12
y 1
(B)
2
1
y −1
(C)
112
−y
(D)
2
1
1
y
−
Miscellaneous Examples
Example 24 Verify that the function y = c1 eax cos bx + c2 eax sin bx, where c1, c2 are
arbitrary constants is a solution of the differential equation
(
)
2
2
2
2
2
0
d y
ady
a
b
y
dx
dx
−
+
+
=
© NCERT
not to be republished
DIFFERENTIAL EQUATIONS
415
Solution The given function is
y = eax [c1 cosbx + c2 sinbx] |
1 | 4906-4909 | 18 The Integrating Factor of the differential equation
22
xdy
y
x
dx −
=
is
(A) e–x
(B) e–y
(C) 1
x
(D) x
19 The Integrating Factor of the differential equation
2
(1
y) dx
yx
dy
−
+
=
( 1
1)
ay
y
is
(A)
12
y 1
(B)
2
1
y −1
(C)
112
−y
(D)
2
1
1
y
−
Miscellaneous Examples
Example 24 Verify that the function y = c1 eax cos bx + c2 eax sin bx, where c1, c2 are
arbitrary constants is a solution of the differential equation
(
)
2
2
2
2
2
0
d y
ady
a
b
y
dx
dx
−
+
+
=
© NCERT
not to be republished
DIFFERENTIAL EQUATIONS
415
Solution The given function is
y = eax [c1 cosbx + c2 sinbx] (1)
Differentiating both sides of equation (1) with respect to x, we get
dy
dx =
1
2
1
2
–
sin
cos
cos
sin
ax
ax
e
bc
bx
bc
bx
c
bx
c
bx e
a
or
dy
dx =
2
1
2
1
[(
)cos
(
)sin
]
eax
bc
ac
bx
ac
bc
bx
+
+
− |
1 | 4907-4910 | The Integrating Factor of the differential equation
22
xdy
y
x
dx −
=
is
(A) e–x
(B) e–y
(C) 1
x
(D) x
19 The Integrating Factor of the differential equation
2
(1
y) dx
yx
dy
−
+
=
( 1
1)
ay
y
is
(A)
12
y 1
(B)
2
1
y −1
(C)
112
−y
(D)
2
1
1
y
−
Miscellaneous Examples
Example 24 Verify that the function y = c1 eax cos bx + c2 eax sin bx, where c1, c2 are
arbitrary constants is a solution of the differential equation
(
)
2
2
2
2
2
0
d y
ady
a
b
y
dx
dx
−
+
+
=
© NCERT
not to be republished
DIFFERENTIAL EQUATIONS
415
Solution The given function is
y = eax [c1 cosbx + c2 sinbx] (1)
Differentiating both sides of equation (1) with respect to x, we get
dy
dx =
1
2
1
2
–
sin
cos
cos
sin
ax
ax
e
bc
bx
bc
bx
c
bx
c
bx e
a
or
dy
dx =
2
1
2
1
[(
)cos
(
)sin
]
eax
bc
ac
bx
ac
bc
bx
+
+
− (2)
Differentiating both sides of equation (2) with respect to x, we get
2
d y2
dx
=
2
1
2
1
[(
) (
sin
)
(
) ( cos
)]
eax
bc
ac
b
bx
ac
bc
b
bx
+
2
1
2
1
[(
) cos
(
) sin
]
ax |
1 | 4908-4911 | The Integrating Factor of the differential equation
2
(1
y) dx
yx
dy
−
+
=
( 1
1)
ay
y
is
(A)
12
y 1
(B)
2
1
y −1
(C)
112
−y
(D)
2
1
1
y
−
Miscellaneous Examples
Example 24 Verify that the function y = c1 eax cos bx + c2 eax sin bx, where c1, c2 are
arbitrary constants is a solution of the differential equation
(
)
2
2
2
2
2
0
d y
ady
a
b
y
dx
dx
−
+
+
=
© NCERT
not to be republished
DIFFERENTIAL EQUATIONS
415
Solution The given function is
y = eax [c1 cosbx + c2 sinbx] (1)
Differentiating both sides of equation (1) with respect to x, we get
dy
dx =
1
2
1
2
–
sin
cos
cos
sin
ax
ax
e
bc
bx
bc
bx
c
bx
c
bx e
a
or
dy
dx =
2
1
2
1
[(
)cos
(
)sin
]
eax
bc
ac
bx
ac
bc
bx
+
+
− (2)
Differentiating both sides of equation (2) with respect to x, we get
2
d y2
dx
=
2
1
2
1
[(
) (
sin
)
(
) ( cos
)]
eax
bc
ac
b
bx
ac
bc
b
bx
+
2
1
2
1
[(
) cos
(
) sin
]
ax bc
ac
bx
ac
bc
bx e
a
+
+
−
=
2
2
2
2
2
1
2
1
2
1
[(
2
) sin
(
2
) cos
]
eax
a c
abc
b c
bx
a c
abc
b c
bx
−
−
+
+
−
Substituting the values of
2
2 ,
d y dy
dx
dx
and y in the given differential equation, we get
L |
1 | 4909-4912 | (1)
Differentiating both sides of equation (1) with respect to x, we get
dy
dx =
1
2
1
2
–
sin
cos
cos
sin
ax
ax
e
bc
bx
bc
bx
c
bx
c
bx e
a
or
dy
dx =
2
1
2
1
[(
)cos
(
)sin
]
eax
bc
ac
bx
ac
bc
bx
+
+
− (2)
Differentiating both sides of equation (2) with respect to x, we get
2
d y2
dx
=
2
1
2
1
[(
) (
sin
)
(
) ( cos
)]
eax
bc
ac
b
bx
ac
bc
b
bx
+
2
1
2
1
[(
) cos
(
) sin
]
ax bc
ac
bx
ac
bc
bx e
a
+
+
−
=
2
2
2
2
2
1
2
1
2
1
[(
2
) sin
(
2
) cos
]
eax
a c
abc
b c
bx
a c
abc
b c
bx
−
−
+
+
−
Substituting the values of
2
2 ,
d y dy
dx
dx
and y in the given differential equation, we get
L H |
1 | 4910-4913 | (2)
Differentiating both sides of equation (2) with respect to x, we get
2
d y2
dx
=
2
1
2
1
[(
) (
sin
)
(
) ( cos
)]
eax
bc
ac
b
bx
ac
bc
b
bx
+
2
1
2
1
[(
) cos
(
) sin
]
ax bc
ac
bx
ac
bc
bx e
a
+
+
−
=
2
2
2
2
2
1
2
1
2
1
[(
2
) sin
(
2
) cos
]
eax
a c
abc
b c
bx
a c
abc
b c
bx
−
−
+
+
−
Substituting the values of
2
2 ,
d y dy
dx
dx
and y in the given differential equation, we get
L H S |
1 | 4911-4914 | bc
ac
bx
ac
bc
bx e
a
+
+
−
=
2
2
2
2
2
1
2
1
2
1
[(
2
) sin
(
2
) cos
]
eax
a c
abc
b c
bx
a c
abc
b c
bx
−
−
+
+
−
Substituting the values of
2
2 ,
d y dy
dx
dx
and y in the given differential equation, we get
L H S =
2
2
2
2
2
1
2
1
2
1
[
2
)sin
(
2
)cos
]
eax
a c
abc
b c
bx
a c
abc
b c
bx
−
−
+
+
−
2
1
2
1
2
[(
)cos
(
)sin
]
aeax
bc
ac
bx
ac
bc
bx
2
2
1
2
(
)
[
cos
sin
]
ax
a
b
e
c
bx
c
bx
=
(
)
2
2
2
2
2
2
1
2
2
1
2
2
2
2
2
2
2
1
2
1
2
1
1
1
2
2
2
sin
(
2
2
2
)cos
ax
a c
abc
b c
a c
abc
a c
b c
bx
e
a c
abc
b c
abc
a c
a c
b c
bx
⎡
⎤
−
−
−
+
+
+
⎢
⎥
⎢
⎥
+
+
−
−
−
+
+
⎣
⎦
=
[0 sin
0cos
]
eax
bx
bx
×
+
= eax × 0 = 0 = R |
1 | 4912-4915 | H S =
2
2
2
2
2
1
2
1
2
1
[
2
)sin
(
2
)cos
]
eax
a c
abc
b c
bx
a c
abc
b c
bx
−
−
+
+
−
2
1
2
1
2
[(
)cos
(
)sin
]
aeax
bc
ac
bx
ac
bc
bx
2
2
1
2
(
)
[
cos
sin
]
ax
a
b
e
c
bx
c
bx
=
(
)
2
2
2
2
2
2
1
2
2
1
2
2
2
2
2
2
2
1
2
1
2
1
1
1
2
2
2
sin
(
2
2
2
)cos
ax
a c
abc
b c
a c
abc
a c
b c
bx
e
a c
abc
b c
abc
a c
a c
b c
bx
⎡
⎤
−
−
−
+
+
+
⎢
⎥
⎢
⎥
+
+
−
−
−
+
+
⎣
⎦
=
[0 sin
0cos
]
eax
bx
bx
×
+
= eax × 0 = 0 = R H |
1 | 4913-4916 | S =
2
2
2
2
2
1
2
1
2
1
[
2
)sin
(
2
)cos
]
eax
a c
abc
b c
bx
a c
abc
b c
bx
−
−
+
+
−
2
1
2
1
2
[(
)cos
(
)sin
]
aeax
bc
ac
bx
ac
bc
bx
2
2
1
2
(
)
[
cos
sin
]
ax
a
b
e
c
bx
c
bx
=
(
)
2
2
2
2
2
2
1
2
2
1
2
2
2
2
2
2
2
1
2
1
2
1
1
1
2
2
2
sin
(
2
2
2
)cos
ax
a c
abc
b c
a c
abc
a c
b c
bx
e
a c
abc
b c
abc
a c
a c
b c
bx
⎡
⎤
−
−
−
+
+
+
⎢
⎥
⎢
⎥
+
+
−
−
−
+
+
⎣
⎦
=
[0 sin
0cos
]
eax
bx
bx
×
+
= eax × 0 = 0 = R H S |
1 | 4914-4917 | =
2
2
2
2
2
1
2
1
2
1
[
2
)sin
(
2
)cos
]
eax
a c
abc
b c
bx
a c
abc
b c
bx
−
−
+
+
−
2
1
2
1
2
[(
)cos
(
)sin
]
aeax
bc
ac
bx
ac
bc
bx
2
2
1
2
(
)
[
cos
sin
]
ax
a
b
e
c
bx
c
bx
=
(
)
2
2
2
2
2
2
1
2
2
1
2
2
2
2
2
2
2
1
2
1
2
1
1
1
2
2
2
sin
(
2
2
2
)cos
ax
a c
abc
b c
a c
abc
a c
b c
bx
e
a c
abc
b c
abc
a c
a c
b c
bx
⎡
⎤
−
−
−
+
+
+
⎢
⎥
⎢
⎥
+
+
−
−
−
+
+
⎣
⎦
=
[0 sin
0cos
]
eax
bx
bx
×
+
= eax × 0 = 0 = R H S Hence, the given function is a solution of the given differential equation |
1 | 4915-4918 | H S Hence, the given function is a solution of the given differential equation Example 25 Form the differential equation of the family of circles in the second
quadrant and touching the coordinate axes |
1 | 4916-4919 | S Hence, the given function is a solution of the given differential equation Example 25 Form the differential equation of the family of circles in the second
quadrant and touching the coordinate axes Solution Let C denote the family of circles in the second quadrant and touching the
coordinate axes |
1 | 4917-4920 | Hence, the given function is a solution of the given differential equation Example 25 Form the differential equation of the family of circles in the second
quadrant and touching the coordinate axes Solution Let C denote the family of circles in the second quadrant and touching the
coordinate axes Let (–a, a) be the coordinate of the centre of any member of
this family (see Fig 9 |
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