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1 | 4618-4621 | 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) Therefore, equation of family C is
x2 + (y – a)2 = a2 or x2 + y2 = 2ay |
1 | 4619-4622 | Let (0, a) be the coordinates of the
centre of any member of the family (see Fig 9 4) Therefore, equation of family C is
x2 + (y – a)2 = a2 or x2 + y2 = 2ay (1)
where, a is an arbitrary constant |
1 | 4620-4623 | 4) Therefore, equation of family C is
x2 + (y – a)2 = a2 or x2 + y2 = 2ay (1)
where, a is an arbitrary constant Differentiating both
sides of equation (1) with respect to x,we get
2
2
dy
x
y dx
= 2
dy
a dx
Fig 9 |
1 | 4621-4624 | Therefore, equation of family C is
x2 + (y – a)2 = a2 or x2 + y2 = 2ay (1)
where, a is an arbitrary constant Differentiating both
sides of equation (1) with respect to x,we get
2
2
dy
x
y dx
= 2
dy
a dx
Fig 9 3
Fig 9 |
1 | 4622-4625 | (1)
where, a is an arbitrary constant Differentiating both
sides of equation (1) with respect to x,we get
2
2
dy
x
y dx
= 2
dy
a dx
Fig 9 3
Fig 9 4
X
X’
Y’
Y
O
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MATHEMATICS
390
or
dy
x
y dx
=
dy
a dx or a =
dy
x
dyy dx
dx |
1 | 4623-4626 | Differentiating both
sides of equation (1) with respect to x,we get
2
2
dy
x
y dx
= 2
dy
a dx
Fig 9 3
Fig 9 4
X
X’
Y’
Y
O
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MATHEMATICS
390
or
dy
x
y dx
=
dy
a dx or a =
dy
x
dyy dx
dx (2)
Substituting the value of a from equation (2) in equation (1), we get
x2 + y2 = 2
dy
x
y dx
y
dy
dx
or
2
2
(
)
dy x
y
dx
=
2
2
2
dy
xy
y dx
or
dy
dx =
2
2
2
–
xy
x
y
This is the required differential equation of the given family of circles |
1 | 4624-4627 | 3
Fig 9 4
X
X’
Y’
Y
O
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MATHEMATICS
390
or
dy
x
y dx
=
dy
a dx or a =
dy
x
dyy dx
dx (2)
Substituting the value of a from equation (2) in equation (1), we get
x2 + y2 = 2
dy
x
y dx
y
dy
dx
or
2
2
(
)
dy x
y
dx
=
2
2
2
dy
xy
y dx
or
dy
dx =
2
2
2
–
xy
x
y
This is the required differential equation of the given family of circles Example 8 Form the differential equation representing the family of parabolas having
vertex at origin and axis along positive direction of x-axis |
1 | 4625-4628 | 4
X
X’
Y’
Y
O
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MATHEMATICS
390
or
dy
x
y dx
=
dy
a dx or a =
dy
x
dyy dx
dx (2)
Substituting the value of a from equation (2) in equation (1), we get
x2 + y2 = 2
dy
x
y dx
y
dy
dx
or
2
2
(
)
dy x
y
dx
=
2
2
2
dy
xy
y dx
or
dy
dx =
2
2
2
–
xy
x
y
This is the required differential equation of the given family of circles Example 8 Form the differential equation representing the family of parabolas having
vertex at origin and axis along positive direction of x-axis Solution Let P denote the family of above said parabolas (see Fig 9 |
1 | 4626-4629 | (2)
Substituting the value of a from equation (2) in equation (1), we get
x2 + y2 = 2
dy
x
y dx
y
dy
dx
or
2
2
(
)
dy x
y
dx
=
2
2
2
dy
xy
y dx
or
dy
dx =
2
2
2
–
xy
x
y
This is the required differential equation of the given family of circles Example 8 Form the differential equation representing the family of parabolas having
vertex at origin and axis along positive direction of x-axis Solution Let P denote the family of above said parabolas (see Fig 9 5) and let (a, 0) be the
focus of a member of the given family, where a is an arbitrary constant |
1 | 4627-4630 | Example 8 Form the differential equation representing the family of parabolas having
vertex at origin and axis along positive direction of x-axis Solution Let P denote the family of above said parabolas (see Fig 9 5) and let (a, 0) be the
focus of a member of the given family, where a is an arbitrary constant Therefore, equation
of family P is
y2 = 4ax |
1 | 4628-4631 | Solution Let P denote the family of above said parabolas (see Fig 9 5) and let (a, 0) be the
focus of a member of the given family, where a is an arbitrary constant Therefore, equation
of family P is
y2 = 4ax (1)
Differentiating both sides of equation (1) with respect to x, we get
2
dy
y dx = 4a |
1 | 4629-4632 | 5) and let (a, 0) be the
focus of a member of the given family, where a is an arbitrary constant Therefore, equation
of family P is
y2 = 4ax (1)
Differentiating both sides of equation (1) with respect to x, we get
2
dy
y dx = 4a (2)
Substituting the value of 4a from equation (2)
in equation (1), we get
y2 =
2
( )
ydy
x
dx
⎛
⎞
⎜
⎟
⎝
⎠
or
2
2
dy
y
−xy dx
= 0
which is the differential equation of the given family
of parabolas |
1 | 4630-4633 | Therefore, equation
of family P is
y2 = 4ax (1)
Differentiating both sides of equation (1) with respect to x, we get
2
dy
y dx = 4a (2)
Substituting the value of 4a from equation (2)
in equation (1), we get
y2 =
2
( )
ydy
x
dx
⎛
⎞
⎜
⎟
⎝
⎠
or
2
2
dy
y
−xy dx
= 0
which is the differential equation of the given family
of parabolas Fig 9 |
1 | 4631-4634 | (1)
Differentiating both sides of equation (1) with respect to x, we get
2
dy
y dx = 4a (2)
Substituting the value of 4a from equation (2)
in equation (1), we get
y2 =
2
( )
ydy
x
dx
⎛
⎞
⎜
⎟
⎝
⎠
or
2
2
dy
y
−xy dx
= 0
which is the differential equation of the given family
of parabolas Fig 9 5
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DIFFERENTIAL EQUATIONS
391
EXERCISE 9 |
1 | 4632-4635 | (2)
Substituting the value of 4a from equation (2)
in equation (1), we get
y2 =
2
( )
ydy
x
dx
⎛
⎞
⎜
⎟
⎝
⎠
or
2
2
dy
y
−xy dx
= 0
which is the differential equation of the given family
of parabolas Fig 9 5
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DIFFERENTIAL EQUATIONS
391
EXERCISE 9 3
In each of the Exercises 1 to 5, form a differential equation representing the given
family of curves by eliminating arbitrary constants a and b |
1 | 4633-4636 | Fig 9 5
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DIFFERENTIAL EQUATIONS
391
EXERCISE 9 3
In each of the Exercises 1 to 5, form a differential equation representing the given
family of curves by eliminating arbitrary constants a and b 1 |
1 | 4634-4637 | 5
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DIFFERENTIAL EQUATIONS
391
EXERCISE 9 3
In each of the Exercises 1 to 5, form a differential equation representing the given
family of curves by eliminating arbitrary constants a and b 1 1
x
y
a
+b
=
2 |
1 | 4635-4638 | 3
In each of the Exercises 1 to 5, form a differential equation representing the given
family of curves by eliminating arbitrary constants a and b 1 1
x
y
a
+b
=
2 y2 = a (b2 – x2)
3 |
1 | 4636-4639 | 1 1
x
y
a
+b
=
2 y2 = a (b2 – x2)
3 y = a e3x + b e– 2x
4 |
1 | 4637-4640 | 1
x
y
a
+b
=
2 y2 = a (b2 – x2)
3 y = a e3x + b e– 2x
4 y = e2x (a + bx)
5 |
1 | 4638-4641 | y2 = a (b2 – x2)
3 y = a e3x + b e– 2x
4 y = e2x (a + bx)
5 y = ex (a cos x + b sin x)
6 |
1 | 4639-4642 | y = a e3x + b e– 2x
4 y = e2x (a + bx)
5 y = ex (a cos x + b sin x)
6 Form the differential equation of the family of circles touching the y-axis at
origin |
1 | 4640-4643 | y = e2x (a + bx)
5 y = ex (a cos x + b sin x)
6 Form the differential equation of the family of circles touching the y-axis at
origin 7 |
1 | 4641-4644 | y = ex (a cos x + b sin x)
6 Form the differential equation of the family of circles touching the y-axis at
origin 7 Form the differential equation of the family of parabolas having vertex at origin
and axis along positive y-axis |
1 | 4642-4645 | Form the differential equation of the family of circles touching the y-axis at
origin 7 Form the differential equation of the family of parabolas having vertex at origin
and axis along positive y-axis 8 |
1 | 4643-4646 | 7 Form the differential equation of the family of parabolas having vertex at origin
and axis along positive y-axis 8 Form the differential equation of the family of ellipses having foci on y-axis and
centre at origin |
1 | 4644-4647 | Form the differential equation of the family of parabolas having vertex at origin
and axis along positive y-axis 8 Form the differential equation of the family of ellipses having foci on y-axis and
centre at origin 9 |
1 | 4645-4648 | 8 Form the differential equation of the family of ellipses having foci on y-axis and
centre at origin 9 Form the differential equation of the family of hyperbolas having foci on x-axis
and centre at origin |
1 | 4646-4649 | Form the differential equation of the family of ellipses having foci on y-axis and
centre at origin 9 Form the differential equation of the family of hyperbolas having foci on x-axis
and centre at origin 10 |
1 | 4647-4650 | 9 Form the differential equation of the family of hyperbolas having foci on x-axis
and centre at origin 10 Form the differential equation of the family of circles having centre on y-axis
and radius 3 units |
1 | 4648-4651 | Form the differential equation of the family of hyperbolas having foci on x-axis
and centre at origin 10 Form the differential equation of the family of circles having centre on y-axis
and radius 3 units 11 |
1 | 4649-4652 | 10 Form the differential equation of the family of circles having centre on y-axis
and radius 3 units 11 Which of the following differential equations has y = c1 ex + c2 e–x as the general
solution |
1 | 4650-4653 | Form the differential equation of the family of circles having centre on y-axis
and radius 3 units 11 Which of the following differential equations has y = c1 ex + c2 e–x as the general
solution (A)
2
2
0
d y
y
dx
+
=
(B)
2
2
0
d y
y
dx
−
=
(C)
2
2
1
0
d y
dx
+ =
(D)
2
2
1
0
d y
dx
− =
12 |
1 | 4651-4654 | 11 Which of the following differential equations has y = c1 ex + c2 e–x as the general
solution (A)
2
2
0
d y
y
dx
+
=
(B)
2
2
0
d y
y
dx
−
=
(C)
2
2
1
0
d y
dx
+ =
(D)
2
2
1
0
d y
dx
− =
12 Which of the following differential equations has y = x as one of its particular
solution |
1 | 4652-4655 | Which of the following differential equations has y = c1 ex + c2 e–x as the general
solution (A)
2
2
0
d y
y
dx
+
=
(B)
2
2
0
d y
y
dx
−
=
(C)
2
2
1
0
d y
dx
+ =
(D)
2
2
1
0
d y
dx
− =
12 Which of the following differential equations has y = x as one of its particular
solution (A)
2
2
d y2
xdy
xy
x
dx
dx
−
+
=
(B)
2
d y2
xdy
xy
x
dx
dx
+
+
=
(C)
2
2
2
0
d y
xdy
xy
dx
dx
(D)
2
2
0
d y
xdy
xy
dx
dx
+
+
=
9 |
1 | 4653-4656 | (A)
2
2
0
d y
y
dx
+
=
(B)
2
2
0
d y
y
dx
−
=
(C)
2
2
1
0
d y
dx
+ =
(D)
2
2
1
0
d y
dx
− =
12 Which of the following differential equations has y = x as one of its particular
solution (A)
2
2
d y2
xdy
xy
x
dx
dx
−
+
=
(B)
2
d y2
xdy
xy
x
dx
dx
+
+
=
(C)
2
2
2
0
d y
xdy
xy
dx
dx
(D)
2
2
0
d y
xdy
xy
dx
dx
+
+
=
9 5 |
1 | 4654-4657 | Which of the following differential equations has y = x as one of its particular
solution (A)
2
2
d y2
xdy
xy
x
dx
dx
−
+
=
(B)
2
d y2
xdy
xy
x
dx
dx
+
+
=
(C)
2
2
2
0
d y
xdy
xy
dx
dx
(D)
2
2
0
d y
xdy
xy
dx
dx
+
+
=
9 5 Methods of Solving First Order, First Degree Differential Equations
In this section we shall discuss three methods of solving first order first degree differential
equations |
1 | 4655-4658 | (A)
2
2
d y2
xdy
xy
x
dx
dx
−
+
=
(B)
2
d y2
xdy
xy
x
dx
dx
+
+
=
(C)
2
2
2
0
d y
xdy
xy
dx
dx
(D)
2
2
0
d y
xdy
xy
dx
dx
+
+
=
9 5 Methods of Solving First Order, First Degree Differential Equations
In this section we shall discuss three methods of solving first order first degree differential
equations 9 |
1 | 4656-4659 | 5 Methods of Solving First Order, First Degree Differential Equations
In this section we shall discuss three methods of solving first order first degree differential
equations 9 5 |
1 | 4657-4660 | Methods of Solving First Order, First Degree Differential Equations
In this section we shall discuss three methods of solving first order first degree differential
equations 9 5 1 Differential equations with variables separable
A first order-first degree differential equation is of the form
dy
dx = F(x, y) |
1 | 4658-4661 | 9 5 1 Differential equations with variables separable
A first order-first degree differential equation is of the form
dy
dx = F(x, y) (1)
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MATHEMATICS
392
If F (x, y) can be expressed as a product g (x) h(y), where, g(x) is a function of x
and h(y) is a function of y, then the differential equation (1) is said to be of variable
separable type |
1 | 4659-4662 | 5 1 Differential equations with variables separable
A first order-first degree differential equation is of the form
dy
dx = F(x, y) (1)
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MATHEMATICS
392
If F (x, y) can be expressed as a product g (x) h(y), where, g(x) is a function of x
and h(y) is a function of y, then the differential equation (1) is said to be of variable
separable type The differential equation (1) then has the form
dy
dx = h (y) |
1 | 4660-4663 | 1 Differential equations with variables separable
A first order-first degree differential equation is of the form
dy
dx = F(x, y) (1)
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MATHEMATICS
392
If F (x, y) can be expressed as a product g (x) h(y), where, g(x) is a function of x
and h(y) is a function of y, then the differential equation (1) is said to be of variable
separable type The differential equation (1) then has the form
dy
dx = h (y) g(x) |
1 | 4661-4664 | (1)
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MATHEMATICS
392
If F (x, y) can be expressed as a product g (x) h(y), where, g(x) is a function of x
and h(y) is a function of y, then the differential equation (1) is said to be of variable
separable type The differential equation (1) then has the form
dy
dx = h (y) g(x) (2)
If h(y) ≠ 0, separating the variables, (2) can be rewritten as
1
( )
h y dy = g(x) dx |
1 | 4662-4665 | The differential equation (1) then has the form
dy
dx = h (y) g(x) (2)
If h(y) ≠ 0, separating the variables, (2) can be rewritten as
1
( )
h y dy = g(x) dx (3)
Integrating both sides of (3), we get
1
( ) dy
∫h y
=
( )
∫g x dx |
1 | 4663-4666 | g(x) (2)
If h(y) ≠ 0, separating the variables, (2) can be rewritten as
1
( )
h y dy = g(x) dx (3)
Integrating both sides of (3), we get
1
( ) dy
∫h y
=
( )
∫g x dx (4)
Thus, (4) provides the solutions of given differential equation in the form
H(y) = G(x) + C
Here, H (y) and G (x) are the anti derivatives of
1
( )
h y and g(x) respectively and
C is the arbitrary constant |
1 | 4664-4667 | (2)
If h(y) ≠ 0, separating the variables, (2) can be rewritten as
1
( )
h y dy = g(x) dx (3)
Integrating both sides of (3), we get
1
( ) dy
∫h y
=
( )
∫g x dx (4)
Thus, (4) provides the solutions of given differential equation in the form
H(y) = G(x) + C
Here, H (y) and G (x) are the anti derivatives of
1
( )
h y and g(x) respectively and
C is the arbitrary constant Example 9 Find the general solution of the differential equation
1
2
dy
x
dx
+y
=
−
, (y ≠ 2)
Solution We have
dy
dx =
1
x2
−+y |
1 | 4665-4668 | (3)
Integrating both sides of (3), we get
1
( ) dy
∫h y
=
( )
∫g x dx (4)
Thus, (4) provides the solutions of given differential equation in the form
H(y) = G(x) + C
Here, H (y) and G (x) are the anti derivatives of
1
( )
h y and g(x) respectively and
C is the arbitrary constant Example 9 Find the general solution of the differential equation
1
2
dy
x
dx
+y
=
−
, (y ≠ 2)
Solution We have
dy
dx =
1
x2
−+y (1)
Separating the variables in equation (1), we get
(2 – y) dy = (x + 1) dx |
1 | 4666-4669 | (4)
Thus, (4) provides the solutions of given differential equation in the form
H(y) = G(x) + C
Here, H (y) and G (x) are the anti derivatives of
1
( )
h y and g(x) respectively and
C is the arbitrary constant Example 9 Find the general solution of the differential equation
1
2
dy
x
dx
+y
=
−
, (y ≠ 2)
Solution We have
dy
dx =
1
x2
−+y (1)
Separating the variables in equation (1), we get
(2 – y) dy = (x + 1) dx (2)
Integrating both sides of equation (2), we get
(2
)
∫−y dy
=
(
1)
x
dx
+
∫
or
2
2
y −y2
=
2
C1
x2
+x
+
or
x2 + y2 + 2x – 4y + 2 C1 = 0
or
x2 + y2 + 2x – 4y + C = 0, where C = 2C1
which is the general solution of equation (1) |
1 | 4667-4670 | Example 9 Find the general solution of the differential equation
1
2
dy
x
dx
+y
=
−
, (y ≠ 2)
Solution We have
dy
dx =
1
x2
−+y (1)
Separating the variables in equation (1), we get
(2 – y) dy = (x + 1) dx (2)
Integrating both sides of equation (2), we get
(2
)
∫−y dy
=
(
1)
x
dx
+
∫
or
2
2
y −y2
=
2
C1
x2
+x
+
or
x2 + y2 + 2x – 4y + 2 C1 = 0
or
x2 + y2 + 2x – 4y + C = 0, where C = 2C1
which is the general solution of equation (1) © NCERT
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DIFFERENTIAL EQUATIONS
393
Example 10 Find the general solution of the differential equation
2
2
11
dy
y
dx
x
=+
+ |
1 | 4668-4671 | (1)
Separating the variables in equation (1), we get
(2 – y) dy = (x + 1) dx (2)
Integrating both sides of equation (2), we get
(2
)
∫−y dy
=
(
1)
x
dx
+
∫
or
2
2
y −y2
=
2
C1
x2
+x
+
or
x2 + y2 + 2x – 4y + 2 C1 = 0
or
x2 + y2 + 2x – 4y + C = 0, where C = 2C1
which is the general solution of equation (1) © NCERT
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DIFFERENTIAL EQUATIONS
393
Example 10 Find the general solution of the differential equation
2
2
11
dy
y
dx
x
=+
+ Solution Since 1 + y2 ≠ 0, therefore separating the variables, the given differential
equation can be written as
2
1
dy
+y
=
2
1
dx
+x |
1 | 4669-4672 | (2)
Integrating both sides of equation (2), we get
(2
)
∫−y dy
=
(
1)
x
dx
+
∫
or
2
2
y −y2
=
2
C1
x2
+x
+
or
x2 + y2 + 2x – 4y + 2 C1 = 0
or
x2 + y2 + 2x – 4y + C = 0, where C = 2C1
which is the general solution of equation (1) © NCERT
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DIFFERENTIAL EQUATIONS
393
Example 10 Find the general solution of the differential equation
2
2
11
dy
y
dx
x
=+
+ Solution Since 1 + y2 ≠ 0, therefore separating the variables, the given differential
equation can be written as
2
1
dy
+y
=
2
1
dx
+x (1)
Integrating both sides of equation (1), we get
2
1
dy
∫+y
=
2
1
dx
+x
∫
or
tan–1 y = tan–1x + C
which is the general solution of equation (1) |
1 | 4670-4673 | © NCERT
not to be republished
DIFFERENTIAL EQUATIONS
393
Example 10 Find the general solution of the differential equation
2
2
11
dy
y
dx
x
=+
+ Solution Since 1 + y2 ≠ 0, therefore separating the variables, the given differential
equation can be written as
2
1
dy
+y
=
2
1
dx
+x (1)
Integrating both sides of equation (1), we get
2
1
dy
∫+y
=
2
1
dx
+x
∫
or
tan–1 y = tan–1x + C
which is the general solution of equation (1) Example 11 Find the particular solution of the differential equation
2
4
dy
dx = −xy
given
that y = 1, when x = 0 |
1 | 4671-4674 | Solution Since 1 + y2 ≠ 0, therefore separating the variables, the given differential
equation can be written as
2
1
dy
+y
=
2
1
dx
+x (1)
Integrating both sides of equation (1), we get
2
1
dy
∫+y
=
2
1
dx
+x
∫
or
tan–1 y = tan–1x + C
which is the general solution of equation (1) Example 11 Find the particular solution of the differential equation
2
4
dy
dx = −xy
given
that y = 1, when x = 0 Solution If y ≠ 0, the given differential equation can be written as
dy2
y = – 4x dx |
1 | 4672-4675 | (1)
Integrating both sides of equation (1), we get
2
1
dy
∫+y
=
2
1
dx
+x
∫
or
tan–1 y = tan–1x + C
which is the general solution of equation (1) Example 11 Find the particular solution of the differential equation
2
4
dy
dx = −xy
given
that y = 1, when x = 0 Solution If y ≠ 0, the given differential equation can be written as
dy2
y = – 4x dx (1)
Integrating both sides of equation (1), we get
y∫dy2
=
4 x dx
− ∫
or
−y1
= – 2x2 + C
or
y =
212
x −C |
1 | 4673-4676 | Example 11 Find the particular solution of the differential equation
2
4
dy
dx = −xy
given
that y = 1, when x = 0 Solution If y ≠ 0, the given differential equation can be written as
dy2
y = – 4x dx (1)
Integrating both sides of equation (1), we get
y∫dy2
=
4 x dx
− ∫
or
−y1
= – 2x2 + C
or
y =
212
x −C (2)
Substituting y = 1 and x = 0 in equation (2), we get, C = – 1 |
1 | 4674-4677 | Solution If y ≠ 0, the given differential equation can be written as
dy2
y = – 4x dx (1)
Integrating both sides of equation (1), we get
y∫dy2
=
4 x dx
− ∫
or
−y1
= – 2x2 + C
or
y =
212
x −C (2)
Substituting y = 1 and x = 0 in equation (2), we get, C = – 1 Now substituting the value of C in equation (2), we get the particular solution of the
given differential equation as
212
1
y
x
=
+ |
1 | 4675-4678 | (1)
Integrating both sides of equation (1), we get
y∫dy2
=
4 x dx
− ∫
or
−y1
= – 2x2 + C
or
y =
212
x −C (2)
Substituting y = 1 and x = 0 in equation (2), we get, C = – 1 Now substituting the value of C in equation (2), we get the particular solution of the
given differential equation as
212
1
y
x
=
+ Example 12 Find the equation of the curve passing through the point (1, 1) whose
differential equation is x dy = (2x2 + 1) dx (x ≠ 0) |
1 | 4676-4679 | (2)
Substituting y = 1 and x = 0 in equation (2), we get, C = – 1 Now substituting the value of C in equation (2), we get the particular solution of the
given differential equation as
212
1
y
x
=
+ Example 12 Find the equation of the curve passing through the point (1, 1) whose
differential equation is x dy = (2x2 + 1) dx (x ≠ 0) © NCERT
not to be republished
MATHEMATICS
394
Solution The given differential equation can be expressed as
dy* =
22
1
*
x
dx
x
or
dy =
1
2x
xdx
⎛
⎞
+
⎜
⎟
⎝
⎠ |
1 | 4677-4680 | Now substituting the value of C in equation (2), we get the particular solution of the
given differential equation as
212
1
y
x
=
+ Example 12 Find the equation of the curve passing through the point (1, 1) whose
differential equation is x dy = (2x2 + 1) dx (x ≠ 0) © NCERT
not to be republished
MATHEMATICS
394
Solution The given differential equation can be expressed as
dy* =
22
1
*
x
dx
x
or
dy =
1
2x
xdx
⎛
⎞
+
⎜
⎟
⎝
⎠ (1)
Integrating both sides of equation (1), we get
∫dy
=
1
2x
xdx
⎛
⎞
+
⎜
⎟
⎝
⎠
∫
or
y = x2 + log |x| + C |
1 | 4678-4681 | Example 12 Find the equation of the curve passing through the point (1, 1) whose
differential equation is x dy = (2x2 + 1) dx (x ≠ 0) © NCERT
not to be republished
MATHEMATICS
394
Solution The given differential equation can be expressed as
dy* =
22
1
*
x
dx
x
or
dy =
1
2x
xdx
⎛
⎞
+
⎜
⎟
⎝
⎠ (1)
Integrating both sides of equation (1), we get
∫dy
=
1
2x
xdx
⎛
⎞
+
⎜
⎟
⎝
⎠
∫
or
y = x2 + log |x| + C (2)
Equation (2) represents the family of solution curves of the given differential equation
but we are interested in finding the equation of a particular member of the family which
passes through the point (1, 1) |
1 | 4679-4682 | © NCERT
not to be republished
MATHEMATICS
394
Solution The given differential equation can be expressed as
dy* =
22
1
*
x
dx
x
or
dy =
1
2x
xdx
⎛
⎞
+
⎜
⎟
⎝
⎠ (1)
Integrating both sides of equation (1), we get
∫dy
=
1
2x
xdx
⎛
⎞
+
⎜
⎟
⎝
⎠
∫
or
y = x2 + log |x| + C (2)
Equation (2) represents the family of solution curves of the given differential equation
but we are interested in finding the equation of a particular member of the family which
passes through the point (1, 1) Therefore substituting x = 1, y = 1 in equation (2), we
get C = 0 |
1 | 4680-4683 | (1)
Integrating both sides of equation (1), we get
∫dy
=
1
2x
xdx
⎛
⎞
+
⎜
⎟
⎝
⎠
∫
or
y = x2 + log |x| + C (2)
Equation (2) represents the family of solution curves of the given differential equation
but we are interested in finding the equation of a particular member of the family which
passes through the point (1, 1) Therefore substituting x = 1, y = 1 in equation (2), we
get C = 0 Now substituting the value of C in equation (2) we get the equation of the required
curve as y = x2 + log |x| |
1 | 4681-4684 | (2)
Equation (2) represents the family of solution curves of the given differential equation
but we are interested in finding the equation of a particular member of the family which
passes through the point (1, 1) Therefore substituting x = 1, y = 1 in equation (2), we
get C = 0 Now substituting the value of C in equation (2) we get the equation of the required
curve as y = x2 + log |x| Example 13 Find the equation of a curve passing through the point (–2, 3), given that
the slope of the tangent to the curve at any point (x, y) is
2x2
y |
1 | 4682-4685 | Therefore substituting x = 1, y = 1 in equation (2), we
get C = 0 Now substituting the value of C in equation (2) we get the equation of the required
curve as y = x2 + log |x| Example 13 Find the equation of a curve passing through the point (–2, 3), given that
the slope of the tangent to the curve at any point (x, y) is
2x2
y Solution We know that the slope of the tangent to a curve is given by dy
dx |
1 | 4683-4686 | Now substituting the value of C in equation (2) we get the equation of the required
curve as y = x2 + log |x| Example 13 Find the equation of a curve passing through the point (–2, 3), given that
the slope of the tangent to the curve at any point (x, y) is
2x2
y Solution We know that the slope of the tangent to a curve is given by dy
dx so,
dy
dx =
y2x2 |
1 | 4684-4687 | Example 13 Find the equation of a curve passing through the point (–2, 3), given that
the slope of the tangent to the curve at any point (x, y) is
2x2
y Solution We know that the slope of the tangent to a curve is given by dy
dx so,
dy
dx =
y2x2 (1)
Separating the variables, equation (1) can be written as
y2 dy = 2x dx |
1 | 4685-4688 | Solution We know that the slope of the tangent to a curve is given by dy
dx so,
dy
dx =
y2x2 (1)
Separating the variables, equation (1) can be written as
y2 dy = 2x dx (2)
Integrating both sides of equation (2), we get
2
∫y dy
=
2x dx
∫
or
3
3
y = x2 + C |
1 | 4686-4689 | so,
dy
dx =
y2x2 (1)
Separating the variables, equation (1) can be written as
y2 dy = 2x dx (2)
Integrating both sides of equation (2), we get
2
∫y dy
=
2x dx
∫
or
3
3
y = x2 + C (3)
*
The notation
dy
dx due to Leibnitz is extremely flexible and useful in many calculation and formal
transformations, where, we can deal with symbols dy and dx exactly as if they were ordinary numbers |
1 | 4687-4690 | (1)
Separating the variables, equation (1) can be written as
y2 dy = 2x dx (2)
Integrating both sides of equation (2), we get
2
∫y dy
=
2x dx
∫
or
3
3
y = x2 + C (3)
*
The notation
dy
dx due to Leibnitz is extremely flexible and useful in many calculation and formal
transformations, where, we can deal with symbols dy and dx exactly as if they were ordinary numbers By
treating dx and dy like separate entities, we can give neater expressions to many calculations |
1 | 4688-4691 | (2)
Integrating both sides of equation (2), we get
2
∫y dy
=
2x dx
∫
or
3
3
y = x2 + C (3)
*
The notation
dy
dx due to Leibnitz is extremely flexible and useful in many calculation and formal
transformations, where, we can deal with symbols dy and dx exactly as if they were ordinary numbers By
treating dx and dy like separate entities, we can give neater expressions to many calculations Refer: Introduction to Calculus and Analysis, volume-I page 172, By Richard Courant,
Fritz John Spinger – Verlog New York |
1 | 4689-4692 | (3)
*
The notation
dy
dx due to Leibnitz is extremely flexible and useful in many calculation and formal
transformations, where, we can deal with symbols dy and dx exactly as if they were ordinary numbers By
treating dx and dy like separate entities, we can give neater expressions to many calculations Refer: Introduction to Calculus and Analysis, volume-I page 172, By Richard Courant,
Fritz John Spinger – Verlog New York © NCERT
not to be republished
DIFFERENTIAL EQUATIONS
395
Substituting x = –2, y = 3 in equation (3), we get C = 5 |
1 | 4690-4693 | By
treating dx and dy like separate entities, we can give neater expressions to many calculations Refer: Introduction to Calculus and Analysis, volume-I page 172, By Richard Courant,
Fritz John Spinger – Verlog New York © NCERT
not to be republished
DIFFERENTIAL EQUATIONS
395
Substituting x = –2, y = 3 in equation (3), we get C = 5 Substituting the value of C in equation (3), we get the equation of the required curve as
3
2
5
y3
=x
+
or
1
2
3
(3
15)
y
x
=
+
Example 14 In a bank, principal increases continuously at the rate of 5% per year |
1 | 4691-4694 | Refer: Introduction to Calculus and Analysis, volume-I page 172, By Richard Courant,
Fritz John Spinger – Verlog New York © NCERT
not to be republished
DIFFERENTIAL EQUATIONS
395
Substituting x = –2, y = 3 in equation (3), we get C = 5 Substituting the value of C in equation (3), we get the equation of the required curve as
3
2
5
y3
=x
+
or
1
2
3
(3
15)
y
x
=
+
Example 14 In a bank, principal increases continuously at the rate of 5% per year In
how many years Rs 1000 double itself |
1 | 4692-4695 | © NCERT
not to be republished
DIFFERENTIAL EQUATIONS
395
Substituting x = –2, y = 3 in equation (3), we get C = 5 Substituting the value of C in equation (3), we get the equation of the required curve as
3
2
5
y3
=x
+
or
1
2
3
(3
15)
y
x
=
+
Example 14 In a bank, principal increases continuously at the rate of 5% per year In
how many years Rs 1000 double itself Solution Let P be the principal at any time t |
1 | 4693-4696 | Substituting the value of C in equation (3), we get the equation of the required curve as
3
2
5
y3
=x
+
or
1
2
3
(3
15)
y
x
=
+
Example 14 In a bank, principal increases continuously at the rate of 5% per year In
how many years Rs 1000 double itself Solution Let P be the principal at any time t According to the given problem,
dp
dt =
5
P
⎛100
⎞×
⎜
⎟
⎝
⎠
or
dp
dt = P
20 |
1 | 4694-4697 | In
how many years Rs 1000 double itself Solution Let P be the principal at any time t According to the given problem,
dp
dt =
5
P
⎛100
⎞×
⎜
⎟
⎝
⎠
or
dp
dt = P
20 (1)
separating the variables in equation (1), we get
P
dp = 20
dt |
1 | 4695-4698 | Solution Let P be the principal at any time t According to the given problem,
dp
dt =
5
P
⎛100
⎞×
⎜
⎟
⎝
⎠
or
dp
dt = P
20 (1)
separating the variables in equation (1), we get
P
dp = 20
dt (2)
Integrating both sides of equation (2), we get
log P =
C1
20
t +
or
P =
C1
et20
e⋅
or
P =
20
C
et
(where
C1
C
e
=
) |
1 | 4696-4699 | According to the given problem,
dp
dt =
5
P
⎛100
⎞×
⎜
⎟
⎝
⎠
or
dp
dt = P
20 (1)
separating the variables in equation (1), we get
P
dp = 20
dt (2)
Integrating both sides of equation (2), we get
log P =
C1
20
t +
or
P =
C1
et20
e⋅
or
P =
20
C
et
(where
C1
C
e
=
) (3)
Now
P = 1000, when t = 0
Substituting the values of P and t in (3), we get C = 1000 |
1 | 4697-4700 | (1)
separating the variables in equation (1), we get
P
dp = 20
dt (2)
Integrating both sides of equation (2), we get
log P =
C1
20
t +
or
P =
C1
et20
e⋅
or
P =
20
C
et
(where
C1
C
e
=
) (3)
Now
P = 1000, when t = 0
Substituting the values of P and t in (3), we get C = 1000 Therefore, equation (3),
gives
P = 1000
t20
e
Let t years be the time required to double the principal |
1 | 4698-4701 | (2)
Integrating both sides of equation (2), we get
log P =
C1
20
t +
or
P =
C1
et20
e⋅
or
P =
20
C
et
(where
C1
C
e
=
) (3)
Now
P = 1000, when t = 0
Substituting the values of P and t in (3), we get C = 1000 Therefore, equation (3),
gives
P = 1000
t20
e
Let t years be the time required to double the principal Then
2000 = 1000
et20
⇒ t = 20 loge2
EXERCISE 9 |
1 | 4699-4702 | (3)
Now
P = 1000, when t = 0
Substituting the values of P and t in (3), we get C = 1000 Therefore, equation (3),
gives
P = 1000
t20
e
Let t years be the time required to double the principal Then
2000 = 1000
et20
⇒ t = 20 loge2
EXERCISE 9 4
For each of the differential equations in Exercises 1 to 10, find the general solution:
1 |
1 | 4700-4703 | Therefore, equation (3),
gives
P = 1000
t20
e
Let t years be the time required to double the principal Then
2000 = 1000
et20
⇒ t = 20 loge2
EXERCISE 9 4
For each of the differential equations in Exercises 1 to 10, find the general solution:
1 1
cos
1
cos
dy
x
dx
x
= +−
2 |
1 | 4701-4704 | Then
2000 = 1000
et20
⇒ t = 20 loge2
EXERCISE 9 4
For each of the differential equations in Exercises 1 to 10, find the general solution:
1 1
cos
1
cos
dy
x
dx
x
= +−
2 2
4
( 2
2)
dy
y
y
dx =
−
− <
<
© NCERT
not to be republished
MATHEMATICS
396
3 |
1 | 4702-4705 | 4
For each of the differential equations in Exercises 1 to 10, find the general solution:
1 1
cos
1
cos
dy
x
dx
x
= +−
2 2
4
( 2
2)
dy
y
y
dx =
−
− <
<
© NCERT
not to be republished
MATHEMATICS
396
3 1(
1)
dy
y
y
dx +
=
≠
4 |
1 | 4703-4706 | 1
cos
1
cos
dy
x
dx
x
= +−
2 2
4
( 2
2)
dy
y
y
dx =
−
− <
<
© NCERT
not to be republished
MATHEMATICS
396
3 1(
1)
dy
y
y
dx +
=
≠
4 sec2 x tan y dx + sec2 y tan x dy = 0
5 |
1 | 4704-4707 | 2
4
( 2
2)
dy
y
y
dx =
−
− <
<
© NCERT
not to be republished
MATHEMATICS
396
3 1(
1)
dy
y
y
dx +
=
≠
4 sec2 x tan y dx + sec2 y tan x dy = 0
5 (ex + e–x) dy – (ex – e–x) dx = 0
6 |
1 | 4705-4708 | 1(
1)
dy
y
y
dx +
=
≠
4 sec2 x tan y dx + sec2 y tan x dy = 0
5 (ex + e–x) dy – (ex – e–x) dx = 0
6 2
2
(1
) (1
)
dy
x
y
dx =
+
+
7 |
1 | 4706-4709 | sec2 x tan y dx + sec2 y tan x dy = 0
5 (ex + e–x) dy – (ex – e–x) dx = 0
6 2
2
(1
) (1
)
dy
x
y
dx =
+
+
7 y log y dx – x dy = 0
8 |
1 | 4707-4710 | (ex + e–x) dy – (ex – e–x) dx = 0
6 2
2
(1
) (1
)
dy
x
y
dx =
+
+
7 y log y dx – x dy = 0
8 5
5
xdy
y
dx = −
9 |
1 | 4708-4711 | 2
2
(1
) (1
)
dy
x
y
dx =
+
+
7 y log y dx – x dy = 0
8 5
5
xdy
y
dx = −
9 sin1
dy
x
dx
−
=
10 |
1 | 4709-4712 | y log y dx – x dy = 0
8 5
5
xdy
y
dx = −
9 sin1
dy
x
dx
−
=
10 ex tan y dx + (1 – ex) sec2 y dy = 0
For each of the differential equations in Exercises 11 to 14, find a particular solution
satisfying the given condition:
11 |
1 | 4710-4713 | 5
5
xdy
y
dx = −
9 sin1
dy
x
dx
−
=
10 ex tan y dx + (1 – ex) sec2 y dy = 0
For each of the differential equations in Exercises 11 to 14, find a particular solution
satisfying the given condition:
11 3
2
(
1) dy
x
x
x
dx
+
+
+
= 2x2 + x; y = 1 when x = 0
12 |
1 | 4711-4714 | sin1
dy
x
dx
−
=
10 ex tan y dx + (1 – ex) sec2 y dy = 0
For each of the differential equations in Exercises 11 to 14, find a particular solution
satisfying the given condition:
11 3
2
(
1) dy
x
x
x
dx
+
+
+
= 2x2 + x; y = 1 when x = 0
12 (2
1)
1
dy
x x
dx
−
= ; y = 0 when x = 2
13 |
1 | 4712-4715 | ex tan y dx + (1 – ex) sec2 y dy = 0
For each of the differential equations in Exercises 11 to 14, find a particular solution
satisfying the given condition:
11 3
2
(
1) dy
x
x
x
dx
+
+
+
= 2x2 + x; y = 1 when x = 0
12 (2
1)
1
dy
x x
dx
−
= ; y = 0 when x = 2
13 cos
dy
a
⎛dx
⎞ =
⎜
⎟
⎝
⎠
(a ∈ R); y = 2 when x = 0
14 |
1 | 4713-4716 | 3
2
(
1) dy
x
x
x
dx
+
+
+
= 2x2 + x; y = 1 when x = 0
12 (2
1)
1
dy
x x
dx
−
= ; y = 0 when x = 2
13 cos
dy
a
⎛dx
⎞ =
⎜
⎟
⎝
⎠
(a ∈ R); y = 2 when x = 0
14 tan
dy
y
x
dx =
; y = 1 when x = 0
15 |
1 | 4714-4717 | (2
1)
1
dy
x x
dx
−
= ; y = 0 when x = 2
13 cos
dy
a
⎛dx
⎞ =
⎜
⎟
⎝
⎠
(a ∈ R); y = 2 when x = 0
14 tan
dy
y
x
dx =
; y = 1 when x = 0
15 Find the equation of a curve passing through the point (0, 0) and whose differential
equation is y′ = ex sin x |
1 | 4715-4718 | cos
dy
a
⎛dx
⎞ =
⎜
⎟
⎝
⎠
(a ∈ R); y = 2 when x = 0
14 tan
dy
y
x
dx =
; y = 1 when x = 0
15 Find the equation of a curve passing through the point (0, 0) and whose differential
equation is y′ = ex sin x 16 |
1 | 4716-4719 | tan
dy
y
x
dx =
; y = 1 when x = 0
15 Find the equation of a curve passing through the point (0, 0) and whose differential
equation is y′ = ex sin x 16 For the differential equation
(
2) (
2)
xydy
x
y
dx =
+
+
, find the solution curve
passing through the point (1, –1) |
1 | 4717-4720 | Find the equation of a curve passing through the point (0, 0) and whose differential
equation is y′ = ex sin x 16 For the differential equation
(
2) (
2)
xydy
x
y
dx =
+
+
, find the solution curve
passing through the point (1, –1) 17 |
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