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1 | 4518-4521 | , the solution obtained from the general
solution by giving particular values to the arbitrary constants is called a particular
solution of the differential equation Example 2 Verify that the function y = e– 3x is a solution of the differential equation
2
2
6
0
d y
dy
y
dx
dx
+
−
=
Solution Given function is y = e– 3x Differentiating both sides of equation with respect
to x , we get
3
3
x
dy
e
dx
−
= − (1)
Now, differentiating (1) with respect to x, we have
2
d y2
dx
= 9 e – 3x
Substituting the values of
2
2 ,
d y dy
dx
dx
and y in the given differential equation, we get
L |
1 | 4519-4522 | Example 2 Verify that the function y = e– 3x is a solution of the differential equation
2
2
6
0
d y
dy
y
dx
dx
+
−
=
Solution Given function is y = e– 3x Differentiating both sides of equation with respect
to x , we get
3
3
x
dy
e
dx
−
= − (1)
Now, differentiating (1) with respect to x, we have
2
d y2
dx
= 9 e – 3x
Substituting the values of
2
2 ,
d y dy
dx
dx
and y in the given differential equation, we get
L H |
1 | 4520-4523 | Differentiating both sides of equation with respect
to x , we get
3
3
x
dy
e
dx
−
= − (1)
Now, differentiating (1) with respect to x, we have
2
d y2
dx
= 9 e – 3x
Substituting the values of
2
2 ,
d y dy
dx
dx
and y in the given differential equation, we get
L H S |
1 | 4521-4524 | (1)
Now, differentiating (1) with respect to x, we have
2
d y2
dx
= 9 e – 3x
Substituting the values of
2
2 ,
d y dy
dx
dx
and y in the given differential equation, we get
L H S = 9 e– 3x + (–3e– 3x) – 6 |
1 | 4522-4525 | H S = 9 e– 3x + (–3e– 3x) – 6 e– 3x = 9 e– 3x – 9 e– 3x = 0 = R |
1 | 4523-4526 | S = 9 e– 3x + (–3e– 3x) – 6 e– 3x = 9 e– 3x – 9 e– 3x = 0 = R H |
1 | 4524-4527 | = 9 e– 3x + (–3e– 3x) – 6 e– 3x = 9 e– 3x – 9 e– 3x = 0 = R H S |
1 | 4525-4528 | e– 3x = 9 e– 3x – 9 e– 3x = 0 = R H S Therefore, the given function is a solution of the given differential equation |
1 | 4526-4529 | H S Therefore, the given function is a solution of the given differential equation Example 3 Verify that the function y = a cos x + b sin x, where, a, b ∈ R is a solution
of the differential equation
2
2
0
d y
y
dx
+
=
Solution The given function is
y = a cos x + b sin x |
1 | 4527-4530 | S Therefore, the given function is a solution of the given differential equation Example 3 Verify that the function y = a cos x + b sin x, where, a, b ∈ R is a solution
of the differential equation
2
2
0
d y
y
dx
+
=
Solution The given function is
y = a cos x + b sin x (1)
Differentiating both sides of equation (1) with respect to x, successively, we get
dy
dx = – a sinx + b cosx
2
d y2
dx
= – a cos x – b sinx
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DIFFERENTIAL EQUATIONS
385
Substituting the values of
2
d y2
dx
and y in the given differential equation, we get
L |
1 | 4528-4531 | Therefore, the given function is a solution of the given differential equation Example 3 Verify that the function y = a cos x + b sin x, where, a, b ∈ R is a solution
of the differential equation
2
2
0
d y
y
dx
+
=
Solution The given function is
y = a cos x + b sin x (1)
Differentiating both sides of equation (1) with respect to x, successively, we get
dy
dx = – a sinx + b cosx
2
d y2
dx
= – a cos x – b sinx
© NCERT
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DIFFERENTIAL EQUATIONS
385
Substituting the values of
2
d y2
dx
and y in the given differential equation, we get
L H |
1 | 4529-4532 | Example 3 Verify that the function y = a cos x + b sin x, where, a, b ∈ R is a solution
of the differential equation
2
2
0
d y
y
dx
+
=
Solution The given function is
y = a cos x + b sin x (1)
Differentiating both sides of equation (1) with respect to x, successively, we get
dy
dx = – a sinx + b cosx
2
d y2
dx
= – a cos x – b sinx
© NCERT
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DIFFERENTIAL EQUATIONS
385
Substituting the values of
2
d y2
dx
and y in the given differential equation, we get
L H S |
1 | 4530-4533 | (1)
Differentiating both sides of equation (1) with respect to x, successively, we get
dy
dx = – a sinx + b cosx
2
d y2
dx
= – a cos x – b sinx
© NCERT
not to be republished
DIFFERENTIAL EQUATIONS
385
Substituting the values of
2
d y2
dx
and y in the given differential equation, we get
L H S = (– a cos x – b sin x) + (a cos x + b sin x) = 0 = R |
1 | 4531-4534 | H S = (– a cos x – b sin x) + (a cos x + b sin x) = 0 = R H |
1 | 4532-4535 | S = (– a cos x – b sin x) + (a cos x + b sin x) = 0 = R H S |
1 | 4533-4536 | = (– a cos x – b sin x) + (a cos x + b sin x) = 0 = R H S Therefore, the given function is a solution of the given differential equation |
1 | 4534-4537 | H S Therefore, the given function is a solution of the given differential equation EXERCISE 9 |
1 | 4535-4538 | S Therefore, the given function is a solution of the given differential equation EXERCISE 9 2
In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a
solution of the corresponding differential equation:
1 |
1 | 4536-4539 | Therefore, the given function is a solution of the given differential equation EXERCISE 9 2
In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a
solution of the corresponding differential equation:
1 y = ex + 1
:
y″ – y′ = 0
2 |
1 | 4537-4540 | EXERCISE 9 2
In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a
solution of the corresponding differential equation:
1 y = ex + 1
:
y″ – y′ = 0
2 y = x2 + 2x + C
:
y′ – 2x – 2 = 0
3 |
1 | 4538-4541 | 2
In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a
solution of the corresponding differential equation:
1 y = ex + 1
:
y″ – y′ = 0
2 y = x2 + 2x + C
:
y′ – 2x – 2 = 0
3 y = cos x + C
:
y′ + sin x = 0
4 |
1 | 4539-4542 | y = ex + 1
:
y″ – y′ = 0
2 y = x2 + 2x + C
:
y′ – 2x – 2 = 0
3 y = cos x + C
:
y′ + sin x = 0
4 y =
2
1
+x
:
y′ =
2
1
xy
x
+
5 |
1 | 4540-4543 | y = x2 + 2x + C
:
y′ – 2x – 2 = 0
3 y = cos x + C
:
y′ + sin x = 0
4 y =
2
1
+x
:
y′ =
2
1
xy
x
+
5 y = Ax
:
xy′ = y (x ≠ 0)
6 |
1 | 4541-4544 | y = cos x + C
:
y′ + sin x = 0
4 y =
2
1
+x
:
y′ =
2
1
xy
x
+
5 y = Ax
:
xy′ = y (x ≠ 0)
6 y = x sin x
:
xy′ = y + x
2
2
x
−y
(x ≠ 0 and x > y or x < – y)
7 |
1 | 4542-4545 | y =
2
1
+x
:
y′ =
2
1
xy
x
+
5 y = Ax
:
xy′ = y (x ≠ 0)
6 y = x sin x
:
xy′ = y + x
2
2
x
−y
(x ≠ 0 and x > y or x < – y)
7 xy = log y + C
:
y′ =
2
1
y
−xy
(xy ≠ 1)
8 |
1 | 4543-4546 | y = Ax
:
xy′ = y (x ≠ 0)
6 y = x sin x
:
xy′ = y + x
2
2
x
−y
(x ≠ 0 and x > y or x < – y)
7 xy = log y + C
:
y′ =
2
1
y
−xy
(xy ≠ 1)
8 y – cos y = x
:
(y sin y + cos y + x) y′ = y
9 |
1 | 4544-4547 | y = x sin x
:
xy′ = y + x
2
2
x
−y
(x ≠ 0 and x > y or x < – y)
7 xy = log y + C
:
y′ =
2
1
y
−xy
(xy ≠ 1)
8 y – cos y = x
:
(y sin y + cos y + x) y′ = y
9 x + y = tan–1y
:
y2 y′ + y2 + 1 = 0
10 |
1 | 4545-4548 | xy = log y + C
:
y′ =
2
1
y
−xy
(xy ≠ 1)
8 y – cos y = x
:
(y sin y + cos y + x) y′ = y
9 x + y = tan–1y
:
y2 y′ + y2 + 1 = 0
10 y =
2
2
a
−x
x ∈ (–a, a) :
x + y dy
dx = 0 (y ≠ 0)
11 |
1 | 4546-4549 | y – cos y = x
:
(y sin y + cos y + x) y′ = y
9 x + y = tan–1y
:
y2 y′ + y2 + 1 = 0
10 y =
2
2
a
−x
x ∈ (–a, a) :
x + y dy
dx = 0 (y ≠ 0)
11 The number of arbitrary constants in the general solution of a differential equation
of fourth order are:
(A) 0
(B) 2
(C) 3
(D) 4
12 |
1 | 4547-4550 | x + y = tan–1y
:
y2 y′ + y2 + 1 = 0
10 y =
2
2
a
−x
x ∈ (–a, a) :
x + y dy
dx = 0 (y ≠ 0)
11 The number of arbitrary constants in the general solution of a differential equation
of fourth order are:
(A) 0
(B) 2
(C) 3
(D) 4
12 The number of arbitrary constants in the particular solution of a differential equation
of third order are:
(A) 3
(B) 2
(C) 1
(D) 0
9 |
1 | 4548-4551 | y =
2
2
a
−x
x ∈ (–a, a) :
x + y dy
dx = 0 (y ≠ 0)
11 The number of arbitrary constants in the general solution of a differential equation
of fourth order are:
(A) 0
(B) 2
(C) 3
(D) 4
12 The number of arbitrary constants in the particular solution of a differential equation
of third order are:
(A) 3
(B) 2
(C) 1
(D) 0
9 4 Formation of a Differential Equation whose General Solution is given
We know that the equation
x2 + y2 + 2x – 4y + 4 = 0 |
1 | 4549-4552 | The number of arbitrary constants in the general solution of a differential equation
of fourth order are:
(A) 0
(B) 2
(C) 3
(D) 4
12 The number of arbitrary constants in the particular solution of a differential equation
of third order are:
(A) 3
(B) 2
(C) 1
(D) 0
9 4 Formation of a Differential Equation whose General Solution is given
We know that the equation
x2 + y2 + 2x – 4y + 4 = 0 (1)
represents a circle having centre at (–1, 2) and radius 1 unit |
1 | 4550-4553 | The number of arbitrary constants in the particular solution of a differential equation
of third order are:
(A) 3
(B) 2
(C) 1
(D) 0
9 4 Formation of a Differential Equation whose General Solution is given
We know that the equation
x2 + y2 + 2x – 4y + 4 = 0 (1)
represents a circle having centre at (–1, 2) and radius 1 unit © NCERT
not to be republished
MATHEMATICS
386
Differentiating equation (1) with respect to x, we get
dy
dx =
1
2
x
y
−+
(y ≠ 2) |
1 | 4551-4554 | 4 Formation of a Differential Equation whose General Solution is given
We know that the equation
x2 + y2 + 2x – 4y + 4 = 0 (1)
represents a circle having centre at (–1, 2) and radius 1 unit © NCERT
not to be republished
MATHEMATICS
386
Differentiating equation (1) with respect to x, we get
dy
dx =
1
2
x
y
−+
(y ≠ 2) (2)
which is a differential equation |
1 | 4552-4555 | (1)
represents a circle having centre at (–1, 2) and radius 1 unit © NCERT
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MATHEMATICS
386
Differentiating equation (1) with respect to x, we get
dy
dx =
1
2
x
y
−+
(y ≠ 2) (2)
which is a differential equation You will find later on [See (example 9 section 9 |
1 | 4553-4556 | © NCERT
not to be republished
MATHEMATICS
386
Differentiating equation (1) with respect to x, we get
dy
dx =
1
2
x
y
−+
(y ≠ 2) (2)
which is a differential equation You will find later on [See (example 9 section 9 5 |
1 | 4554-4557 | (2)
which is a differential equation You will find later on [See (example 9 section 9 5 1 |
1 | 4555-4558 | You will find later on [See (example 9 section 9 5 1 )]
that this equation represents the family of circles and one member of the family is the
circle given in equation (1) |
1 | 4556-4559 | 5 1 )]
that this equation represents the family of circles and one member of the family is the
circle given in equation (1) Let us consider the equation
x2 + y2 = r2 |
1 | 4557-4560 | 1 )]
that this equation represents the family of circles and one member of the family is the
circle given in equation (1) Let us consider the equation
x2 + y2 = r2 (3)
By giving different values to r, we get different members of the family e |
1 | 4558-4561 | )]
that this equation represents the family of circles and one member of the family is the
circle given in equation (1) Let us consider the equation
x2 + y2 = r2 (3)
By giving different values to r, we get different members of the family e g |
1 | 4559-4562 | Let us consider the equation
x2 + y2 = r2 (3)
By giving different values to r, we get different members of the family e g x2 + y2 = 1, x2 + y2 = 4, x2 + y2 = 9 etc |
1 | 4560-4563 | (3)
By giving different values to r, we get different members of the family e g x2 + y2 = 1, x2 + y2 = 4, x2 + y2 = 9 etc (see Fig 9 |
1 | 4561-4564 | g x2 + y2 = 1, x2 + y2 = 4, x2 + y2 = 9 etc (see Fig 9 1) |
1 | 4562-4565 | x2 + y2 = 1, x2 + y2 = 4, x2 + y2 = 9 etc (see Fig 9 1) Thus, equation (3) represents a family of concentric
circles centered at the origin and having different radii |
1 | 4563-4566 | (see Fig 9 1) Thus, equation (3) represents a family of concentric
circles centered at the origin and having different radii We are interested in finding a differential equation
that is satisfied by each member of the family |
1 | 4564-4567 | 1) Thus, equation (3) represents a family of concentric
circles centered at the origin and having different radii We are interested in finding a differential equation
that is satisfied by each member of the family The
differential equation must be free from r because r is
different for different members of the family |
1 | 4565-4568 | Thus, equation (3) represents a family of concentric
circles centered at the origin and having different radii We are interested in finding a differential equation
that is satisfied by each member of the family The
differential equation must be free from r because r is
different for different members of the family This
equation is obtained by differentiating equation (3) with
respect to x, i |
1 | 4566-4569 | We are interested in finding a differential equation
that is satisfied by each member of the family The
differential equation must be free from r because r is
different for different members of the family This
equation is obtained by differentiating equation (3) with
respect to x, i e |
1 | 4567-4570 | The
differential equation must be free from r because r is
different for different members of the family This
equation is obtained by differentiating equation (3) with
respect to x, i e ,
2x + 2y dy
dx = 0 or x + y dy
dx = 0 |
1 | 4568-4571 | This
equation is obtained by differentiating equation (3) with
respect to x, i e ,
2x + 2y dy
dx = 0 or x + y dy
dx = 0 (4)
which represents the family of concentric circles given by equation (3) |
1 | 4569-4572 | e ,
2x + 2y dy
dx = 0 or x + y dy
dx = 0 (4)
which represents the family of concentric circles given by equation (3) Again, let us consider the equation
y = mx + c |
1 | 4570-4573 | ,
2x + 2y dy
dx = 0 or x + y dy
dx = 0 (4)
which represents the family of concentric circles given by equation (3) Again, let us consider the equation
y = mx + c (5)
By giving different values to the parameters m and c, we get different members of
the family, e |
1 | 4571-4574 | (4)
which represents the family of concentric circles given by equation (3) Again, let us consider the equation
y = mx + c (5)
By giving different values to the parameters m and c, we get different members of
the family, e g |
1 | 4572-4575 | Again, let us consider the equation
y = mx + c (5)
By giving different values to the parameters m and c, we get different members of
the family, e g ,
y = x
(m = 1, c = 0)
y =
3 x
(m =
3 , c = 0)
y = x + 1
(m = 1, c = 1)
y = – x
(m = – 1, c = 0)
y = – x – 1
(m = – 1, c = – 1) etc |
1 | 4573-4576 | (5)
By giving different values to the parameters m and c, we get different members of
the family, e g ,
y = x
(m = 1, c = 0)
y =
3 x
(m =
3 , c = 0)
y = x + 1
(m = 1, c = 1)
y = – x
(m = – 1, c = 0)
y = – x – 1
(m = – 1, c = – 1) etc ( see Fig 9 |
1 | 4574-4577 | g ,
y = x
(m = 1, c = 0)
y =
3 x
(m =
3 , c = 0)
y = x + 1
(m = 1, c = 1)
y = – x
(m = – 1, c = 0)
y = – x – 1
(m = – 1, c = – 1) etc ( see Fig 9 2) |
1 | 4575-4578 | ,
y = x
(m = 1, c = 0)
y =
3 x
(m =
3 , c = 0)
y = x + 1
(m = 1, c = 1)
y = – x
(m = – 1, c = 0)
y = – x – 1
(m = – 1, c = – 1) etc ( see Fig 9 2) Thus, equation (5) represents the family of straight lines, where m, c are parameters |
1 | 4576-4579 | ( see Fig 9 2) Thus, equation (5) represents the family of straight lines, where m, c are parameters We are now interested in finding a differential equation that is satisfied by each
member of the family |
1 | 4577-4580 | 2) Thus, equation (5) represents the family of straight lines, where m, c are parameters We are now interested in finding a differential equation that is satisfied by each
member of the family Further, the equation must be free from m and c because m and
Fig 9 |
1 | 4578-4581 | Thus, equation (5) represents the family of straight lines, where m, c are parameters We are now interested in finding a differential equation that is satisfied by each
member of the family Further, the equation must be free from m and c because m and
Fig 9 1
© NCERT
not to be republished
DIFFERENTIAL EQUATIONS
387
X
X’
Y
Y’
y = x+1
y = x
y = –x
y = –x–1
y =
x
3
O
c are different for different members of the family |
1 | 4579-4582 | We are now interested in finding a differential equation that is satisfied by each
member of the family Further, the equation must be free from m and c because m and
Fig 9 1
© NCERT
not to be republished
DIFFERENTIAL EQUATIONS
387
X
X’
Y
Y’
y = x+1
y = x
y = –x
y = –x–1
y =
x
3
O
c are different for different members of the family This is obtained by differentiating equation (5) with
respect to x, successively we get
dy
dx =m
, and
2
2
0
d y
dx
= |
1 | 4580-4583 | Further, the equation must be free from m and c because m and
Fig 9 1
© NCERT
not to be republished
DIFFERENTIAL EQUATIONS
387
X
X’
Y
Y’
y = x+1
y = x
y = –x
y = –x–1
y =
x
3
O
c are different for different members of the family This is obtained by differentiating equation (5) with
respect to x, successively we get
dy
dx =m
, and
2
2
0
d y
dx
= (6)
The equation (6) represents the family of straight
lines given by equation (5) |
1 | 4581-4584 | 1
© NCERT
not to be republished
DIFFERENTIAL EQUATIONS
387
X
X’
Y
Y’
y = x+1
y = x
y = –x
y = –x–1
y =
x
3
O
c are different for different members of the family This is obtained by differentiating equation (5) with
respect to x, successively we get
dy
dx =m
, and
2
2
0
d y
dx
= (6)
The equation (6) represents the family of straight
lines given by equation (5) Note that equations (3) and (5) are the general
solutions of equations (4) and (6) respectively |
1 | 4582-4585 | This is obtained by differentiating equation (5) with
respect to x, successively we get
dy
dx =m
, and
2
2
0
d y
dx
= (6)
The equation (6) represents the family of straight
lines given by equation (5) Note that equations (3) and (5) are the general
solutions of equations (4) and (6) respectively 9 |
1 | 4583-4586 | (6)
The equation (6) represents the family of straight
lines given by equation (5) Note that equations (3) and (5) are the general
solutions of equations (4) and (6) respectively 9 4 |
1 | 4584-4587 | Note that equations (3) and (5) are the general
solutions of equations (4) and (6) respectively 9 4 1 Procedure to form a differential equation that will represent a given
family of curves
(a)
If the given family F1 of curves depends on only one parameter then it is
represented by an equation of the form
F1 (x, y, a) = 0 |
1 | 4585-4588 | 9 4 1 Procedure to form a differential equation that will represent a given
family of curves
(a)
If the given family F1 of curves depends on only one parameter then it is
represented by an equation of the form
F1 (x, y, a) = 0 (1)
For example, the family of parabolas y2 = ax can be represented by an equation
of the form f (x, y, a) : y2 = ax |
1 | 4586-4589 | 4 1 Procedure to form a differential equation that will represent a given
family of curves
(a)
If the given family F1 of curves depends on only one parameter then it is
represented by an equation of the form
F1 (x, y, a) = 0 (1)
For example, the family of parabolas y2 = ax can be represented by an equation
of the form f (x, y, a) : y2 = ax Differentiating equation (1) with respect to x, we get an equation involving
y′, y, x, and a, i |
1 | 4587-4590 | 1 Procedure to form a differential equation that will represent a given
family of curves
(a)
If the given family F1 of curves depends on only one parameter then it is
represented by an equation of the form
F1 (x, y, a) = 0 (1)
For example, the family of parabolas y2 = ax can be represented by an equation
of the form f (x, y, a) : y2 = ax Differentiating equation (1) with respect to x, we get an equation involving
y′, y, x, and a, i e |
1 | 4588-4591 | (1)
For example, the family of parabolas y2 = ax can be represented by an equation
of the form f (x, y, a) : y2 = ax Differentiating equation (1) with respect to x, we get an equation involving
y′, y, x, and a, i e ,
g (x, y, y′, a) = 0 |
1 | 4589-4592 | Differentiating equation (1) with respect to x, we get an equation involving
y′, y, x, and a, i e ,
g (x, y, y′, a) = 0 (2)
The required differential equation is then obtained by eliminating a from equations
(1) and (2) as
F(x, y, y′) = 0 |
1 | 4590-4593 | e ,
g (x, y, y′, a) = 0 (2)
The required differential equation is then obtained by eliminating a from equations
(1) and (2) as
F(x, y, y′) = 0 (3)
(b)
If the given family F2 of curves depends on the parameters a, b (say) then it is
represented by an equation of the from
F2 (x, y, a, b) = 0 |
1 | 4591-4594 | ,
g (x, y, y′, a) = 0 (2)
The required differential equation is then obtained by eliminating a from equations
(1) and (2) as
F(x, y, y′) = 0 (3)
(b)
If the given family F2 of curves depends on the parameters a, b (say) then it is
represented by an equation of the from
F2 (x, y, a, b) = 0 (4)
Differentiating equation (4) with respect to x, we get an equation involving
y′, x, y, a, b, i |
1 | 4592-4595 | (2)
The required differential equation is then obtained by eliminating a from equations
(1) and (2) as
F(x, y, y′) = 0 (3)
(b)
If the given family F2 of curves depends on the parameters a, b (say) then it is
represented by an equation of the from
F2 (x, y, a, b) = 0 (4)
Differentiating equation (4) with respect to x, we get an equation involving
y′, x, y, a, b, i e |
1 | 4593-4596 | (3)
(b)
If the given family F2 of curves depends on the parameters a, b (say) then it is
represented by an equation of the from
F2 (x, y, a, b) = 0 (4)
Differentiating equation (4) with respect to x, we get an equation involving
y′, x, y, a, b, i e ,
g (x, y, y′, a, b) = 0 |
1 | 4594-4597 | (4)
Differentiating equation (4) with respect to x, we get an equation involving
y′, x, y, a, b, i e ,
g (x, y, y′, a, b) = 0 (5)
But it is not possible to eliminate two parameters a and b from the two equations
and so, we need a third equation |
1 | 4595-4598 | e ,
g (x, y, y′, a, b) = 0 (5)
But it is not possible to eliminate two parameters a and b from the two equations
and so, we need a third equation This equation is obtained by differentiating
equation (5), with respect to x, to obtain a relation of the form
h (x, y, y′, y″, a, b) = 0 |
1 | 4596-4599 | ,
g (x, y, y′, a, b) = 0 (5)
But it is not possible to eliminate two parameters a and b from the two equations
and so, we need a third equation This equation is obtained by differentiating
equation (5), with respect to x, to obtain a relation of the form
h (x, y, y′, y″, a, b) = 0 (6)
Fig 9 |
1 | 4597-4600 | (5)
But it is not possible to eliminate two parameters a and b from the two equations
and so, we need a third equation This equation is obtained by differentiating
equation (5), with respect to x, to obtain a relation of the form
h (x, y, y′, y″, a, b) = 0 (6)
Fig 9 2
© NCERT
not to be republished
MATHEMATICS
388
The required differential equation is then obtained by eliminating a and b from
equations (4), (5) and (6) as
F (x, y, y′, y″) = 0 |
1 | 4598-4601 | This equation is obtained by differentiating
equation (5), with respect to x, to obtain a relation of the form
h (x, y, y′, y″, a, b) = 0 (6)
Fig 9 2
© NCERT
not to be republished
MATHEMATICS
388
The required differential equation is then obtained by eliminating a and b from
equations (4), (5) and (6) as
F (x, y, y′, y″) = 0 (7)
�Note The order of a differential equation representing a family of curves is
same as the number of arbitrary constants present in the equation corresponding to
the family of curves |
1 | 4599-4602 | (6)
Fig 9 2
© NCERT
not to be republished
MATHEMATICS
388
The required differential equation is then obtained by eliminating a and b from
equations (4), (5) and (6) as
F (x, y, y′, y″) = 0 (7)
�Note The order of a differential equation representing a family of curves is
same as the number of arbitrary constants present in the equation corresponding to
the family of curves Example 4 Form the differential equation representing the family of curves y = mx,
where, m is arbitrary constant |
1 | 4600-4603 | 2
© NCERT
not to be republished
MATHEMATICS
388
The required differential equation is then obtained by eliminating a and b from
equations (4), (5) and (6) as
F (x, y, y′, y″) = 0 (7)
�Note The order of a differential equation representing a family of curves is
same as the number of arbitrary constants present in the equation corresponding to
the family of curves Example 4 Form the differential equation representing the family of curves y = mx,
where, m is arbitrary constant Solution We have
y = mx |
1 | 4601-4604 | (7)
�Note The order of a differential equation representing a family of curves is
same as the number of arbitrary constants present in the equation corresponding to
the family of curves Example 4 Form the differential equation representing the family of curves y = mx,
where, m is arbitrary constant Solution We have
y = mx (1)
Differentiating both sides of equation (1) with respect to x, we get
dy
dx = m
Substituting the value of m in equation (1) we get
dy
y
x
dx
or
dy
x dx – y = 0
which is free from the parameter m and hence this is the required differential equation |
1 | 4602-4605 | Example 4 Form the differential equation representing the family of curves y = mx,
where, m is arbitrary constant Solution We have
y = mx (1)
Differentiating both sides of equation (1) with respect to x, we get
dy
dx = m
Substituting the value of m in equation (1) we get
dy
y
x
dx
or
dy
x dx – y = 0
which is free from the parameter m and hence this is the required differential equation Example 5 Form the differential equation representing the family of curves
y = a sin (x + b), where a, b are arbitrary constants |
1 | 4603-4606 | Solution We have
y = mx (1)
Differentiating both sides of equation (1) with respect to x, we get
dy
dx = m
Substituting the value of m in equation (1) we get
dy
y
x
dx
or
dy
x dx – y = 0
which is free from the parameter m and hence this is the required differential equation Example 5 Form the differential equation representing the family of curves
y = a sin (x + b), where a, b are arbitrary constants Solution We have
y = a sin(x + b) |
1 | 4604-4607 | (1)
Differentiating both sides of equation (1) with respect to x, we get
dy
dx = m
Substituting the value of m in equation (1) we get
dy
y
x
dx
or
dy
x dx – y = 0
which is free from the parameter m and hence this is the required differential equation Example 5 Form the differential equation representing the family of curves
y = a sin (x + b), where a, b are arbitrary constants Solution We have
y = a sin(x + b) (1)
Differentiating both sides of equation (1) with respect to x, successively we get
dy
dx = a cos (x + b) |
1 | 4605-4608 | Example 5 Form the differential equation representing the family of curves
y = a sin (x + b), where a, b are arbitrary constants Solution We have
y = a sin(x + b) (1)
Differentiating both sides of equation (1) with respect to x, successively we get
dy
dx = a cos (x + b) (2)
2
d y2
dx
= – a sin(x + b) |
1 | 4606-4609 | Solution We have
y = a sin(x + b) (1)
Differentiating both sides of equation (1) with respect to x, successively we get
dy
dx = a cos (x + b) (2)
2
d y2
dx
= – a sin(x + b) (3)
Eliminating a and b from equations (1), (2) and (3), we get
2
d y2
y
dx
+
= 0 |
1 | 4607-4610 | (1)
Differentiating both sides of equation (1) with respect to x, successively we get
dy
dx = a cos (x + b) (2)
2
d y2
dx
= – a sin(x + b) (3)
Eliminating a and b from equations (1), (2) and (3), we get
2
d y2
y
dx
+
= 0 (4)
which is free from the arbitrary constants a and b and hence this the required differential
equation |
1 | 4608-4611 | (2)
2
d y2
dx
= – a sin(x + b) (3)
Eliminating a and b from equations (1), (2) and (3), we get
2
d y2
y
dx
+
= 0 (4)
which is free from the arbitrary constants a and b and hence this the required differential
equation © NCERT
not to be republished
DIFFERENTIAL EQUATIONS
389
Example 6 Form the differential equation
representing the family of ellipses having foci on
x-axis and centre at the origin |
1 | 4609-4612 | (3)
Eliminating a and b from equations (1), (2) and (3), we get
2
d y2
y
dx
+
= 0 (4)
which is free from the arbitrary constants a and b and hence this the required differential
equation © NCERT
not to be republished
DIFFERENTIAL EQUATIONS
389
Example 6 Form the differential equation
representing the family of ellipses having foci on
x-axis and centre at the origin Solution We know that the equation of said family
of ellipses (see Fig 9 |
1 | 4610-4613 | (4)
which is free from the arbitrary constants a and b and hence this the required differential
equation © NCERT
not to be republished
DIFFERENTIAL EQUATIONS
389
Example 6 Form the differential equation
representing the family of ellipses having foci on
x-axis and centre at the origin Solution We know that the equation of said family
of ellipses (see Fig 9 3) is
2
2
2
2
x
y
a
+b
= 1 |
1 | 4611-4614 | © NCERT
not to be republished
DIFFERENTIAL EQUATIONS
389
Example 6 Form the differential equation
representing the family of ellipses having foci on
x-axis and centre at the origin Solution We know that the equation of said family
of ellipses (see Fig 9 3) is
2
2
2
2
x
y
a
+b
= 1 (1)
Differentiating equation (1) with respect to x, we get
2
2
2
2
0
x
y dy
dx
a
+b
=
or
y
dy
x
⎛dx
⎞
⎜
⎟
⎝
⎠ =
2
−ab2 |
1 | 4612-4615 | Solution We know that the equation of said family
of ellipses (see Fig 9 3) is
2
2
2
2
x
y
a
+b
= 1 (1)
Differentiating equation (1) with respect to x, we get
2
2
2
2
0
x
y dy
dx
a
+b
=
or
y
dy
x
⎛dx
⎞
⎜
⎟
⎝
⎠ =
2
−ab2 (2)
Differentiating both sides of equation (2) with respect to x, we get
2
2
2
xdy
y
y
d y
dy
dx
x
dx
dx
x
= 0
or
2
2
2
–
d y
dy
dy
xy
x
y
dx
dx
dx
= 0 |
1 | 4613-4616 | 3) is
2
2
2
2
x
y
a
+b
= 1 (1)
Differentiating equation (1) with respect to x, we get
2
2
2
2
0
x
y dy
dx
a
+b
=
or
y
dy
x
⎛dx
⎞
⎜
⎟
⎝
⎠ =
2
−ab2 (2)
Differentiating both sides of equation (2) with respect to x, we get
2
2
2
xdy
y
y
d y
dy
dx
x
dx
dx
x
= 0
or
2
2
2
–
d y
dy
dy
xy
x
y
dx
dx
dx
= 0 (3)
which is the required differential equation |
1 | 4614-4617 | (1)
Differentiating equation (1) with respect to x, we get
2
2
2
2
0
x
y dy
dx
a
+b
=
or
y
dy
x
⎛dx
⎞
⎜
⎟
⎝
⎠ =
2
−ab2 (2)
Differentiating both sides of equation (2) with respect to x, we get
2
2
2
xdy
y
y
d y
dy
dx
x
dx
dx
x
= 0
or
2
2
2
–
d y
dy
dy
xy
x
y
dx
dx
dx
= 0 (3)
which is the required differential equation Example 7 Form the differential equation of the family
of circles touching the x-axis at origin |
1 | 4615-4618 | (2)
Differentiating both sides of equation (2) with respect to x, we get
2
2
2
xdy
y
y
d y
dy
dx
x
dx
dx
x
= 0
or
2
2
2
–
d y
dy
dy
xy
x
y
dx
dx
dx
= 0 (3)
which is the required differential equation Example 7 Form the differential equation of the family
of circles touching the x-axis at origin Solution Let C denote the family of circles touching
x-axis at origin |
1 | 4616-4619 | (3)
which is the required differential equation Example 7 Form the differential equation of the family
of circles touching the x-axis at origin Solution Let C denote the family of circles touching
x-axis at origin Let (0, a) be the coordinates of the
centre of any member of the family (see Fig 9 |
1 | 4617-4620 | Example 7 Form the differential equation of the family
of circles touching the x-axis at origin Solution Let C denote the family of circles touching
x-axis at origin Let (0, a) be the coordinates of the
centre of any member of the family (see Fig 9 4) |
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