NCERT Solutions Class 12 Mathematics Part II . 674 Pages (668-1342). Free Flip-Book by Study Innovations

Class XII Chapter 9 – Differential Equations Maths

Question 7:
Answer
The given differential equation is:

This equation is the form of a linear differential equation as:

The general solution of the given differential equation is given by the relation,
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Class XII Chapter 9 – Differential Equations Maths

Substituting the value of in equation (1), we get:

This is the required general solution of the given differential equation.
Question 8:
Answer

This equation is a linear differential equation of the form:

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Class XII Chapter 9 – Differential Equations Maths

The general solution of the given differential equation is given by the relation,

Question 9:
Answer

This equation is a linear differential equation of the form:
The general solution of the given differential equation is given by the relation,

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Class XII Chapter 9 – Differential Equations Maths

Question 10:
Answer

This is a linear differential equation of the form:
The general solution of the given differential equation is given by the relation,

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Class XII Chapter 9 – Differential Equations Maths

Question 11:
Answer

This is a linear differential equation of the form:
The general solution of the given differential equation is given by the relation,

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Class XII Chapter 9 – Differential Equations Maths

Question 12:
Answer

This is a linear differential equation of the form:
The general solution of the given differential equation is given by the relation,

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Class XII Chapter 9 – Differential Equations Maths

Question 13:
Answer
The given differential equation is
This is a linear equation of the form:

The general solution of the given differential equation is given by the relation,

Now,
Therefore,

Substituting C = –2 in equation (1), we get:
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Class XII Chapter 9 – Differential Equations Maths

Hence, the required solution of the given differential equation is
Question 14:
Answer

This is a linear differential equation of the form:
The general solution of the given differential equation is given by the relation,

Now, y = 0 at x = 1.
Therefore,

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Class XII Chapter 9 – Differential Equations Maths
Substituting in equation (1), we get:

This is the required general solution of the given differential equation.
Question 15:

Answer
The given differential equation is
This is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

Now,
Therefore, we get:

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Class XII Chapter 9 – Differential Equations Maths

Substituting C = 4 in equation (1), we get:

This is the required particular solution of the given differential equation.

Question 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.
Answer
Let F (x, y) be the curve passing through the origin.

At point (x, y), the slope of the curve will be
According to the given information:

This is a linear differential equation of the form:
The general solution of the given differential equation is given by the relation,

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Class XII Chapter 9 – Differential Equations Maths

Substituting in equation (1), we get:

The curve passes through the origin.
Therefore, equation (2) becomes:
1=C

Substituting C = 1 in equation (2), we get:
⇒
Hence, the required equation of curve passing through the origin is
Question 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.
Answer
Let F (x, y) be the curve and let (x, y) be a point on the curve. The slope of the tangent

to the curve at (x, y) is
According to the given information:

This is a linear differential equation of the form:

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Class XII Chapter 9 – Differential Equations Maths

The general equation of the curve is given by the relation,

Therefore, equation (1) becomes:

The curve passes through point (0, 2).
Therefore, equation (2) becomes:
0 + 2 – 4 = Ce0
⇒–2=C

⇒C=–2
Substituting C = –2 in equation (2), we get:

This is the required equation of the curve.
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Class XII Chapter 9 – Differential Equations Maths

Question 18: is

The integrating factor of the differential equation
A. e–x
B. e–y
C.

D. x
Answer
The given differential equation is:

This is a linear differential equation of the form:

The integrating factor (I.F) is given by the relation,

Hence, the correct answer is C.
Question 19:
The integrating factor of the differential equation.

is
A.

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Class XII Chapter 9 – Differential Equations Maths
B.

C.

D.

Answer
The given differential equation is:

This is a linear differential equation of the form:
The integrating factor (I.F) is given by the relation,

Hence, the correct answer is D.

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Class XII Chapter 9 – Differential Equations Maths

Miscellaneous Solutions

Question 1:
For each of the differential equations given below, indicate its order and degree (if
defined).
(i)

(ii)

(iii)
Answer
(i) The differential equation is given as:

The highest order derivative present in the differential equation is . Thus, its order is

two. The highest power raised to is one. Hence, its degree is one.

(ii) The differential equation is given as:

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Class XII Chapter 9 – Differential Equations Maths

The highest order derivative present in the differential equation is . Thus, its order is
one. The highest power raised to is three. Hence, its degree is three.
(iii) The differential equation is given as:

The highest order derivative present in the differential equation is . Thus, its order is
four.
However, the given differential equation is not a polynomial equation. Hence, its degree
is not defined.

Question 2:
For each of the exercises given below, verify that the given function (implicit or explicit)
is a solution of the corresponding differential equation.
(i)

(ii)

(iii)

(iv)

Answer
(i)
Differentiating both sides with respect to x, we get:

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Class XII Chapter 9 – Differential Equations Maths

Again, differentiating both sides with respect to x, we get:

Now, on substituting the values of and in the differential equation, we get:

⇒ L.H.S. ≠ R.H.S.
Hence, the given function is not a solution of the corresponding differential equation.
(ii)
Differentiating both sides with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

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Class XII Chapter 9 – Differential Equations Maths

Now, on substituting the values of and in the L.H.S. of the given differential
equation, we get:

Hence, the given function is a solution of the corresponding differential equation.
(iii)
Differentiating both sides with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

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Class XII Chapter 9 – Differential Equations Maths

Substituting the value of in the L.H.S. of the given differential equation, we get:

Hence, the given function is a solution of the corresponding differential equation.
(iv)
Differentiating both sides with respect to x, we get:

Substituting the value of in the L.H.S. of the given differential equation, we get:
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Class XII Chapter 9 – Differential Equations Maths

Hence, the given function is a solution of the corresponding differential equation.

Question 3:
Form the differential equation representing the family of curves given by

where a is an arbitrary constant.

Answer

Differentiating with respect to x, we get:

From equation (1), we get:
On substituting this value in equation (3), we get:

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Class XII Chapter 9 – Differential Equations Maths

Hence, the differential equation of the family of curves is given as

Question 4: is the general solution of differential
, where c is a parameter.
Prove that
equation
Answer

This is a homogeneous equation. To simplify it, we need to make the substitution as:

Substituting the values of y and in equation (1), we get:

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Class XII Chapter 9 – Differential Equations Maths

Integrating both sides, we get:

Substituting the values of I1 and I2 in equation (3), we get:

Therefore, equation (2) becomes:
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Class XII Chapter 9 – Differential Equations Maths

Hence, the given result is proved.
Question 5:
Form the differential equation of the family of circles in the first quadrant which touch
the coordinate axes.
Answer
The equation of a circle in the first quadrant with centre (a, a) and radius (a) which
touches the coordinate axes is:

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Class XII Chapter 9 – Differential Equations Maths

Differentiating equation (1) with respect to x, we get:
Substituting the value of a in equation (1), we get:

Hence, the required differential equation of the family of circles is

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Class XII Chapter 9 – Differential Equations Maths
Question 6:

Find the general solution of the differential equation
Answer

Integrating both sides, we get: is given by

Question 7:
Show that the general solution of the differential equation
(x + y + 1) = A (1 – x – y – 2xy), where A is parameter
Answer

Integrating both sides, we get:
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Class XII Chapter 9 – Differential Equations Maths

Hence, the given result is proved.
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Class XII Chapter 9 – Differential Equations Maths
Question 8:

Find the equation of the curve passing through the point whose differential

equation is,
Answer
The differential equation of the given curve is:

Integrating both sides, we get:

The curve passes through point

On substituting in equation (1), we get:

Hence, the required equation of the curve is
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Class XII Chapter 9 – Differential Equations Maths

Question 9:
Find the particular solution of the differential equation

, given that y = 1 when x = 0

Answer

Integrating both sides, we get:

Substituting these values in equation (1), we get:

Now, y = 1 at x = 0.
Therefore, equation (2) becomes:

Substituting in equation (2), we get:

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Class XII Chapter 9 – Differential Equations Maths

This is the required particular solution of the given differential equation.

Question 10:
Solve the differential equation

Answer

Differentiating it with respect to y, we get:

From equation (1) and equation (2), we get:
Integrating both sides, we get:

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Class XII Chapter 9 – Differential Equations Maths

Question 11: , given that

Find a particular solution of the differential equation
y = – 1, when x = 0 (Hint: put x – y = t)
Answer

Substituting the values of x – y and in equation (1), we get:
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Class XII Chapter 9 – Differential Equations Maths

Integrating both sides, we get:

Now, y = –1 at x = 0.
Therefore, equation (3) becomes:
log 1 = 0 – 1 + C

⇒C=1

Substituting C = 1 in equation (3) we get:

This is the required particular solution of the given differential equation.

Question 12:
Solve the differential equation

Answer

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Class XII Chapter 9 – Differential Equations Maths

This equation is a linear differential equation of the form
The general solution of the given differential equation is given by,

Question 13: ,
Find a particular solution of the differential equation

given that y = 0 when
Answer
The given differential equation is:

This equation is a linear differential equation of the form

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Class XII Chapter 9 – Differential Equations Maths

The general solution of the given differential equation is given by,

Now,
Therefore, equation (1) becomes:

Substituting in equation (1), we get:

This is the required particular solution of the given differential equation.
Question 14:

Find a particular solution of the differential equation , given that y = 0
when x = 0
Answer

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Class XII Chapter 9 – Differential Equations Maths

Integrating both sides, we get:

Substituting this value in equation (1), we get:

Now, at x = 0 and y = 0, equation (2) becomes:
Substituting C = 1 in equation (2), we get:

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Class XII Chapter 9 – Differential Equations Maths

This is the required particular solution of the given differential equation.

Question 15:
The population of a village increases continuously at the rate proportional to the number
of its inhabitants present at any time. If the population of the village was 20000 in 1999
and 25000 in the year 2004, what will be the population of the village in 2009?
Answer
Let the population at any instant (t) be y.
It is given that the rate of increase of population is proportional to the number of
inhabitants at any instant.

Integrating both sides, we get:
log y = kt + C … (1)
In the year 1999, t = 0 and y = 20000.
Therefore, we get:
log 20000 = C … (2)
In the year 2004, t = 5 and y = 25000.
Therefore, we get:

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Class XII Chapter 9 – Differential Equations Maths

In the year 2009, t = 10 years.
Now, on substituting the values of t, k, and C in equation (1), we get:

Hence, the population of the village in 2009 will be 31250.
Question 16:

The general solution of the differential equation is
A. xy = C
B. x = Cy2
C. y = Cx
D. y = Cx2
Answer
The given differential equation is:

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Class XII Chapter 9 – Differential Equations Maths

Integrating both sides, we get:

Hence, the correct answer is C. is
Question 17:

The general solution of a differential equation of the type
A.
B.
C.
D.
Answer
The integrating factor of the given differential equation
The general solution of the differential equation is given by,

Hence, the correct answer is C.
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Class XII Chapter 9 – Differential Equations Maths
is
Question 18:

The general solution of the differential equation
A. xey + x2 = C
B. xey + y2 = C
C. yex + x2 = C
D. yey + x2 = C
Answer
The given differential equation is:

This is a linear differential equation of the form
The general solution of the given differential equation is given by,

Hence, the correct answer is C.

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Class XII Chapter 10 – Vector Algebra Maths

Exercise 10.1
Question 1:
Represent graphically a displacement of 40 km, 30° east of north.
Answer

Here, vector represents the displacement of 40 km, 30° East of North.

Question 2:
Classify the following measures as scalars and vectors.
(i) 10 kg (ii) 2 metres north-west (iii) 40°
(iv) 40 watt (v) 10–19 coulomb (vi) 20 m/s2
Answer
(i) 10 kg is a scalar quantity because it involves only magnitude.
(ii) 2 meters north-west is a vector quantity as it involves both magnitude and direction.
(iii) 40° is a scalar quantity as it involves only magnitude.
(iv) 40 watts is a scalar quantity as it involves only magnitude.
(v) 10–19 coulomb is a scalar quantity as it involves only magnitude.
(vi) 20 m/s2 is a vector quantity as it involves magnitude as well as direction.

Question 3:
Classify the following as scalar and vector quantities.
(i) time period (ii) distance (iii) force

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Class XII Chapter 10 – Vector Algebra Maths

(iv) velocity (v) work done
Answer
(i) Time period is a scalar quantity as it involves only magnitude.
(ii) Distance is a scalar quantity as it involves only magnitude.
(iii) Force is a vector quantity as it involves both magnitude and direction.
(iv) Velocity is a vector quantity as it involves both magnitude as well as direction.
(v) Work done is a scalar quantity as it involves only magnitude.

Question 4:
In Figure, identify the following vectors.

(i) Coinitial (ii) Equal (iii) Collinear but not equal
Answer

(i) Vectors and are coinitial because they have the same initial point.

(ii) Vectors and are equal because they have the same magnitude and direction.
(iii) Vectors and are collinear but not equal. This is because although they are
parallel, their directions are not the same.

Question 5:
Answer the following as true or false.
(i) and are collinear.
(ii) Two collinear vectors are always equal in magnitude.
(iii) Two vectors having same magnitude are collinear.
(iv) Two collinear vectors having the same magnitude are equal.
Answer
(i) True.

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Class XII Chapter 10 – Vector Algebra Maths

Vectors and are parallel to the same line.
(ii) False.
Collinear vectors are those vectors that are parallel to the same line.
(iii) False.

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Class XII Chapter 10 – Vector Algebra Maths
Exercise 10.2

Question 1:
Compute the magnitude of the following vectors:

Answer
The given vectors are:

Question 2:
Write two different vectors having same magnitude.
Answer

Hence, are two different vectors having the same magnitude. The vectors are

different because they have different directions.

Question 3:

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Class XII Chapter 10 – Vector Algebra Maths

Write two different vectors having same direction.
Answer

The direction cosines of are the same. Hence, the two vectors have the same
direction.

Question 4:

Find the values of x and y so that the vectors are equal
Answer

The two vectors will be equal if their corresponding components are

equal.

Hence, the required values of x and y are 2 and 3 respectively.

Question 5:
Find the scalar and vector components of the vector with initial point (2, 1) and terminal
point (–5, 7).
Answer
The vector with the initial point P (2, 1) and terminal point Q (–5, 7) can be given by,

Hence, the required scalar components are –7 and 6 while the vector components are
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Class XII Chapter 10 – Vector Algebra Maths

Question 6: .
.
Find the sum of the vectors
Answer
The given vectors are

Question 7: .

Find the unit vector in the direction of the vector
Answer

The unit vector in the direction of vector is given by .

Question 8:
Find the unit vector in the direction of vector , where P and Q are the points
(1, 2, 3) and (4, 5, 6), respectively.
Answer
The given points are P (1, 2, 3) and Q (4, 5, 6).

Hence, the unit vector in the direction of is

.
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Class XII Chapter 10 – Vector Algebra Maths

Question 9: and , find the unit vector in the direction
For given vectors,
of the vector and .
Answer
The given vectors are

Hence, the unit vector in the direction of is

. which has magnitude 8 units.

Question 10:
Find a vector in the direction of vector
Answer

Hence, the vector in the direction of vector which has magnitude 8 units is
given by,

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Class XII Chapter 10 – Vector Algebra Maths

Question 11: are collinear.

Show that the vectors
Answer

.

Hence, the given vectors are collinear.

Question 12:
Find the direction cosines of the vector
Answer

Hence, the direction cosines of

Question 13:
Find the direction cosines of the vector joining the points A (1, 2, –3) and
B (–1, –2, 1) directed from A to B.
Answer
The given points are A (1, 2, –3) and B (–1, –2, 1).

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Class XII Chapter 10 – Vector Algebra Maths

Hence, the direction cosines of are

Question 14: is equally inclined to the axes OX, OY, and OZ.

Show that the vector
Answer

Therefore, the direction cosines of

Now, let α, β, and γbe the angles formed by with the positive directions of x, y, and z
axes.
Then, we have

Hence, the given vector is equally inclined to axes OX, OY, and OZ.

Question 15:
Find the position vector of a point R which divides the line joining two points P and Q

whose position vectors are respectively, in the ration 2:1

(i) internally

(ii) externally

Answer

The position vector of point R dividing the line segment joining two points

P and Q in the ratio m: n is given by:

i. Internally:

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Class XII Chapter 10 – Vector Algebra Maths

ii. Externally:

Position vectors of P and Q are given as:
(i) The position vector of point R which divides the line joining two points P and Q
internally in the ratio 2:1 is given by,

(ii) The position vector of point R which divides the line joining two points P and Q
externally in the ratio 2:1 is given by,

Question 16:
Find the position vector of the mid point of the vector joining the points P (2, 3, 4) and Q
(4, 1, – 2).
Answer
The position vector of mid-point R of the vector joining points P (2, 3, 4) and Q (4, 1, –
2) is given by,

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Class XII Chapter 10 – Vector Algebra Maths

Question 17:

Show that the points A, B and C with position vectors, ,

respectively form the vertices of a right angled triangle.
Answer
Position vectors of points A, B, and C are respectively given as:

Hence, ABC is a right-angled triangle.
Question 18:
In triangle ABC which of the following is not true:

A.
B.
C.
D.

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Class XII Chapter 10 – Vector Algebra Maths
Answer

On applying the triangle law of addition in the given triangle, we have:

From equations (1) and (3), we have:

Hence, the equation given in alternative C is incorrect.
The correct answer is C.

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