Class 12

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 10 Vector Algebra Ex 10.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3

Question 1.
Find the angle between two vectors \(\vec{a}\) and \(\vec{b}\) with magnitudes √3 and 2, respectively, having \(\vec{a}\) . \(\vec{b}\) = √6.
Solution.
It is given that, |\(\vec{a}\)| = √3, |\(\vec{b}\)| = 2 and
\(\vec{a}\) . \(\vec{b}\) = √6

∴ √6 = √3 × 2 × cos θ

⇒ cos θ = \(\frac{\sqrt{6}}{\sqrt{3} \times 2}\)

⇒ cos θ = \(\frac{1}{\sqrt{2}}\)

⇒ θ = \(\frac{\pi}{4}\)
Hence, the angle between the given vectors \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{4}\).

Question 2.
Find the angle between the vectors î – 2 ĵ + 3k̂ and 3 î – 2 ĵ + k̂.
Solution.
The given vectors are \(\vec{a}\) = î – 2 ĵ + 3k̂ and \(\vec{b}\) = 3 î – 2 ĵ + k̂

Question 3.
Find the projection of the vector î – ĵ on the vector î + ĵ.
Solution.
Let \(\vec{a}\) = î – ĵ and \(\vec{b}\) = î + ĵ
Now, projection of vector a on S is given by
\(\frac{1}{|\vec{b}|}(\vec{a} \cdot \vec{b})=\frac{1}{\sqrt{1+1}}\{1.1+(-1)(1)\}=\frac{1}{\sqrt{2}}(1-1)=0\)
Hence, the projection of vector \(\vec{a}\) and \(\vec{b}\) is 0.

Question 4.
Find the projection of the vector î + 3ĵ + k̂ on the vector 7î – ĵ + 8k̂.
Solution.
Let \(\vec{a}\) = i + 3j + 7k and \(\vec{b}\) = 7i – j + 8k
Now, projection of vector a and b is given by
\(\frac{1}{|\vec{b}|}(\vec{a} \cdot \vec{b})=\frac{1}{\sqrt{7^{2}+(-1)^{2}+8^{2}}}\{1(7)+3(-1)+7(8)\}\)

= \(\frac{7-3+56}{\sqrt{49+1+64}}=\frac{60}{\sqrt{114}}\)

Question 5.
Show that each of the given three vectors is a unit vector \(\frac{1}{7}\) (2î + 3ĵ + 8k̂), \(\frac{1}{7}\) (3î- 6ĵ + 2k̂), \(\frac{1}{7}\) (6î + 2k̂ – 3k̂)
Also, show that they are mutually perpendicular to each other.
Solution.

Question 6.
Find |\(|\overrightarrow{\boldsymbol{a}}|\)| and |\(|\overrightarrow{\boldsymbol{b}}|\)|, if \((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})\) = 8 and \(|\vec{a}|=8|\vec{b}|\).
Solution.

Question 7.
Evaluate the product \((3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})\).
Solution.

Question 8.
Find the magnitude of two vectors \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\) having the same magnitude and such that the angle between them is 60° and their scalar product is \(\frac{1}{2}\).
Solution.
Let θ be the angle between the vectors \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\)

It is given that |\(\overrightarrow{\boldsymbol{b}}\)| = |\(\overrightarrow{\boldsymbol{b}}\)|, \(\overrightarrow{\boldsymbol{a}}\) . \(\overrightarrow{\boldsymbol{b}}\) = \(\frac{1}{2}\) and θ = 60°

We know that \(\overrightarrow{\boldsymbol{a}}\) . \(\overrightarrow{\boldsymbol{b}}\) = |\(\overrightarrow{\boldsymbol{a}}\) . \(\overrightarrow{\boldsymbol{b}}\)| cos θ

Question 9.
Find |\(\vec{x}\)|, if for a unit vector \(\vec{a}\), (\(\vec{x}\) – \(\vec{a}\)) . (\(\vec{x}\) + \(\vec{a}\)) = 12.
Solution.

Question 10.
If \(\vec{a}\) = 2î + 2 ĵ + 3k̂, \(\vec{b}\) = – î + 2 ĵ + k̂ and \(\vec{c}\) = 3î + ĵ are such that \(\vec{a}\) + \(\vec{b}\) is perpendicular to \(\vec{c}\), then find the value of λ.
Solution.

⇒ (2 – λ)3 + (2 + 2λ)1 + (3 + λ)0 = 0
⇒ 6 – 3λ + 2 + 2λ = 0
⇒ – λ + 8 = 0
⇒ λ = 8
Hence, the required value of λ is 8.

Question 11.
Show that |\(\vec{a}\)| \(\vec{b}\) + |\(\vec{b}\)| \(\vec{a}\) is perpendicular to |\(\vec{a}\)| \(\vec{b}\) – |\(\vec{b}\)| \(\vec{a}\), for any two non-zero vectors a and b.
Solution.

Question 12.
If \(\vec{a}\) . \(\vec{a}\) = 0 and \(\vec{a}\) . \(\vec{b}\) = 0, then what can be concluded about the vector 6?
Solution.

Question 13.
If \(\vec{a}\), \(\vec{a}\), \(\vec{c}\) are unit vectors such that \(\vec{a}\) + \(\vec{a}\) + \(\vec{c}\) = 0, find the value of \(\vec{a}\) . \(\vec{b}\) + \(\vec{b}\) . \(\vec{c}\) + \(\vec{c}\) . \(\vec{a}\).
Solution.

Question 14.
If either vector \(\vec{a}=\overrightarrow{0}\) or \(\overrightarrow{\boldsymbol{b}}=\overrightarrow{\mathbf{0}}\), then \(\overrightarrow{\boldsymbol{a}} \cdot \overrightarrow{\boldsymbol{b}}\) = 0. But the converse need not be true. Justify your answer with an example.
Solution.

Question 15.
If the vertices A, B, C of a triangle ABC are (1, 2, 3), (- 1, 0, 0), (0, 1, 2), respectively, then find ∠ABC [∠ABC is the angle
between the vector \(\overrightarrow{\boldsymbol{B A}}\) and \(\overrightarrow{\boldsymbol{B C}}\)].
Solution.

Question 16.
Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, – 1) are collinear.
Solution.

Question 17.
Show that the vectors 2î – ĵ + k̂, î – 3ĵ – 5k̂ and 3î – 4ĵ – 4k̂ form the vertices of a right angled triangle.
Solution.

Question 18.
If \(\overrightarrow{\boldsymbol{a}}\) is a non-zero vector of magnitude ‘a’ and λ, is a non-zero scalar, then λ \(\overrightarrow{\boldsymbol{a}}\) is unit vector if
(A) λ = 1
(B) λ = – 1
(C) a = |λ|
(D) a = \(\frac{1}{|\lambda|}\)
Solution.

Hence, vector λ \(\overrightarrow{\boldsymbol{a}}\) is a unit vector if a = \(\frac{1}{|\lambda|}\).
The correct answer is (D).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 10 Vector Algebra Ex 10.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.2

Question 1.
Compute the magnitude of the following vectors
\(\overrightarrow{\boldsymbol{a}}=\hat{\boldsymbol{i}}+\hat{\boldsymbol{j}}+\hat{\boldsymbol{k}}\); \(\overrightarrow{\boldsymbol{b}}=\mathbf{2} \hat{\boldsymbol{i}}-\mathbf{7} \hat{j}-\mathbf{3} \hat{\boldsymbol{k}}\); \(\vec{c}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}\)
Solution.
The given vectors are

Question 2.
Write two different vectors having same magnitude.
Solution.
Consider the vectors \(\vec{a}=\hat{i}+2 \hat{j}+\hat{k}\) and \(\vec{b}=\hat{i}+\hat{j}+2 \hat{k}\)
\(|\vec{a}|=\sqrt{1^{2}+2^{2}+1^{2}}=\sqrt{6}\),
\([|\vec{b}|=\sqrt{1^{2}+1^{2}+2^{2}}=\sqrt{6}/latex]
Thus, vectors [latex]\vec{a}\) and \(\vec{a}\) have the same magnitude.

Question 3.
Write two different vectors having same direction.
Solution.

The direction cosines of \(\vec{p}\) and \(\vec{q}\) are the same.
Hence, the two vectors have the same direction.

Question 4.
Find the values of x and y so that the vectors \(2 \hat{i}+\mathbf{3} \hat{j}\) and \(x \hat{i}+y \hat{j}\) are equal.
Solution.
The two vectors \(2 \hat{i}+\mathbf{3} \hat{j}\) and \(x \hat{i}+y \hat{j}\) 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).
Solution.
The vector with the initial point P(2,1) and terminal point Q(- 5, 7) can be given by,
\(\overline{P Q}=(-5-2) \hat{i}+(7-1) \hat{j}\)

⇒ \(\overline{P Q}=7 \hat{i}+6 \hat{j}\)
Hence, the required scalar components are -7 and 6 while the vector components are – 7î and 6ĵ.

Question 6.
Find the sum of the vectors a = î – 2ĵ + k̂, b = – 2î + 4ĵ + 5k̂ and c = î – 6ĵ – 7k̂.
Solution.
The given vectors are a = î – 2ĵ + k̂, b = – 2î + 4ĵ + 5k̂ and c = î – 6ĵ – 7k̂.
∴ \(\vec{a}+\vec{b}+\vec{c}\) = (1 – 2 + 1) î + (- 2 + 4 – 6) ĵ + (1 + 5 – 7) k̂.
0 . î – 4 ĵ – 1 . k̂ = – 4 ĵ – k̂ .

Question 7.
Find the unit vector in the direction of the vector \(\vec{a}\) = î + ĵ + 2k̂.
Solution.
The unit vector â in the direction of vector \(\vec{a}\) = î + ĵ + 2k̂ is given by

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

Question 9.
For given vectors, \(\vec{a}\) = 2i – j + 2 k and \(\vec{b}\) = – i + j k, find the unit vector in the direction of the vector \(\vec{a}\) + \(\vec{b}\).
Solution.
The given vectors are \(\vec{a}\) = 2 î – ĵ + 2 k̂ and \(\vec{b}\) = – î + ĵ – k̂
\(\vec{a}\) = 2 î – ĵ + 2 k̂
\(\vec{b}\) = – î + ĵ – k̂
.-. \(\vec{a}\) + \(\vec{a}\) = (2 – 1) î +(- 1 + 1) ĵ + (2 – 1) k̂
= 1 î + 0 ĵ + 1 k̂
|\(\vec{a}\) + \(\vec{b}\)| = \(\sqrt{1^{2}+1^{2}}=\sqrt{2}\)
Hence, the unit vector in the direction of (\(\vec{a}\) + \(\vec{b}\)) is \(\frac{(\vec{a}+\vec{b})}{|\vec{a}-\vec{b}|}=\frac{\hat{i}+\hat{k}}{\sqrt{2}}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}\).

Question 10.
Find a vector in the direction of vector 5î – ĵ + 2k̂ which has magnitude 8 units.
Solution.
Let \(\vec{a}\) = – 5î – ĵ + 2k̂
∴ |\(\vec{a}\)| = \(\sqrt{5^{2}+(-1)^{2}+2^{2}}=\sqrt{25+1+4}=\sqrt{30}\)

⇒ \(\hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{5 \hat{i}-\hat{j}+2 \hat{k}}{\sqrt{30}}\)

Hence, the vector in the direction of vector 5î – ĵ + 2k̂ which has magnitude 8 units is given by,

Question 11.
Show that the vectors 2î – 3ĵ + 4k̂ and – 4î + 6ĵ – 8k̂ are collinear.
Solution.
Let \(\vec{a}\) = 2î – 3ĵ + 4k̂ and \(\vec{b}\) = – 4î + 6ĵ – 8k̂.

It is observed that \(\vec{b}\) = – 4î + 6ĵ – 8k̂
= -2 (2î – 3ĵ + 4k̂) = – 2 \(\vec{a}\)
\(\vec{b}\) = λ \(\vec{a}\), where λ = – 2.
Hence, the given vectors are collinear.

Question 12.
Find the direction cosines of the vector î + 2 ĵ + 3 k̂.
Solution.
Let |\(\vec{a}\)| = î + 2 ĵ + 3 k̂
|\(\vec{a}\)| = \(\sqrt{1^{2}+2^{2}+3^{2}}=\sqrt{1+4+9}=\sqrt{14}\)
Hence the direction cosines of \(\vec{a}\) are \(\left(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right)\).

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.
Solution.
The given points are A(1, 2, – 3) and B(- 1, – 2, 1).

Question 14.
Show that the vector î + ĵ + k̂ is equally inclined to the axes OX, OY and OZ.
Solution.
Let \(\vec{a}\) = î + ĵ + k̂
Then, |\(\vec{a}\)| = \(\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}\)

Therefore, the direction cosines of \(\vec{a}\) are \(\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\).
Now, let α, β and γ be the angles formed by \(\vec{a}\) with the positive directions of x, y and z axes.

Then, we have cos α = \(\frac{1}{\sqrt{3}}\), cos β = \(\frac{1}{\sqrt{3}}\), cos γ = \(\frac{1}{\sqrt{3}}\).

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 î + 2ĵ – k̂ and – î + ĵ + k̂ respectively, in the ratio 2 : 1
(i) internally
(ii) externally
Solution.
The position vector of point R dividing the line segment joining two points P and Q in the ratio m : n is given by
Case I:
Internally \(\frac{m \vec{b}+n \vec{a}}{m+n}\)

Case II:
Externally \(\frac{m \vec{b}-n \vec{a}}{m-n}\)
Position vectors of P and Q are given as \(\overrightarrow{O P}\) = î + 2 ĵ – k̂ and \(\overrightarrow{O Q}\) = – î + ĵ + k̂

(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
\(\overrightarrow{O R}\) = \(\frac{2(-\hat{i}+\hat{j}+\hat{k})+1(\hat{i}+2 \hat{j}-\hat{k})}{2+1}=\frac{(-2 \hat{i}+2 \hat{j}+2 \hat{k})+(\hat{i}+\hat{j}-\hat{k})}{3}\)

= \(\frac{-\hat{i}+4 \hat{j}+\hat{k}}{3}=-\frac{1}{3} \hat{i}+\frac{4}{3} \hat{j}+\frac{1}{3} \hat{k}\)

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
\(\overrightarrow{O R}\) = \(\frac{2(-\hat{i}+\hat{j}+\hat{k})-1(\hat{i}+2 \hat{j}-\hat{k})}{2-1}\)
= (- 2î + 2ĵ + 2k̂) – (î + 2 ĵ – k̂)
= – 3 î + 3 k̂.

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).
Solution.
The position vector of mid-point R of the vector joining points P(2, 3, 4) and Q(4, 1, – 2) given by
\(\overrightarrow{O R}\) = \(\frac{(2 \hat{i}+3 \hat{j}+4 \hat{k})+(4 \hat{i}+\hat{j}-2 \hat{k})}{2}\)

= \(\frac{(2+4) \hat{i}+(3+1) \hat{j}+(4-2) \hat{k}}{2}\)

= \(\frac{6 \hat{i}+4 \hat{j}+2 \hat{k}}{2}\)

= 3î + 3ĵ + k̂

Question 17.
Show that the points A, B and C with position vectors, a = 3î – 4ĵ – 4k̂, b = 2î – j + k̂ and c = î – 3ĵ – 5k̂, respectively form the vertices of a right angled triangle.
Solution.
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) \(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}=\overrightarrow{0}\)
(B) \(\overrightarrow{\boldsymbol{A B}}+\overrightarrow{\boldsymbol{B C}}-\overrightarrow{\boldsymbol{A C}}=\overrightarrow{\mathbf{0}}\)
(C) \(\overrightarrow{\boldsymbol{A B}}+\overrightarrow{B C}-\overrightarrow{C A}=\overrightarrow{0}\)
(D) \(\overrightarrow{\mathbf{A B}}-\overrightarrow{C B}+\overrightarrow{C A}=\overrightarrow{0}\)
Solution.
On applying the triangle law of addition in the given triangle, we have
\(\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C}\) ………….(i)

⇒ \(\overrightarrow{A B}+\overrightarrow{B C}=-\overrightarrow{C A}\)

⇒ \(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}=\overrightarrow{0}\) …………(ii)

∴ The equation given in alternative A is true.
\(\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C}\)

⇒ \(\overrightarrow{A B}+\overrightarrow{B C}-\overrightarrow{A C}=\overrightarrow{0}\)
∴ The equation given in alternative B is true.
From equation (ii), we have
\(\overrightarrow{A B}-\overrightarrow{C B}+\overrightarrow{C A}=\overrightarrow{0}\)
∴ The equation given in alternative D is true.
Now, consider the equation given in alternative C.
\(\overrightarrow{A B}+\overrightarrow{B C}-\overrightarrow{C A}=\overrightarrow{0}\)

⇒ \(\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{C A}\)………………..(iii)
From equations (i) and (iii), we have AC – CA
⇒ \(\overrightarrow{A C}=\overrightarrow{C A}\)

⇒ \(\overrightarrow{A C}=-\overrightarrow{A C}\)

⇒ \(\overrightarrow{A C}+\overrightarrow{A C}=\overrightarrow{0}\)

⇒ \(2 \overrightarrow{A C}=\overrightarrow{0}\)

⇒ \(\overrightarrow{A C}=\overrightarrow{0}\) which is not true.
Hence, the equation given in alternative C is incorrect.
The correct answer is (C).

Question 19.
If a and B are two collinear vectors, then which of the following are incorrect
(A) \(\overrightarrow{\boldsymbol{b}}\) = X \(\overrightarrow{\boldsymbol{a}}\), for some scalar X
(B) \(\overrightarrow{\boldsymbol{a}}\) = ± \(\overrightarrow{\boldsymbol{b}}\)
(C) The respective components of a and B are proportional
(D) Both the vectors \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\) have same direction, but different magnitudes.
Solution.

Thus, the respective components of \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\) are proportional.
However, vector \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\) can have different directions.
Hence, the statement given in D is incorrect.
The correct answer is (D).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter Differential Equations Ex 9.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6

Direction (1 – 12): For each of the given differential equation, find the general solution.

Question 1.
\(\frac{d y}{d x}\) + 2y = sin x
Solution.
The given differential equation is \(\frac{d y}{d x}\) + 2y = sin x
This is in the form of \(\frac{d y}{d x}\) + Py = Q (where P = 2 and Q = sin x).
Now, I.F = e^{∫ P dx} = e^{∫ 2 dx} = e^2x.
The solution of the given differential equation is given by the relation,
y (I.F.) = ∫ (Q × I.F.) dx + C
⇒ ye^2x = ∫ sin x. e^2x dx + C …………….(i)
Let I = ∫ sin x. e^2x dx
⇒ I = sin x . ∫ e^2x dx – ∫ (\(\frac{d}{d x}\) (sin x) . ∫ e^2x dx) dx

Question 2.
\(\frac{d y}{d x}\) + 3y = e^{– 2x}
Solution.
The given differential equation is
\(\frac{d y}{d x}\) + 3y= e^{– 2x}
This is in the form of \(\frac{d y}{d x}\) + Py = Q (where P = 3 and Q = e^{– 2x})
Now, I.F.= e^{∫ P dx} = e^{∫ 3 dx} = e^3x
The solution of the given differential equation is given by the relation,
y (I.F.) = ∫ (Q × I.F.) dx + C
⇒ y e^3x = ∫ (e^{– 2x} × e^3x) + C
⇒ y e^3x = ∫e^x dx + C
⇒ y e^3x = e^x + C
⇒ ye^3x = e^{– 2x} + Ce^{– 3x}
This is the required general solution of the given differential equation.

Question 3.
\(\frac{d y}{d x}\) + \(\frac{y}{x}\) = x²
Solution.
The given differential equation is
\(\frac{d y}{d x}\) + \(\frac{y}{x}\) = x²
This is in the form of \(\frac{d y}{d x}\) + Py = Q (where P = \(\frac{1}{x}\) and Q = x²)
Now, I.F. = e^{∫ P dx}
= e^{∫ \(\frac{1}{x}\) dx}
= e^{log x} = x
The solution of the given differential equation is given by the relation
y (I.F.) = ∫ (Q × I.F.) dx + C
y(x) = ∫ (x² . x) dx + C
xy = ∫ x³ dx + C
⇒ xy = \(\frac{x^{4}}{4}\) + C
This is the required general solution of the given differential equation.

Question 4.
\(\frac{d y}{d x}\) + (sec x) y = tan x 0 ≤ x < 2
Solution.
The given differential equation is
\(\frac{d y}{d x}\) + (sec x) y = tan x
This is in the form of \(\frac{d y}{d x}\) + Py = Q (where P = sec x and Q = tan x)
Now, I.F = e^{∫ P dx}
= e^{∫ sec x}
= e^{log (sec x + tan x)}
= sec x + tan x
The general solution of the given differential equation is given by the relation
y (I.F.) = ∫ (Q × I.F.) dx + C
⇒ y (sec x + tan x) = ∫ tan x (sec x + tan x) dx + C
⇒ y (sec x + tan x) = ∫ sec x tan x dx + ∫ tan² x dx + C
⇒ y (sec x + tan x) = sec x + ∫ (sec² x – 1) dx + C
⇒ y (sec x + tan x) = sec x + tan x – x + C

Question 5.
cos² x \(\frac{d v}{d x}\) + y = tan x (0 ≤ x < \(\frac{\pi}{2}\))
Solution.
Given, cos² x \(\frac{d y}{d x}\) + y = tan x
On dividing by cos² x on both sides, we get
\(\frac{d y}{d x}+\frac{1}{\cos ^{2} x} y=\frac{\tan x}{\cos ^{2} x}\)
The equation is of the form \(\frac{d y}{d x}\) + Py = Q
(where P = \(\frac{1}{\cos ^{2} x}\) = sec² x, Q = \(\frac{\tan x}{\cos ^{2} x}\))
∴ I.F. = e^{∫ P dx}
= e^{∫ sec2 x}
= e^{tan x}
The general solution of the given differential equation is given by
y (I.F.) = ∫ (Q × I.F.) dx + C
i.e., y × e^{tan x} = ∫ \(\frac{\tan x}{\cos ^{2} x}\) e^{tan x} dx + C
= ∫ tan x . e sec² x + c
Put tan x = t, sec² x dx = dt
∴ ye^{tan x} = ∫ t e^t dt + C
Integrating by parts taking t as a first function.
= ye^{tan x} = te^t – ∫ 1 . e^t dt + C
= te^t – e^t + C
= tan x e^{tan x} – e^{tan x} + C (Putting t = tan x)
or y = tan x – 1 + Ce^{– tan x}

Question 6.
x \(\frac{d y}{d x}\) + 2y = x² log x
Solution.
The given differential equation is x \(\frac{d y}{d x}\) + 2y = x² log x
\(\frac{d y}{d x}\) + \(\frac{2}{dx}\) y = x log x
This equation is in the form of a linear differential equation as
\(\frac{d y}{d x}\) + Py = Q (where P = \(\frac{2}{x}\) and Q = x log X)
Now, I.F. = e^{∫ P dx}
= e^{∫ \(\frac{2}{x}\) dx}
= e^{2 log x}
= e^{log x2}
= x²
The general solution of the given differential equation is given by the relation,
y (I.F.) = ∫ (Q × I.F.) dx + C
y . x² = ∫ (x log x . x²) dx + C
x² y = ∫ (x³ log x) dx + C

Question 7.
x log x \(\frac{d y}{d x}\) + y = log x
Solution.
The given differential equation is x log x \(\frac{d y}{d x}\) + y = \(\frac{2}{x}\) log x

⇒ \(\frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x^{2}}\)
This equation is the form of a linear differential equation as

\(\frac{d y}{d x}\) + Py = Q, (where P = \(\frac{1}{x \log x}\) and Q = \(\frac{2}{x^{2}}\))

Now, I.F. = e^{∫ P dx}
= e^{∫ \(\frac{1}{x \log x}\) dx}
= e^{log (log x)} = log x

The general solution of the given differential equation is given by the relation,
y(I.F.) = ∫ (Q × I.F.) dx + C

Substiting the value of ∫ (\(\frac{2}{x^{2}}\) log x) in equation (i), we get
y log x = – \(\frac{2}{x}\) (1 + log x) + C
This is the required general solution of the given differential equation.

Question 8.
(1 + x²) dy + 2xy dx = cot x dx (x ≠ 0)
Solution.
Given, (1 + x²) dy + 2xy dx = cot x dx

⇒ \(\frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{\cot x}{1+x^{2}}\)

This equation is a linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q (where P = \(\frac{2 x}{1+x^{2}}\) and Q = \(\frac{\cot x}{1+x^{2}}\))

Now, I.F. = e^{∫ P dx}
= e^{\(\frac{2 x}{1+x^{2}}\) dx}
= e^{log (1 + x2)}
= 1 + x²
The general solution of the given differential equation is given by the relation,
y × I.F. = ∫ (Q × l.F.) dx
⇒ y (1 + x²) = ∫ [\(\frac{\cot x}{1+x^{2}}\) (1 + x²)] dx + C
⇒ y (1 + x²) = ∫ cot x dx + C
⇒ y (1 + x²) = log |sin x| + C

Question 9.
x \(\frac{d y}{d x}\) + y – x + xy cot x = 0,(x ≠ 0)
Solution.
Given, x \(\frac{d y}{d x}\) + y – x + xy cot x = 0
⇒ x \(\frac{d y}{d x}\) + y (1 + x cot x) = x
⇒ \(\frac{d y}{d x}\) + (\(\frac{1}{x}\) + cot x) y = 1
This is a linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q
(where P = \(\frac{1}{x}\) + cot x and Q = 1)
Now, I.F. = e^{∫ P dx}
= e^{(\(\frac{1}{x}\) + cot x) dx}
= e^{log x + log (sin x)}
= e^{log (x sin x)}
= x sin x
The general solution of the given differential equation is given by the relation,
y × I.F. = ∫ (Q × I.F.) dx
⇒ y (x sin x) = ∫ (1 × x sin x) dx + C
⇒ y (x sin x) = ∫ (x sin x) dx + C
⇒ y (x sin x) = x ∫ sin x dx – ∫ [\(\frac{d}{d x}\) (x) . ∫ sin x dx] + C
⇒ y (x sin x) = x(- cos x) – ∫ 1 .(- cos x) dx + C
⇒ y (x sin x) = – x cos x + sin x + C

⇒ y = \(\frac{-x \cos x}{x \sin x}+\frac{\sin x}{x \sin x}+\frac{C}{x \sin x}\)

y = – cot x + \(\frac{1}{x}+\frac{C}{x \sin x}\)

Question 10.
(x + y) \(\frac{d y}{d x}\) = 1
Solution.
Given, (x + y) \(\frac{d y}{d x}\) = 1
⇒ \(\frac{d y}{d x}=\frac{1}{x+y}\)

⇒ \(\frac{d y}{d x}\) = x + y

⇒ \(\frac{d y}{d x}\) – x = y
This is a linear differential equation of the form
\(\frac{d x}{d y}\) + P₁x = Q₁ (where P₁ = – 1 and Q₁ = y)
Now, I.F. = e^{∫ P dx}
= e^{∫ – dy}
= e^{– y}
The general solution of the given differentia1 equation is given by the relation,
x (I.F.) = ∫ (Q × I.F.) dy + C
⇒ xe^{– y} = ∫ (y . e^{– y}) dy + C
⇒ xe^{– y} = y ∫ e^{– y} dy – ∫ [\(\frac{d}{d y}\) (y) ∫ e^{– y} dy] dy + C
⇒ xe^{– y} = y(- e^{– y}) – ∫ (- e^{– y}) dy + C
⇒ xe^{– y} = – ye^{– y} + ∫ e^{– y} dy + C
⇒ xe^{– y} = – ye^{– y} – e^{– y} + C
⇒ x = – y – 1 + C e^y
⇒ x + y + 1 = Ce^y

Question 11.
y dx + (x – y²) dy = 0
Solution.
Given y dx + (x – y²) dy = 0
⇒ y dx = (y² – x) dy
⇒ \(\frac{d x}{d y}\) = \(\frac{y^{2}-x}{y}=y-\frac{x}{y}\)

⇒ \(\frac{d x}{d y}+\frac{x}{y}\) = y
This is a linear differential equation of the form
\(\frac{d x}{d y}\) + P₁x = Q₁ (where P₁ = \(\frac{1}{y}\) and Q₁ = y)
Now, I.F. = e^{∫ P₁ x dx}
= e^{∫ \(\frac{1}{y}\) dy}
= e^{log y} = y
The general solution of the given differential equation is given by the relation,
⇒ x × I.F.= ∫ (Q₁ × I.F.) dy + C
⇒ xy = ∫ (y.y) dy + C
⇒ xy = ∫ y² dx + C
⇒ xy = \(\frac{y^{3}}{3}\) + C
⇒ x = \(\frac{y^{2}}{3}+\frac{C}{y}\).

Question 12.
(x + 3y²) \(\frac{d y}{d x}\) = y (y > 0)
Solution.
Given, (x + 3y²) \(\frac{d y}{d x}\) = y

⇒ \(\frac{d y}{d x}=\frac{y}{x+3 y^{2}}\)

⇒ \(\frac{d x}{d y}=\frac{x+3 y^{2}}{y}=\frac{x}{y}+3 y\)

⇒ \(\frac{d x}{d y}-\frac{x}{y}\) = 3y
This is a linear differential equation of the form
\(\frac{d x}{d y}\) + P₁ x = Q₁
(where P₁ = – \(\frac{1}{y}\) and Q₁ = 3y)
Now, I.F. = e^{∫ P₁ dx}
= e^{– ∫ \(\frac{d y}{y}\)}
= e^{– log y}
= e^{log (\(\frac{1}{y}\))}
= \(\frac{1}{y}\)
The general solution of the given differential equation is given by the relation.
x × I.F.= ∫ (Q₁ × I.F.) dy + C
⇒ x × \(\frac{1}{y}\) = ∫ (3y × \(\frac{1}{y}\)) dy + C
⇒ \(\frac{x}{y}\) = 3y + C
⇒ x = 3y² + Cy.

Direction (13 – 15): For each of the differential equation, find a particular solution satisfying the given condition.

Question 13.
\(\frac{d y}{d x}\) + 2y tan x = sin x; y = 0 when x = \(\frac{\pi}{3}\)
Solution.
The given differential equation is \(\frac{d y}{d x}\) + 2y tan x = sin x
This is a linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q (where P = 2 tan x and Q = sin x)
Now, I.F. = e^{∫ P dx}
= e^{∫ 2 tan x dx}
= e^{2 log |sec x|}
= e^{log (sec2 x)}
= sec² x
The general solution of the given differential equation is given by the relation,
x (I.F.) = ∫ (Q × I.F.) dx + C
⇒ y (sec² x) = ∫ (sin x . sec² x) dx + C
⇒ y sec² x = ∫ (sec x . tan x) dx + C
⇒ y sec² x = sec x + C ……………(i)
Now, y = 0 at x = \(\frac{\pi}{3}\)
Therefore,
0 × sec² \(\frac{\pi}{3}\) = sec\(\frac{\pi}{3}\) + C
⇒ 0 = 2 + C
⇒ C = – 2
Substituting C = – 2 in equation (i), we get
y sec² x = sec x – 2
y = cos x – 2 cos² x
Hence, the required solution of the given differential equation is y = cos x – 2 cos² x.

Question 14.
(1 + x²) \(\frac{d y}{d x}\) + 2xy = \(\frac{1}{1+x^{2}}\); y = 0 when x = 1
Solution.
Given, (1 + x²) \(\frac{d y}{d x}\) + 2xy = \(\frac{1}{1+x^{2}}\)

⇒ \(\frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}}\)
This is a linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q (where P = \(\frac{2 x}{1+x^{2}}\) and Q = \(\frac{1}{\left(1+x^{2}\right)^{2}}\))
Now, I.F. = e^{∫ P dx}
= e^{\(\int \frac{2 x d x}{1+x^{2}}\)}
= e^{log (1 + x2)} = 1 + x²
The general solution of the given differential equation is given by the relation,
x (I.F.) = ∫ (Q × I.F.)dx + C
⇒ y(1 + x²) = \(\left[\frac{1}{\left(1+x^{2}\right)^{2}} \cdot\left(1+x^{2}\right)\right]\) dx + C
⇒ y (1 + x²) = ∫ \(\frac{1}{1+x^{2}}\) dx + C
y (1 + x²) = tan^-1 x + C ……………..(i)
Now, y = 0 at x = 1
Therefore, 0 = tan^-1 1 + C
⇒ C = – \(\frac{\pi}{4}\)
Substituting C = – \(\frac{\pi}{4}\) in equation (i), we get
y (1 + x²) = tan^-1 x – \(\frac{\pi}{4}\)
This is the required general solution of the given differential equation.

Question 15.
\(\frac{d y}{d x}\) – 3y cot x = sin 2x; y = 2 when x = \(\frac{\pi}{2}\).
Solution.
The given differential equation is \(\frac{d y}{d x}\) – 3y cot x = sin 2x
This is a linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q (where P = – 3 cot x and Q = sin 2x)
Now, I.F. = e^{∫ P dx}
= e^{– 3 ∫ cot x dx}
= e^{– 3 log |sin x|}
= e^{log \(\left|\frac{1}{\sin ^{3} x}\right|\)}

= \(\frac{1}{\sin ^{3} x}\)

The general solution of the given differential equation is given by the relation,
y(I.F.) = ∫ (Q × I.F.) dx + C
⇒ \(y \cdot \frac{1}{\sin ^{3} x}=\int\left[\sin 2 x \cdot \frac{1}{\sin ^{3} x}\right] d x+C\)

⇒ y cosec³ x = 2 ∫ (cot x cosec x) dx + C
⇒ y cosec³ x = 2 cosec x + c
⇒ y = \(-\frac{2}{\operatorname{cosec}^{2} x}+\frac{3}{\operatorname{cosec}^{3} x}\)
y = – 2 sin² x + C sin³ x ………..(i)
Now, y = 2 at x = \(\frac{\pi}{2}\)
Therefore, we get
2 = – 2 + C
⇒ C = 4
Substituting C = 4 in equation (i), we get
y = – 2 sin² x + 4 sin³ x
y = 4 sin³ x – 2 sin² x
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.
Solution.
Let F(x, y) be the curve passing through the origin.
At point (x, y), the slope of the curve will be \(\frac{d y}{d x}\).
According to the given information,
\(\frac{d y}{d x}\) = x + y
⇒ \(\frac{d y}{d x}\) – y = x
This is a linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q (where P = – 1 and Q = x)
Now, I.F. = e^{∫ P dx}
= e^{∫ (- 1)} dx
= e^{– x}
The general solution of the given differential equation is given by the relation,
y(I.F.) = ∫ (Q × I.F.) dx + C
ye^{– x} = ∫ xe^{– x} dx + C ……………(i)
Now, ∫ xe^{– x} dx = x ∫ e^{– x} dx – ∫ \(\left[\frac{d}{d x}(x) \cdot \int e^{-x} d x\right]\) dx
= – xe^{– x} – ∫ – e^{– x} dx
= – xe^{– x} + (- e^{– x})
= – e^{– x} (x + 1)
Substituting in equation (i), we get
ye^{– x} = – e^{– x} (x + 1) + C
⇒ y = – (x + 1) + Ce^x
⇒ x + y + 1 = Ce^x ………….(ii)
The curve passes through the origin.
Therefore, equation (ii) becomes
1 = C
Substituting C = 1 in equation (ii), we get
⇒ x + y + 1 = e^x
Hence, the required equation of curve passing through the origin is x + y+ 1 = e^x.

Question 17.
Find the equation of a curve passing through the point (0, 2) given that the sum of the coordiDntes of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.
Solution.
According to question, we have x + y = \(\frac{d y}{d x}\) + 5
or \(\frac{d y}{d x}\) + (- 1) y = x – 5
It is a linear differential equation of the form \(\frac{d y}{d x}\) + Py = Q
∴ P = – 1 andQ = x – 5 and
I.F. = e^{∫ P dx}
l.F. = e^{– x}
The general equation of the curve is given by
y. I.F = ∫ Q × I.F dx + C
⇒ y . e^{– x} = ∫ (x – 5) e^{– x} dx + C
⇒ y . e^{– x} = (x – 5) ∫ e^{– x} dx – ∫ [\(\frac{d}{d x}\) (x – 5) . ∫ e^{– x} dx] dx + C
⇒ y . e^{– x} = (x – 5) (- e^{– x}) – ∫ (- e^{– x}) dx + C
⇒ y . e^{– x} = (5 – x) e^{– x} – e^{– x} + C ……………(ii)
The curve passes through the point (0, 2), therefore
2 e^{– 0} = (5 – 0) e^{– 0} – e⁰ + C
⇒ 2 = 5 – 1 + C
⇒ C = 2 – 4 = – 2
On putting the value of C in equation (ii), we get
e^{– x} y = (5 – x) e^{– x} – e^{– x} – 2
⇒ y = 4 – x – 2e^{– x}
which is the required equation of the curve in reference.

Question 18.
The integrating factor of the differential equation x \(\frac{d y}{d x}\) – y = 2x²
(A) e^{– x}
(B) e^{– y}
(C) \(\frac{1}{x}\)
(D) x
Solution.
The given differential equation is x \(\frac{d y}{d x}\) – y = 2x²
⇒ \(\frac{d y}{d x}\) – \(\frac{y}{x}\) = 2x
This is a linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q
(where P = – \(\frac{1}{x}\) and Q = 2x)
The integrating factor (1.F.) is given by the relation e^{∫ P dx}
∴ I.F. = e^{∫ \(\frac{1}{x}\) dx}
= e^{∫ – log x}
= e^{log (x-1)}
= x^-1
= \(\frac{1}{x}\)
Hence, the correct answer is (C).

Question 19.
The integrating factor of the differential equation (1 – y²) \(\frac{d x}{d y}\) + yx = ay (- 1 < < 1) is
(A) \(\frac{1}{y^{2}-1}\)

(B) \(\frac{1}{\sqrt{y^{2}-1}}\)

(C) \(\frac{1}{1-y^{2}}\)

(D) \(\frac{1}{\sqrt{1-y^{2}}}\)
Solution.
The given differential equation is (1 – y²) \(\frac{d x}{d y}\) + yx = ay

⇒ \(\frac{d x}{d y}+\frac{y x}{1-y^{2}}=\frac{a y}{1-y^{2}}\)

This is a linear differential equation of the form
\(\frac{d x}{d y}\) + P₁ x = Q₁
(where P₁ = \(\frac{y}{1-y^{2}}\) and Q₁ = \(\frac{a y}{1-y^{2}}\))
The integrating factor (I.Fj is given by the relation e^{∫ P₁ dx}
∴ I.F.= e^{∫ P₁ dy}

= \(e^{\int \frac{y}{1-y^{2}} d y}=e^{-\frac{1}{2} \log \left(1-y^{2}\right)}=e^{\log \left[\frac{1}{\sqrt{1-y^{2}}}\right]}=\frac{1}{\sqrt{1-y^{2}}}\)
Hence, the correct answer is (D).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 10 Vector Algebra Ex 10.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.1

Question 1.
Represent graphically a displacement of 40 km, 30° east of north.
Solution.
The displacement is 30° east of north so, we have to draw a straight line making 30° with north.

Here, vector \(\overrightarrow{O P}\) 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/sec²
Solution.
(i) 10 kg is a scalar quantity because it involves only magnitude.
(ii) 2 metres 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/sec² 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
(iv) velocity
(v) work done
Solution.
(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, (a square) identify the following vectors.
(i) Coinitial
(ii) Equal
(iii) Collinear but not equal
Solution.

(i) Vectors \(\vec{a}\) and \(\vec{d}\) are coinitial because they have the same initial point.
(ii) Vectors \(\vec{b}\) and \(\vec{d}\) are equal because they have the same magnitude and direction.
(iii) Vectors \(\vec{a}\) and \(\vec{c}\) 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) \(\vec{a}\) and – \(\vec{a}\) are collinear.
Solution.
True

(ii) Two collinear vectors are always equal in magnitude.
Solution.
False

(iii) Two vectors having same magnitude are collinear.
Solution.
False

(iv) Two collinear vectors having the same magnitude are equal.
Solution.
False.

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 9 Differential Equations Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 9 Differential Equations Miscellaneous Exercise

Question 1.
For each of the differential equations given below, indicate its order and degree (if defined).
(i) \(\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}\) – 6y = log x

(ii) \(\left(\frac{d y}{d x}\right)^{3}-4\left(\frac{d y}{d x}\right)^{2}\) + 7y = sin x

(iii) \(\frac{d^{4} y}{d x^{4}}-\sin \left(\frac{d^{3} y}{d x^{3}}\right)\) = 0
Solution.
(i) The given differential equation is \(\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}\) – 6y = log x

\(\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}\) – 6y – log x = 0
The highest order derivative present in the differential equation is \(\frac{d^{2} y}{d x^{2}}\).
Thus, its order is 2.
The highest power raised to \(\frac{d^{2} y}{d x^{2}}\) is one.
Hence, its degree is 1.

(ii) The given differential equation is \(\) + 7y = sin x

\(\left(\frac{d y}{d x}\right)^{3}-4\left(\frac{d y}{d x}\right)^{2}\) + 7y – sin x = 0
The highest order derivative present in the differential equation is \(\frac{d y}{d x}\).
Thus, its order is 1.
The highest power raised to \(\frac{d y}{d x}\) is three.
Hence, its degree is 3.

(iii) The given differential equation is \(\frac{d^{4} y}{d x^{4}}-\sin \left(\frac{d^{3} y}{d x^{3}}\right)\) = o
The highest order denvative present in the differential equation is \(\frac{d^{4} y}{d x^{4}}\).
Thus, its order is 4.
However, the given differential equation is not a polynomial equation.
Hence, its degree is not defined.

Question 2.
For each of the questions given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
(i) y = ae^x + e^-x + x² : \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}\) – xy + x² – 2 = 0

(ii) y = e^x (a cos x + sin x) : \(\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0\)

(iii) y = x sin 3x : \(x \frac{d^{2} y}{d x^{2}}\) + 9y – 6 cos 3x = 0

(iv) x² = 2y² log y : (x² + y²) \(\frac{d y}{d x}\) – xy = 0
Solution.
(i) Given, y = ae^x + be^{– x} + x²
Differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}=a \frac{d}{d x}\left(e^{x}\right)+b \frac{d}{d x}\left(e^{-x}\right)+\frac{d}{d x}\left(x^{2}\right)\)

\(\frac{d y}{d x}\) = ae^x – be^{– x} + x²
Again, differentiating both sides w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = ae^x + be^{– x} + x²
Now, on substituting the values of \(\frac{d y}{d x}\) and \(\frac{d^{2} y}{d x^{2}}\) in the differential equation, we get
L.H.S. = x\(\frac{d^{2} y}{d x^{2}}\) + 2 \(\frac{d y}{d x}\) – xy + x² – 2
= x (ae^x + be^{– x} + 2) + 2 (ae^x – be^{– x} + 2x) – x (ae^x + be^{– x} + x²) + x² – 2

= (axe^x + bxe^{– x} + 2x) + (2ae^x – 2be^{– x} + 4x) – (axe^x + bxe^-x + x³) + x² – 2

= 2ae^x – 2be^-x + x² + 6x – 2 ≠ 0
= L.H.S. ≠ R.H.S.
Hence, the given function is not a solution of the corresponding differential equation.

(ii) Given, y = e^x (a cos x + b sin x)
= a e^x cos x + b e^x sin x
Differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = a . \(\frac{d}{d x}\) (e^x cos x) + b . \(\frac{d}{d x}\) (e^x sin x)

⇒ \(\frac{d y}{d x}\) = a (e^x cos x – e^x sin x) + b . (e^x sin x + e^x cos x)

⇒ \(\frac{d y}{d x}\) = (a + b) e^x cos x + (b – a) e^x sin x
Again, differentiating both sides w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = (a + b) . \(\frac{d}{d x}\) . (e^x cos x) + (b – a) \(\frac{d}{d x}\) (e^x sin x)

\(\frac{d^{2} y}{d x^{2}}\) = (a + b) . [e^x cos x – e^x sin x + (b – a) e^x sin x + e^x cos x]

\(\frac{d^{2} y}{d x^{2}}\) = e^x [(a + b) (cos x – sin x) + (b – a) (sin x + cos x)]

\(\frac{d^{2} y}{d x^{2}}\) = e^x [a cos x – a sin x + b cos x – b sin x + b sin x + b cos x – a sin x – a cos x]

\(\frac{d^{2} y}{d x^{2}}\) = [2e^x (b cos x – a sin x)] d2 dy
Now, on substituting the values of \(\frac{d^{2} y}{d x^{2}}\) and \(\frac{d y}{d x}\) in the L.H.S. of the given differential equation, we get

\(\frac{d^{2} y}{d x^{2}}\) + 2 \(\frac{d y}{d x}\) + 2y = 2 e^x (b cos x – a sin x) – 2 e^x [(a + b) cos x + (b – a) sin x] + 2 e^x (a cos x + b sin x)
= e^x [(2b c0s x – 2a sin x) – (2a cos x + 2b cos x) – 2b sin x – 2a sin x) + (2a cos x + 2b sin x)]
= e^x [(2b – 2a – 2h + 2a) cos x] + e^x [(- 2a – 2b + 2a + 2b) sin x]
=0
Hence, the given function is a solution of the corresponding differential equation.

(iii) Given, y = x sin 3x
Differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (x sin 3x)
= sin 3x + x . cos 3x . 3
= \(\frac{d y}{d x}\) = sin 3x + 3x cos 3x
Again differentiating both sides w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) (sin 3x) + 3 \(\frac{d}{d x}\) (x cos 3x)

\(\frac{d^{2} y}{d x^{2}}\) = 3 cos 3x + 3 [cos 3x + x (- sin3x). 3]

\(\frac{d^{2} y}{d x^{2}}\) = 6 cos 3x – 9 x sin 3x

Substituting the value of \(\frac{d^{2} y}{d x^{2}}\) in the L.H.S. of the given differential equation, we get

\(\frac{d^{2} y}{d x^{2}}\) + 9y – 6 cos 3x = (6 cos 3x – 9 x sin 3x) + 9x sin 3x – 6 cos 3x = 0
Hence, the given function is a solution of the corresponding differential equation.

(iv) Given, x² = 2y² log y
Differentiating both sides w.r.t. x, we get
2x = 2 . \(\frac{d}{d x}\)
= [y² log y]

⇒ x = [2y . log y . \(\frac{d y}{d x}\) + y² . \(\frac{1}{y}\) . \(\frac{d y}{d x}\)]

x = \(\frac{d y}{d x}\) (2y log y + y)

⇒ \(\frac{d y}{d x}\) = \(\frac{x}{y(1+2 \log y)}\)
Substituting the value of \(\frac{d y}{d x}\) in the L.H.S. of the given differential equation, we get

(x² + y²) \(\frac{d y}{d x}\) – xy = (2y² log y + y²) . \(\frac{x}{y(1+2 \log y)}\) – xy

= y² (1 + 2 log y) . \(\frac{x}{y(1+2 \log y)}\) – xy

= xy – xy = 0

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 (x – a)² + 2y² = a², where a is an arbitrary constant.
Solution.
Given family of curves (x – a)² + 2y² = a²
⇒ x² + a² – 2ax + 2y² = a²
⇒ 2y² = 2ax – x²
Differentiating with respect w.r.t. x, we get
2y \(\frac{d y}{d x}\) = \(\frac{2 a-2 x}{2}\)

Question 4.
Prove that x² – y² = c (x² + y²)² is the general solution of differential equation (x³ – 3xy²)dx = (y² – 3x²y) dy, where c is a parameter.
Solution.
The differential equation is (x³ – 3xy²)dx = (y³ – 3x²y) dy
∴ \(\frac{d y}{d x}=\frac{x^{3}-3 x y^{2}}{y^{3}-3 x^{2} y}\) ……………(i)
This is a homogeneous equation.
Let y = vx
Differentiating w.r.t. x, we get

x² – y² = c (x² + y²)²
Hence, x² – y² = c (x² + y²)² is the solution of the differential equation.
(x³ – 3xy²)dx = (y³ – 3x²y) dy where c is a parameter.

Question 5.
Form the differential equation of the family of circles in the first quadrant which touch the coordinates axes.
Solution.
The equation of a circle in the first quadrant with centre (a, a) and radius
(a) which touches the coordinate axis is
(x – a)² + (y – a)² = a²

Differentiating eq (i) w.r.t x, we get
2 (x – a) + 2 (y – a) \(\frac{d y}{d x}\) = 0
⇒ (x – a) + (y – a) y’ = 0
⇒ x – a + yy’ – ay’ = 0
⇒ x + yy’ – a (1 – y’) = 0
⇒ a = \(\frac{x+y y^{\prime}}{1+y^{\prime}}\)
Substituting the value of a in equation (i), we get
\(\left[x-\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right)\right]^{2}+\left[y-\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right)\right]^{2}+\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right)^{2}\)

⇒ \(\left[\frac{(x-y) y^{\prime}}{\left(1+y^{\prime}\right)}\right]^{2}+\left[\frac{y-x}{1+y^{\prime}}\right]^{2}+\left[\frac{x+y y^{\prime}}{1+y^{\prime}}\right]^{2}\)

⇒ (x – y)² . y’² + (x – y) = (x +yy’)

⇒ (x – y)² [1 + (y’)²] = (x + yy’)²
Hence, the required differential equation of the family of circles is (x – y)² [1 + (y’)²] = (x + yy’)^2.

Question 6.
Find the general solution of the differential equation \(\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}\) = 0
Solution.
Given, differential equation is \(\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}\) = 0
⇒ \(\frac{d y}{d x}=-\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}\)

⇒ \(\frac{d y}{\sqrt{1-y^{2}}}=\frac{d x}{\sqrt{1-x^{2}}}\)
Integrating both sides, we get
sin^-1 y = – sin^-1 x + C
⇒ sin^-1 x + sin^-1 y = C.

Question 7.
Show that the general solution of the differential equation \(\frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}\) is given by (x + y + 1) = A (1 – x – y – 2xy) where A is parameter.
Solution.

Hence, the given result is proved.

Question 8.
Find the equation of the curve passing through the point (o, \(\frac{\pi}{4}\)) whose differential equation is sin x cos y dx + cos x sin y dy = 0.
Solution.
The differential equation of the given curve is sin x cos y dx + cos x sin y dy = 0
⇒ \(\frac{\sin x \cos y d x+\cos x \sin y d y}{\cos x \cos y}=0\)

⇒ tan x dx + tan y dy = 0
Integrating both sides, we get
log (sec x) + log (sec y) = log C
log (sec x sec y) = log C
⇒ sec x . sec y = C
The curve passes through point (0, \(\frac{\pi}{4}\))
∴ 1 × √2 = C
⇒ C = √2
On substituting the value of C in equation (i), we get
sec x . sec y = √2
sec x . \(\frac{1}{cos y}\) = √2

⇒ cos y = \(\frac{\sec x}{\sqrt{2}}\)

Hence, the required equation of the curve is cos y = \(\frac{\sec x}{\sqrt{2}}\).

Question 9.
Find the particular solution of the differential equation (1 + e^2x) dy + (1 + y²) e^x dx = 0, given that y = 1 when x = 0.
Solution.
The differential equation is
(1 + e^2x) dy + (1 + y²) e^x dx = 0
On separating the variables, we get

\(\frac{d y}{1+y^{2}}+\frac{e^{x} d x}{1+e^{2 x}}\) = 0

Integrating both sides, we get
\(\int \frac{d y}{1+y^{2}}+\int \frac{e^{x}}{1+e^{2 x}} d x\) = C

Put e^x = t,
e^x dx = dt
∴ tan^-1 y + \(\int \frac{d t}{1+t^{2}}\) = C
tan^-1 y + tan^-1 t = C
i.e., tan^-1 y + tan^-1 e^x = C
Put y = 1, x = 0
∴ tan^-1 1 + tan^-1 1 = C or
2 tan^-1 1 = C
2 × \(\frac{\pi}{4}\) = C
∴ C = \(\frac{\pi}{2}\)
The particular solution of the given differential equation is
tan^-1 y + tan^-1 e^x = \(\frac{\pi}{2}\)

Question 10.
Solve the differential equation \(y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^{2}\right) d y\) (y ≠ 0)
Solution.
Given differential equation is \(y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^{2}\right) d y\)

From equation (î) and equation (ii), we get
\(\frac{d z}{d y}\) = 1
⇒ dz = dy
Integrating both sides, we get
z = y + C
⇒ e^{\(\frac{x}{y}\)} = y + C.

Question 11.
Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = – 1, when x = 0.
(Hint: put x – y = t).
Solution.
Given differential equation is (x – y) (dx + dy) = dx – dy

Integrating both sides, we get
t + |log t| = 2x + C
⇒ (x – y) + log(x – y) = 2x + C
log |x – y| = x + y + C …………….(iii)
Now, y = – 1 at x = 0.
Therefore, equation (iii) becomes
log 1 = 0 – 1 + C
⇒ C = 1
Substituting C = 1 in equation (iii) we get
log |x – y| = x + y + 1
This is the required particular solution of the given differential equation.

Question 12.
Solve the differential equation \(\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{d y}\) = 1 (x ≠ 0)
Solution.

Question 13.
Find a particular solution of the differential equation \(\frac{d y}{d x}\) + y cot x = 4x cosec x (x ≠ 0),
given that y = 0 when x = \(\frac{\pi}{2}\).
Solution.
The given differential equation is \(\frac{d y}{d x}\) + y cot x = 4x cosec x
This equation is a linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q, (where P = cot x and Q = 4x cosec x)
Now, I.F. = e^{∫ P dx}
= e^{∫ cot x dx}
= e^{log |sin x|}
= sin x
The general solution of the given differential equation is given by
y (I.F.) = ∫ (Q × I.F.) dx + C
⇒ y sin x = ∫ (4x cosec x . sin x)dx + C
⇒ y sin x = 4 ∫ x dx + C
⇒ y sin x = 4\(\frac{x^{1+1}}{1+1}\) + C
⇒ y sin x = 4 . \(\frac{x^{2}}{2}\) + C
⇒ y sin x = 2x² + C …………..(i)
Now, y = 0 at x = \(\frac{\pi}{2}\)
Therefore, equation (i) becomes
0 = 2 × \(\frac{\pi^{2}}{4}\) + C

= C = \(\frac{\pi^{2}}{2}\)
Substituting C = – \(\frac{\pi^{2}}{2}\) in equation (i), we get
y sin x = 2x – \(\frac{\pi^{2}}{2}\).
This is the required particular solution of the given differential equation.

Question 14.
Find a particular solution of the differential equation (x + 1) \(\frac{d y}{d x}\) = 2e^{– y} – 1, given that y = 0 when x = 0.
Solution.
Given differential equation is (x + 1) \(\frac{d y}{d x}\) = 2e^{– y} – 1

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?
Solution.
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.
∴ \(\frac{d y}{d t}\) ∝ y
⇒ \(\frac{d y}{d t}\) = ky (k is a constant)
⇒ \(\frac{d y}{y}\) = k dt
Integrating both sides, we get
log y = kt + C …………..(i)
In the year 1999, t = 0 and y = 20000.
Therefore, we get
log 20000 C …………..(ii)
In the year 2004, t = 5 and y = 25000.
Therefore, we get
log 25000 = k.5 + C
⇒ log 25000 = 5k + log 20000
⇒ 5k = \(\log \left(\frac{25000}{20000}\right)=\log \left(\frac{5}{4}\right)\)
⇒ k = \(\frac{1}{5} \log \left(\frac{5}{4}\right)\) …………(i)
In the year 2009, t = 10 years.
Now, on substituting the values oft, k, and C in equation (i), we get
log y = 10 × \(\frac{1}{5} \log \left(\frac{5}{4}\right)\) + log (20000)
⇒ log y = log \(\left[20000 \times\left(\frac{5}{4}\right)^{2}\right]\)
⇒ y = 20000 × \(\frac{5}{4}\) × \(\frac{5}{4}\)
⇒ y = 31250
Hence, the population of the village in 2009 will be 31250.

Question 16.
The general solution of the differential equation \(\frac{y d x-x d y}{y}\) = 0
(A) xy = C
(B) x = Cy²
(C) y = Cx
(D) y = Cx²
Sol.
The given differential equation is \(\frac{y d x-x d y}{y}\) = 0
⇒ \(\frac{y d x-x d y}{y}\) = 0

⇒ \(\frac{1}{x} d x-\frac{1}{y} d x\) = 0
Integrating both sides, we get
log |x| – log |y| = log k

log |\(\frac{x}{y}\)| = log k
⇒ \(\frac{x}{y}\) = k
⇒ y = \(\frac{1}{k}\) x
⇒ y = Cx, where C = \(\frac{1}{k}\)
Hence, the correct answer is (C).

Question 17.
The general solution of a differential equation of the type \(\frac{d y}{d x}\) + P₁x = Q₁ is
(A) \(y e^{\int P_{1} d y}=\int\left(Q_{1} e^{\int P_{1} d y}\right) d y\) + C

(B) \(y \cdot e^{j P_{1} d x}=\int\left(Q_{1} e^{\int P_{1} d x}\right) d x\) + C

(C) \(x e^{\int P_{1} d y}=\int\left(Q_{1} e^{\int P_{1} d y}\right) d y\) + C

(D) \(x e^{\int P_{1} d x}=\int\left(Q_{1} e^{\int P_{1} d x}\right) d y\) + C
Solution.
The integrating factor of the given differential equation \(\frac{d y}{d x}\) + P₁x = Q₁ is e^{∫ P dx}
The general solution of the differential equation is given by
x (I.F.) = ∫ (Q × I.F.) dy + C
x . e^{∫ P dx} = ∫ (Q₁ e^{∫ P dx} ) dy + C
Hence, the correct answer is (C).

Question 18.
The general solution of the differential equation e^x dy + (ye^x +2x) dx = 0 is
(A) xe^y + x² = C
(B) xe^y + y² = C
(C) ye^x + x² = C
(D) ye^y + x² = C
Solution.
The given differential equation is e^x dy + (ye^x +2x) dx = 0
⇒ e^x \(\frac{d y}{d x}\) + y e^x + 2x = 0
⇒ \(\frac{d y}{d x}\) + y = – 2x e^{– x}
This is a linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q, (where P = 1 and Q = – 2xe^{– x})

Now, I.F.= e^{∫ P dx}
= e^{∫ 1 dx}
= e^x
The general solution of the given differential equation is given by
y(I.F.) = ∫ (Q × I.F.)dx + C
⇒ ye^x = ∫ (- 2xe^{– x} . e^x) dx + C
⇒ ye^x = – ∫ 2x dx + C
⇒ ye^x = – x² + C
⇒ ye^x + x² = C
Hence, the correct answer is (C).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 8 Application of Integrals Ex 8.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2

Question 1.
Find the area of the circle 4x² + 4y² = 9 which is interior to the parabola x² = 4y.
Solution.
The required area is represented by the shaded area OBCDO.

Solving the given equation of circle, 4x² + 4 y² = 9 and parabola x² = 4y, we get the point of intersection as B \(\left(\sqrt{2}, \frac{1}{2}\right)\) and D \(-\left(\sqrt{2}, \frac{1}{2}\right)\).

It can be observed that the required area is symmetrical about y-axis.
∴ Area of OBCDO = 2 × Area of OBCO
We draw BM perpendicular to OA.
Therefore, the coordiantes of M are (√2, 0).
Therefore, Area of OBCO = Area of OMBCO – Area of OMBO

Question 2.
Find the area bounded by curves (x – 1)² + y² = 1 and x² + y² = 1.
Solution.

The area bounded by the curves, (x – 1)² + y² = 1 and x² + y² = 1 represented by the shaded area as given in the figure.

On solving these equations, (x – 1)² + y² = 1 and x² + y² = 1, we get the point of intersection as A \(\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\) and B \(\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)\)

It can be observed that the required area is symmetrical about x-axis.

∴ Area of OBCAO = 2 × Area of OCAO
We join AB, which intersects OC at M, such that AM is perpendicular to OC.
The coordinates of M are (\(\frac{1}{2}\), o)
⇒ Area of OCAO = Area of OMAO + Area of MCAM

Question 3.
Find the area of the region bounded by the curves y = x² + 2, y = x, x = 0 and x = 3.
Solution.
The area bounded by the curves, y = x², y – x, x – 0 and x = 3, is represented by the shaded area OCBAO as
Area of OCBAO = Area of ODBAO – Area of ODCO

Area of OCBAO = Area of ODBAO – Area of ODCO
= \(\int_{0}^{3}\) (x² + 2) dx – \(\int_{0}^{3}\) x dx

= \(\left[\frac{x^{3}}{3}+2 x\right]_{0}^{3}-\left[\frac{x^{2}}{2}\right]_{0}^{3}=[9+6]-\left[\frac{9}{2}\right]\)

= \(15-\frac{9}{2}=\frac{21}{2}\) sq. unit.

Question 4.
Using integration, find the area of region bounded by the triangle whose vertices are (- 1, 0), (1, 3) and (3, 2).
Solution.

BL and CM are drawn perpendicular to x-axis.
It can be observed in the given figure that,
Area of ∆ACB = Area of ALBA +Area of BLMCB – Area of AMCA …………(i)
Equation of the line joining points (x₁, y₁) and (x₂, y₂) is
y – y₁ = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) (x – x₁)
Equation of line segment AB is
y – 0 = \(\frac{3-0}{1+1}\) (x + 1)

⇒ y = \(\frac{3}{2}\) (x + 1)

∴ Area of ALBA = \(\int_{-1}^{1} \frac{3}{2}(x+1)\) dx

= \(\frac{3}{2}\left[\frac{x^{2}}{2}+x\right]_{-1}^{1}=\frac{3}{2}\left[\frac{1}{2}+1-\frac{1}{2}+1\right]\)
= 3 unit

Therefore, from equation (i), we get
Area of ∆MBC = (3 + 5 – 4) = 4 sq. unit.

Question 5.
Using integration, find the area of the triangular region whose sides have the equations y = 2x + 1, y = 3x + 1 and x = 4.
Solution.

The equations of sides of the triangle y are y = 2x + 1, y = 3x + 1 and x = 4.
On solving these equations, we get the vertices of triangle as A(0, 1), B(4, 13) and C(4, 9).
It can be observed that,
Area of ∆ACB = Area of OLBAO – Area of OLCAO
= \(\) (3x + 1) dx – \(\) (2x + 1) dx

= \(\left[\frac{3 x^{2}}{2}+x\right]_{0}^{4}\) – \(\left[\frac{2 x^{2}}{2}+x\right]_{0}^{4}\)
=(24 + 4) – (16 + 4)
= 28 – 20 = 8 sq. unit.

Direction (6 – 7): Choose the correct answer:

Question 6.
Smaller area enclosed by the circle x² + y² = 4 and the line x + y = 2 is
(A) 2 (π – 2)
(B) π – 2
(C) 2π – 1
(D) 2 (π + 2)
Solution.
The smaller area enclosed by the circle, x² + y² – 4 and the line, x + y = 2, is represented by the shaded area ACBA as given in the figure.
It can be observed that,

Question 7.
Area lying between the curves y² = 4x and y = 2x is
(A) \(\frac{2}{3}\)

(B) \(\frac{1}{3}\)

(C) \(\frac{1}{4}\)

(D)\(\frac{3}{4}\)
Solution.
The area lying between the curve, y² = 4x and y = 2x, is represented by the shaded area OBAO as given in the figure.
The points of intersection of these, curves are O (0, 0) and A(1, 2).
We draw AC perpendicular to x-axis such that the coordinates of C are (1, 0).
∴ Area of OBAO = Area of ∆OCA – Area of OCABO
= \(\int_{0}^{1}\) 2x dx – \(\int_{0}^{1}\) 2 √x dx

= \(2\left[\frac{x^{2}}{2}\right]_{0}^{1}-2\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1}\)

= |1 – \(\frac{4}{3}\)|

= |- \(\frac{1}{3}\)| = \(\frac{1}{3}\)

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter Differential Equations Ex 9.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.5

Direction (1 – 10) :
In each question show that the given differential equation is homogeneous and solve each equation.

Question 1.
(x² + xy) dy = (x² + y²) dx
Solution.
The given differential equation, (x² + xy) dy = (x² + y²) dx can be written as

Question 2.
y’ = \(\frac{x+y}{x}\)
Solution.
The given differential equation is y’ = \(\frac{x+y}{x}\)
⇒ \(\frac{d y}{d x}=\frac{x+y}{x}\) ……………..(i)

Let F(x, y) = \(\frac{x+y}{x}\)

Now, F(λx, λy)= \(\frac{\lambda x+\lambda y}{\lambda x}=\frac{x+y}{x}\) = λ⁰ . F(x, y)
Thus, the given equation is a homogeneous equation.
Let y = vx
Differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
Substiting the values of y and \(\frac{d y}{d x}\) in equation (i), we get
v + \(x \frac{d v}{d x}=\frac{x+v x}{x}\)
⇒ v + x \(\frac{d v}{d x}\) = 1 + v
x \(\frac{d v}{d x}\) = 1
⇒ dv = \(\frac{d x}{x}\)
Integrating both sides, we get
v = log x + C
⇒ \(\frac{y}{x}\) = log x + C
⇒ y = x log x + Cx
This is the required solution of the given differential equation.

Question 3.
(x – y) dy – (x + y) dx = 0
Solution.
The given differential equation is (x – y) dy – (x + y) dx = 0
⇒ \(\frac{d y}{d x}=\frac{x+y}{x-y}\)

Let F(x, y) = \(\frac{x+y}{x-y}\)

This is the required solution of the given differential equation.

Question 4.
(x² – y²) dx + 2xy dy = 0
Sol.
The given differential equation is (x² – y²) dx + 2xy dy = 0

Therefore, the given differential equation is a homogeneous equation.
Let y = vx
Differentiating both sides w.r.t. x, we get
⇒ \(\frac{d}{d x}\) (y) = \(\frac{d}{d x}\) (vx)

⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)

Substituting the values of y and \(\frac{d y}{d x}\) in equation (i), we get

Question 5.
x² \(\frac{d y}{d x}\) = x² – 2y² + xy
Solution.
The given differential equation is
x² \(\frac{d y}{d x}\) = x² – 2y² + xy ………………(i)

Question 6.
x dy – y dx = \(\sqrt{x^{2}+y^{2}}\) dx
Solution.
Given, x dy – y dx = \(\sqrt{x^{2}+y^{2}}\) dx
x dy = [y + \(\sqrt{x^{2}+y^{2}}\)] dx

Question 7.
\(\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y d x=\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x d y\)
Solution.
The given differential equation is

Therefore, the given differential equation is a homogeneous equation.
Let y = vx
Differentiating both sides w.r.t. x, we get

This is the required solution of the given differential equation.

Question 8.
x \(\frac{d y}{d x}\) – y + x sin \(\left(\frac{y}{x}\right)\) = 0
Solution.
Given, x \(\frac{d y}{d x}\) – y + x sin \(\left(\frac{y}{x}\right)\) = 0

Question 9.
y dx + x log \(\left(\frac{y}{x}\right)\) dy – 2x dy = 0
Solution.
Given, y dx + x log \(\left(\frac{y}{x}\right)\) dy – 2x dy = 0

Question 10.
\(\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y\) = 0
Solution.
Given, \(\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y\) = 0

Therefore, the given differential equation is a homogeneous equation.
Let x = vy
Differentiating both sides w.r.t. x, we get
⇒ \(\frac{d}{d y}\) (x) = \(\frac{d}{d y}\) (vy)

⇒ \(\frac{d x}{d y}\) = v + y \(\frac{d v}{d y}\)
Substituting the values of x and latex]\frac{d x}{d y}[/latex] in equation (i), we get

Integrating both sides w.r.t. x, we get

⇒ log (v + e^v) = – log y + log C = log \(\left(\frac{C}{y}\right)\)

⇒ \(\left[\frac{x}{y}+e^{\frac{x}{y}}\right]=\frac{C}{y}\)

⇒ x + y \(e^{\frac{x}{y}}\) = C
This is the required solution of the given differential equation.

Direction (11 – 15):
For each of the differential equation, find the particular solution satisfying the given condition.

Question 11.
(x + y) dy + (x – y) dy = 0; y = 1 when x = 1
Solution.
Given, (x + y) dy + (x – y)dx = O

Integrating both sides, we get
\(\frac{1}{2}\) log(1 + v²) + tan^{– 1} u = – log x + k
⇒ log (1 + v²) + 2 tan^{– 1} y = – 2 log x + 2k
⇒ log [(1 + v²) . x²] + 2 tan^{– 1} v = 2k
⇒ log [(1 + \(\frac{y^{2}}{x^{2}}\)) . x²] + 2 tan^{– 1} \(\frac{y}{x}\) = 2k
⇒ log (x² + y²) + 2 tan^{– 1} \(\frac{y}{x}\) = 2k
Now, y = 1 at x = 1
⇒ log 2 + 2 tan^{– 1}1 = 2k
⇒ log 2 + 2 × \(\frac{\pi}{4}\) = 2k
⇒ \(\frac{\pi}{2}\) + log 2 = 2k
Substituting the value of 2k in equation (ii), we get
log (x² + y²) + 2 tan^{– 1} (\(\frac{y}{x}\)) + log 2
This is the required solution of the given differential equation.

Question 12.
x² dy + (xy + y²)dx = 0; y = 1 when x = 1
Solution.
Given, x² dy + (xy + y²)dx = 0

Substituting C² = \(\frac{1}{3}\) in equation (ii), we get

\(\frac{x^{2} y}{y+2 x}=\frac{1}{3}\)

⇒ y + 2x = 3x²y
This is the required solution of the given differential equation.

Question 13.
\(\left[x \sin ^{2}\left(\frac{y}{x}-y\right)\right]\) dx + x dy = 0; y = \(\frac{\pi}{4}\) when x = 1.
Solution.
Given, \(\left[x \sin ^{2}\left(\frac{y}{x}-y\right)\right]\) dx + x dy = 0

When x = 1, then y = \(\frac{\pi}{4}\), therefore
log (1) – cot \(\frac{\pi}{4}\) = C
⇒ 0 – 1 = C
⇒ C = – 1
Putting the value of C in equation (i), we get
log x – cot \(\frac{y}{x}\) = – 1
log |x| – cot \(\frac{y}{x}\) = – log e
cot (\(\frac{y}{x}\)) = log |ex|
This is the required solution of the given differential euqation.

Question 14.
\(\frac{d y}{d x}\) – \(\frac{y}{x}\) + cosec (\(\frac{y}{x}\)) = 0; y = 0 when x = 1
Solution.
Given, \(\frac{d y}{d x}\) – \(\frac{y}{x}\) + cosec (\(\frac{y}{x}\)) = 0 …………..(i)
Which is homogeneous
Let y = vx
Differentiating both sides w.r.t. x,we get

When x = 1 then y = 0 therefore
log |1| – cos 0 = C
⇒ o – 1 = C,
⇒ c = – 1
Putting the value of C in equation (iii), we get
⇒ log |x| – cos (\(\frac{y}{x}\)) = – 1
⇒ log |x| + 1 = cos \(\frac{y}{x}\)
⇒ log |x| + log e = cos (\(\frac{y}{x}\))
⇒ log |ex| = cos (\(\frac{y}{x}\))
This is the required solution of the given differential equation.

Question 15.
2xy + y² – 2x² \(\frac{d y}{d x}\) = 0; y = 2 when x = 1.
Solution.
Given, 2xy + y² – 2x² \(\frac{d y}{d x}\) = 0

When x = 1 then y = 2
⇒ – 1 = log (1) + C
⇒ C = – 1
Substituting C = – 1 in equation (ii), we get
– \(\frac{2 x}{y}\) = log |x| – 1

\(\frac{2 x}{y}\) = 1 – log |x|

y = \(\frac{2 x}{1-\log |x|}\), (x ≠ 0, x ≠ e)
This is the required solution of the given differential equation.

Questio 16.
A homogeneous differential equation of the form \(\frac{d y}{d x}\) = h (\(\frac{x}{y}\)) can be solved by making the substitution
(A) y = vx
(B) v = yx
(C) x = vy
(D) x = v
Solution.
For solving the homogeneous equation of the form \(\frac{d y}{d x}\) = h (\(\frac{x}{y}\)), we need to make the substitution as x = vy.
Hence, the correct answer is (C).

Question 17.
Which of the following is a homogeneous differential equation?
(A) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0
(B) (xy) dx – (x³ + y³ )dy = 0
(C) (x² + 2y²) dx + 2xy dy = 0
(D) y² dx + (x² – xy – y²) dy = 0
Solution.
Function F(x, y) is said to be the homogeneous function of degree n, if
F(λx, λy) = λⁿ F(x, y) for any non-zero constant (λ).
Consider the equation given in alternative D.
y² dx + (x² – xy – y²) dy = 0

Hence, the differential equation given in alternative D is a homogenous equation.

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 7 Integrals Ex 7.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6

Direction (1 – 22): Integrate the functions:

Question 1.
x sin x
Solution.
Let I = ∫ x sin x dx
Taking x as first function and sin x as second function and integrating by parts, we get
I = x ∫ sin x dx – ∫ {(\(\frac{d}{d x}\) (x)) ∫ sin x dx} dx
= x (- cos x) – ∫ 1 . (- cos x) dx
= – x cos x + sin x + C

Question 2.
x sin 3x
Solution.
Let I = ∫ x sin 3x dx
Taking x as first function and sin3x as second function and integrating by parts, we get

Question 3.
x² e^x
Solution.
Let I = ∫ x² e^x dx
Taking x² as first function and e^x as second function and integrating by parts,we get
I = x² ∫ e^x dx – ∫ {(\(\frac{d}{d x}\) x²) ∫ e^x dx} dx
= x² e^x – ∫ 2x . e^x dx
= x² e^x – 2 ∫ x . e^x dx
Again integrating by parts, we get
= x² e^x – 2 [x . ∫ e^x dx – ∫ {(\(\frac{d}{d x}\) x) ∫ e^x dx} dx]
= x² e^x – 2 [x e^x – e^x dx]
= x² e^x – 2 [x e^x – e^x]
= x² e^x – 2x e^x + 2 e^x + C
= e^x (x² – 2x + 2) + C

Question 4.
x log x
Solution.
Let I = ∫ x log x dx
Taking log x as first function and x as second function and integrating by parts, we get

Question 5.
x log 2x
Solution.
Let I = ∫ x log 2x dx
Taking log 2x as first function and x as second function and integrating by parts, we get

Question 6.
x² log x
Solution.
Let I = ∫ x² log x dx
Taking log x as first function and x² as second function and integrating by parts, we get

Question 7.
x sin^-1 x
Solution.
Let I = ∫ x sin^-1 x dx
Taking sin^-1 x as first function and x as second function and integrating by parts, we get

Question 8.
x tan^-1 x
Solution.
Let I = ∫ x tan^-1 x dx
Taking tan^-1 x as first function and x as second function and integrating by parts, we get

Question 9.
x cos^-1 x
Solution.
Let I = ∫ x cos^-1 x dx
Taking cos^-1 x as first function and x as second function and integrating by parts, we get

Question 10.
(sin ^-1 x)²
Solution.
Let I = ∫ (sin ^-1 x)² dx
Taking (sin ^-1 x)² as first function and x as second function and integrating by parts, we get

Question 11.
\(\frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}}\)
Solution.
Let I = ∫ \(\frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}}\) dx

= \(\frac{-1}{2} \int \frac{-2 x}{\sqrt{1-x^{2}}} \cdot \cos ^{-1} x\) dx

Taking cos x as first function and \(\left(\frac{-2 x}{\sqrt{1-x^{2}}}\right)\) as second function and integrating by parts, we get

Question 12.
x sec² x
Solution.
Let I = ∫ x sec² x dx
Taking x as first function and sec² x as second function and integrating by parts, we get
I = x ∫ sec² x dx – ∫ [(\(\frac{d}{d x}\) (x)) ∫ sec² x dx] dx
= x tan x – ∫ 1 . tan x dx
= x tan x + log |cos x| + C

Question 13.
tan^-1 x
Solution.
Let I = ∫ 1 . tan^-1 x dx
Taking tan^-1 x as first function and 1 as second function and integrating by parts, we get

Question 14.
x (log x)²
Solution.
Let I = ∫ x (log x)² dx
Taking (log x)² as first function and x as second function and integrating by parts, we get

Question 15.
(x² + 1) log x
Solution.
Let I = ∫ (x² + 1) log x dx
= ∫ (log x) . (x² + 1) dx
Taking (log x) as first function and (x² + 1) as second function and integrating by parts, we get

Question 16.
e^x (sin x + cos x)
Solution.
Let I = ∫ e^x (sin x + cos x) dx
Let f(x) = sin x
⇒ f’(x) = cos x
∴ I = ∫ ex {f(x) + f'(x)}dx
We know that, ∫ex {f(x) + f’(x)}dx = e^x f(x) + C
∴ I = e^x sin x + C

Question 17.
\(\frac{x e^{x}}{(1+x)^{2}}\)
Solution.

Question 18.
e^x \(\left(\frac{1+\sin x}{1+\cos x}\right)\)
Solution.

Question 19.
e^x \(\left(\frac{1}{x}-\frac{1}{x^{2}}\right)\)
Solution.
Let I = ∫ e^x \(\left(\frac{1}{x}-\frac{1}{x^{2}}\right)\) dx
Put \(\frac{e^{x}}{x}\) = t; i.e., \(\frac{1}{x}\) = t, so that,

\(\left[e^{x} \cdot\left(-\frac{1}{x^{2}}\right)+\frac{1}{x} \cdot e^{x}\right]\) dx = dt

I = ∫ dt = t + C = \(\frac{e^{x}}{x}\) + C

Question 20.
\(\frac{(x-3) e^{x}}{(x-1)^{3}}\)
Solution.

Question 21.
e^2x sin x
Solution.
Let I = ∫ e^2x sin x dx …………….(i)
Integrating by parts, we get

Question 22.
sin^-1 \(\left(\frac{2 x}{1+x^{2}}\right)\)
Solution.
Let x = tan θ so that dx = sec² θ dθ
∴ sin^-1 \(\left(\frac{2 x}{1+x^{2}}\right)\) = sin^-1 \(\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)\)
= sin^-1 (sin 2θ) = 2θ

∴ ∫ sin^-1 \(\left(\frac{2 x}{1+x^{2}}\right)\) dx = ∫ 2θ . secsup>2 θ dθ
= 2 ∫ θ . sec² θ dθ
Integrating by parts, we get
\(2\left[\theta \cdot \int \sec ^{2} \theta d \theta-\int\left\{\left(\frac{d}{d \theta} \theta\right) \int \sec ^{2} \theta d \theta\right\} d \theta\right]\) = 2 [θ . tan θ – ∫ tan θ dθ]

= 2 [θ tan θ + log |cos θ|] + C

= 2 [x tan^-1 x + log \(\left|\frac{1}{\sqrt{1+x^{2}}}\right|\)] + C
= 2x tan^-1 x + 2 log (1 + x²)^{–\(\frac{1}{2}\)} + C
= 2x tan^-1 x + 2 [- \(\frac{1}{2}\) log (1 + x²)] + C
= 2x tan^-1 x – log (1 + x²) + C

Direction (23 – 24): Choose the correct answer in the given question.

Question 23.
∫ x² e^x3 dx equals
(A) \(\frac{1}{3}\) e^x3 + C
(B) \(\frac{1}{3}\) e^x2 + C
(C) \(\frac{1}{2}\) e^x3 + C
(D) \(\frac{1}{2}\) e^x2 + C
Sol.
Let I = ∫ x² e^x3 dx
Also, let x³ = t
⇒ 3x² dx = dt
⇒ I = \(\frac{1}{3}\) ∫ e^t dt
= \(\frac{1}{3}\) (e^t) + C
= \(\frac{1}{3}\) e^x3 + C
Hence, the correct answer is (A).

Question 24.
∫ e^x sec x (1 + tan x) dx equals
(A) e^x cos x + C
(B) e^x sec x + C
(C) e^x sin x + C
(D) e^x tan x + C
Sol.
∫ e^x secx(1 + tan x) dx
Let I = ∫ e^x sec x (1 + tan x) dx
= ∫ e^x (sec x + sec x tan x) dx
Also, let sec x = f(x). sec x tan x = f’(x)
We know that, ∫ e^x {f(x) + f’(x)}dr = e^x f(x) + C
∴ I = e^x sec x + C
Hence, the correct answer is (B). Textbook Exercise Questions and Answers.

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter Differential Equations Ex 9.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.4

Direction (1 -1 0): For each of the differential equations in given questions, find the general solution.

Question 1.
\(\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}\)
Solution.

⇒ y = 2 tan \(\frac{x}{2}\) – x + C
This is the required general solution of the given differential equation.

Question 2.
\(\frac{d y}{d x}=\sqrt{4-y^{2}}\) (- 2 < y < 2)
Solution.
Given, \(\frac{d y}{d x}=\sqrt{4-y^{2}}\)
Separating the variables, we get
⇒ \(\frac{d y}{\sqrt{4-y^{2}}}\) = dx
⇒ sin^-1 \(\frac{y}{2}\) = x + C
Now, integrating both sides of this equation, we get
∫ \(\frac{d y}{\sqrt{4-y^{2}}}\) = ∫ dx
\(\frac{y}{2}\) = sin(x + C)
⇒ y = 2 sin (x + C)
This is the required general solution of the given differential equation.

Question 3.
\(\frac{d y}{d x}\) + y = 1 (y ≠ 1)
Solution.
Given, \(\frac{d y}{d x}\) + y = 1
⇒ dy + y dx = dx
⇒ dy = (1 – y) dx
Separating the variables, we get
⇒ \(\frac{d y}{1-y}\) = dx
Now, integrating both sides, we get
∫ \(\frac{d y}{1-y}\) = ∫ dx
⇒ log(l – y) = x + log C
⇒ – log C – log (1 – y) = x
⇒ log C(1 – y) = – x
⇒ C (1 – y) = e^-x
⇒ 1 – y = \(\frac{1}{C}\) e^-x
⇒ y = 1 – \(\frac{1}{C}\) e^-x
⇒ y = 1 + A e^-x (where A = Question )
This is the required general solution of the given differential equation.

Question 4.
sec² x tan y dx + sec² y tan x dy = 0
Solution.
Given, sec² x tan y dx + sec² y tan x dy = 0

⇒ \(\frac{\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y}{\tan x \tan y}\) = 0

⇒ \(\frac{\sec ^{2} x}{\tan x} d x+\frac{\sec ^{2} y}{\tan y} d y\) = 0

Integrating bothsides, we get

⇒ \(\int \frac{\sec ^{2} x}{\tan x} d x+\int \frac{\sec ^{2} y}{\tan y} d y\) = 0
⇒ log | tan x | + log | tan y | = log C
⇒ log | tan x tan y | = log C
⇒ tan x tan y = C
Which is the required solution, where x ≠ odd multiple of \(\frac{\pi}{2}\) and x ∈ R.

Question 5.
(e^x + e^{– x}) dy – (e^x – e^{– x}) dx = 0
Solution.
Given, (e^x + e^{– x}) dy – (e^x – e^{– x}) dx = 0
⇒ (e^x + e^{– x}) dy = (e^x – e^{– x}) dx
⇒ dy = \(\left[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right]\) dx
Integrating both sides, we get
∫ dy = ∫ \(\left[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right]\) dx + C

⇒ y = ∫ \(\left[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right]\) dx + C …………….(i)

Let (e^x + e^{– x}) = t
Differentiating both sides w.r.t. x, we get
\(\frac{d}{d x}\) (e^x + e^{– x}) = \(\frac{d t}{d x}\)
⇒ e^x – e^{– x} = \(\frac{d t}{d x}\)
⇒ (e^x – e^{– x}) dx = dt
Substituting this value in equation (i), we get
y = ∫ \(\frac{1}{t}\) dt + C
⇒ y = log (t) + C
⇒ y = log (e^x + e^{– x}) + C
This is the required general solution of the given differential equation.

Question 6.
\(\frac{d y}{d x}\) = (1 + x²) (1 + y²)
Solution.
Given, \(\frac{d y}{d x}\) = (1 + x²) (1 + y²)
⇒ \(\frac{d y}{1+y^{2}}\) = (1 + x²) dx
Integrating both sides, we get
∫ \(\frac{d y}{1+y^{2}}\) = ∫ (1 + x²) dx
⇒ tan^-1 y = ∫ dx + ∫ x² dx
⇒ tan^-1 y = x + \(\frac{x^{3}}{3}\) + C
This is the required general solution of the given differential equation.

Question 7.
y log y dx – x dy = 0
Solution.
Given, y log y dx – x dy = 0
⇒ y log y dx = x dy
⇒ \(\frac{d y}{y \log y}=\frac{d x}{x}\)
Integrating both sides, we get
⇒ \(\int \frac{d y}{y \log y}=\int \frac{d x}{x}\)
Let log y = t
Differentiating w.r.t. y, we get
⇒ \(\frac{d}{d y}\) (log y) = \(\frac{d t}{d y}\)

⇒ \(\frac{1}{y}=\frac{d t}{d y}\)

⇒ \(\frac{1}{y}\) dy = dt
Substituting this value in equation (i), we get
\(\int \frac{d t}{t}=\int \frac{d x}{x}\)

⇒ log t = log x + log C
⇒ log (log y) = log Cx
⇒ log y = Cx
⇒ y = e^Cx

Question 8.
x⁵ \(\frac{d y}{d x}\) = – y⁵
Solution.

⇒ x^{– 4} + y^{– 4} = – 4 k
⇒ x^{– 4} + y^{– 4} = C (C = – 4k)
This is the required general solution of the given differential equation.

Question 9.
\(\frac{d y}{d x}\) = sin^{– 1} x
Solution.
Given, \(\frac{d y}{d x}\) = sin^{– 1} x
⇒ dy = sin^{– 1} x dx
Integrating both sides, we get
∫ dy = ∫ sin^{– 1} x dx
⇒ y = ∫ (sin^{– 1} x . 1) dx
⇒ y = sin^{– 1} x . ∫ (1) dx – ∫ \(\left[\left(\frac{d}{d x}\left(\sin ^{-1} x\right) \cdot \int(1) d x\right)\right]\) dx

⇒ y = sin^{– 1} x . x – ∫ \(\left(\frac{1}{\sqrt{1-x^{2}}} \cdot x\right)\) dx

⇒ y = x sin^{– 1} x + ∫ \(\frac{-x}{\sqrt{1-x^{2}}}\) dx
Let 1 – x² = t
Differentiating w.r.t x, we get
⇒ \(\frac{d}{d x}\) (1 – x²) = \(\frac{d t}{d x}\)
⇒ – 2x = \(\frac{d t}{d x}\)
⇒ x dx = – \(\frac{1}{2}\) dt
Substituting this value in equation (i), we get
y = x sin^-1 x + ∫ \(\frac{1}{2 \sqrt{t}}\) dt

⇒ y = x sin^-1 x + \(\frac{1}{2} \cdot \int(t)^{-\frac{1}{2}}\) dt

⇒ y = x sin^-1 x + \(\frac{1}{2} \cdot \frac{t^{\frac{1}{2}}}{\frac{1}{2}}\) + C

⇒ y = x sin^-1 x + √t + C

⇒ y = x sin^-1 x + √1 – x² + C

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

Question 10.
e^x tan y dx + (1 – e^x) sec² y dy = 0
Solution.
Given, e^x tan y dx + (1 – e^x) sec² y dy = 0
(1 – e^x)sec² y dy = – e^x tan y dx
Separating the variables, we get
\(\frac{\sec ^{2} y}{\tan y} d y=\frac{-e^{x}}{1-e^{x}} d x\) ………………(i)
Integrating both sides, we get
\(\int \frac{\sec ^{2} y}{\tan y} d y=\int \frac{-e^{x}}{1-e^{x}} d x\)
Let tan y = u
Differentiating w.r.t. y, we get

Substituting the value of \(\int \frac{\sec ^{2} y}{\tan y}\) dy and \(\int \frac{-e^{x}}{1-e^{x}}\) dx in equation (i), we get
⇒ log (tan y) = log(1 – e^x) + logC
⇒ log (tan y) = log [C(1 – e^x)]
⇒ tan y = C (1 – e^x)
This is the required general solution of the given differential equation.

Direction (11 – 14): For each of the differential equation in given questions, find a particular solution satisfying the given condition.
Question 11.
(x³ + x² + x + 1)\(\frac{d y}{d x}\) = 2x² + x y = 1 when x = 0.
Solution.

Question 12.
x (x² – 1) \(\frac{d y}{d x}\) = 1; y = 0 when x = 2
Solution.

Question 13.
cos (\(\frac{d y}{d x}\)) = a, (a ∈ R); y = 1 when x = 0
Solution.
Given, cos (\(\frac{d y}{d x}\)) = a
⇒ \(\frac{d y}{d x}\) = cos^-1 a
⇒ dy = cos^-1 a dx
Integrating both sides, we get
∫ dy = ∫ cos^-1 a dx
⇒ y = cos^-1 a . x + C
⇒ y = x cos^-1 a + C
Now, y = 1 when x = 0
⇒ 1 = 0 . cos^-1 a + C
⇒ C = 1
Substituting C = 1 in equation (i), we get
y = x cos^-1 a + C
⇒ \(\frac{y-1}{x}\) = cos^-1 a
⇒ cos (\(\frac{y-1}{x}\)) = a.

Question 14.
\(\frac{d y}{d x}\) = y tan x; y = 1 when x = 0
Solution.
Given, \(\frac{d y}{d x}\) = y tan x
⇒ \(\frac{d y}{y}\) = tan x dx
Integrating both sides, we get
∫ \(\frac{d y}{y}\) = ∫ tan x dx
⇒ log y = log (sec x) + log C
⇒ log y = log (C sec x)
⇒ y = C sec x
Now, y = 1 when x = 0
⇒ 1 + C × sec 0
⇒ 1 = C × 1
Substituting C = 1 in equation (j), we get
y = sec x

Question 15.
Find the equation of curve passing through the point (0, 0) and whose differential equation is y’ = e^x sin x.
Solution.
Differential equation is
y’ = e^x sin x or \(\frac{d y}{d x}\) = e^x sin x
∴ dy = e^x sin x dx
Integrating both sides, we get
∫ dy = ∫ e^x sin x dx
Integrating by parts taking e^x as the first function
y= e^x (- cos x) – ∫ e^x (- cos x) dx
= – e^x cos x + ∫ e^x cos x dx
Again integrating by parts taking e^x as first function
= – e^x cosx + e^x sin x – ∫ e^x sin x dx
or 2y = – e^x cos x + e^x sec + C
y = \(\frac{e^{x}}{2}\) (- cos x + sin x) + C ………….(i)
Put x = 0, y = 0
On substituting C = \(\frac{1}{2}\) in equation (i), we get
y = \(\frac{e^{x}}{2}\) (sin x – cos x) + \(\frac{1}{2}\)
⇒ 2y = e^x (sin x – cos x) + 1
⇒ 2y – 1 = e^x (sin x – cos x)
Hence, the required equation of the curve is 2y – 1 = e^x (sin x – cos x).

Question 16.
For the differential equation xy \(\frac{d y}{d x}\) = (x + 2)(y + 2), find the solution curve passing through the point (1, – 1).
Solution.
The differential equation is xy \(\frac{d y}{d x}\) = (x + 2) (y +2)
or xy dy = (x + 2) (y + 2) dx
Dividing by x (y +2)
\(\frac{y}{y+2} \frac{d y}{d x}=\frac{x+2}{x} d x\)
Integrating both sides, we get
\(\int y \frac{d y}{y+2}=\int \frac{x+2}{x} d x\)

\(\int \frac{y+2-2}{x+2} d y=\int\left(1+\frac{2}{x}\right) d x\) \(\int\left(1-\frac{2}{y+2}\right) d y=\int\left(1+\frac{2}{x}\right) d x\)

y – 2 log (y + 2) = x + 2 log x + C
y – x – C = log [x² (y + 2)²]
The curve passes through (1, -1)
∴ – 1 – 2 log 1 = 1 + 2 log 1 + C [∵ log 1 = 0]
– 1 = 1 + C
∴ C = – 2
Substituting C = – 2 in equation (i), we get
y – x + 2 = log [x² (y + 2)²]
This is the required solution of the given curve.

Question 17.
Find the equation of a curve passing through the point (0, -2) given that at any point (x, y) on the curve, the product of the slope of its tangents and y-coordinate of the point is equal to the x-coordinate of the point.
Sol.
Let x and y be the x-coordinate and y-coordinate of the curve respectively.
We know that the slope of a tangent to the curve in the coordinate axis is given by the \(\frac{d y}{d x}\).
According to the given information, we get
y . \(\frac{d y}{d x}\) = x dx
y dy = x dx
Integratin both sides, we get
∫ y dy = ∫ x dx
⇒ \(\frac{y^{2}}{2}=\frac{x^{2}}{2}\) + C
⇒ y² – x² = 2C …………..(i)
Now, the curve passes through the point (0, – 2).
∴ (- 2)² – 0² = 2C
⇒ 2C = 4
Substituting 2C = 4 in equation (i), we get
y² – x² = 4
This is the required equation of the curve.

Question 18.
At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (- 4, – 3). Find the equation of the curve given that it passes through (- 2, 1).
Solution.
It is given that (x, y) is the point of contact of the curve and its tangent.
The slope (m₁) of the line segment joining (x, y) and (- 4, – 3) is \(\frac{y+3}{x+4}\).
We know that the slope of the tangent to the curve is given by the relation, \(\frac{d y}{d x}\)
∴ slope (m₂) of the tangent = \(\frac{d y}{d x}\)
According to the question,
m₂ = 2m
⇒ \(\frac{d y}{d x}=\frac{2(y+3)}{x+4}\)

⇒ \(\frac{d y}{y+3}=\frac{2 d x}{x+4}\)

Integrating both sides, we get

\(\int \frac{d y}{y+3}=2 \int \frac{d y}{x+4}\)
⇒ log (y + 3) = 2log (x + 4) + log C
⇒ log (y + 3) = log C (x + 4)²
⇒ y + 3 = C (x + 4)² ………………(i)
This is the general equation of the curve.
It is given that it passes through point (- 2, 1)
⇒ 1 + 3 = C(- 2 + 4)²
⇒ 4 = 4C
⇒ C = 1
Substituting C = 1 in equation (i), we get
y + 3 = C(x + 4)
This is the required equation of the curve.

Question 19.
The volume of spherical balloon being inflated changes at a constant rate, If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.
Solution.
Let the rate of change of the volume of the balloon be k (where k is a constant).
⇒ \(\frac{d V}{d t}\) = k

⇒ \(\frac{d}{d t}\left(\frac{4}{3} \pi r^{3}\right)\) = k [Volume of sphere = \(\frac{4}{3}\) πr³]

⇒ \(\frac{4}{3}\) π . 3r² . \(\frac{d r}{d t}\) = k

⇒ 4πr² dr = k dt
Integrating both sides, we get
4π ∫ r² dr = k ∫ dt
⇒ 4π . \(\frac{r^{3}}{3}\) = kt + C
⇒ 4πr³ = 3 (kt + C) ……………(i)
Now, at t = 0, r = 3
⇒ 4π × 3³ = 3 (k × 0 + C)
⇒ 108 π = 3C
⇒ C = 36π
At t = 3, r = 6;
⇒ 4π × 6³ = 3(k × 3 + C)
⇒ 864π = 3 (3k + 36π)
⇒ 3k = – 288π – 36π = 252π
k = 84π
Substituting the values of k and C in equation (i), we get
4πr³ = 3 [84π + 36π]
⇒ 4πr³ = 4π (63t + 27)
⇒ r³ = (63t + 27)
⇒ r = (63π + 27)\(\frac{1}{3}\)
Thus, the radius of the balloon after t seconds is (63t + 27)\(\frac{1}{3}\).

Question 20.
In a bank, principal increases continuously at the rate of r% per year. Find the value of r if loo double itself in 10 years
Sol.
Let p, t and r represent the principal, time and rate of interest respectively.
It is given that the principal increases continuously at the rate r% per year.

It is given that when t = 0, p = 100
⇒ 100 = e^k ………….(ii)
Now, if t = 10, then p = 2 × 100 = 200
Therefore, equation (i) becomes
200 = \(e^{\frac{r}{10}+k}\)
200 = \(e^{\frac{r}{10}}\) . e^k
⇒ 200 = \(e^{\frac{r}{10}}\) . 100 [From equation (ii)]
⇒ \(e^{\frac{r}{10}}\) = 2
⇒ \(\frac{r}{10}\) = log_e 2
⇒ \(\frac{r}{10}\) = 0.6931
⇒ r = 6.931
Hence, the value of r is 6.93%.

Question 21.
In a bank, principal increases continuously at the rate of 5% per year. An amount oR 1000 is deposited with this bank, how much will it worth after 10 years (e^0.5 = 1.648).
Solution.
Let p and t be the principal and time respectively.
It is given that the principal increases continuously at the rate of 5% per year.

Now, if t = 0 then p = 1000
⇒ 1000 = e^c ………….(ii)
At t = 10, equation (i) becomes
⇒ p = \(e^{\frac{1}{2}+C}\)
⇒ p = e^0.5 x e^C
⇒ p = 1.648 × 1000
⇒ p = 1648
Hence, after lo years the amount will worth ₹ 1648.

Question 22.
In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000,11 the rate of growth of bacteria is proportional to the number present?
Solution.
Let y be the number of bacteria at any instant t.
It is given that the rate of growth of the bacteria is proportional to the number present.
∴ \(\frac{d y}{d t}\) ∝ y

\(\frac{d y}{d t}\) = ky (where k is a constant)

⇒ \(\frac{d y}{y}\) = k dt

Integrating both sides, we get
∫ \(\frac{d y}{y}\) = k ∫ dt
⇒ logy = kt + C
Let y₀ be the number of bacteria at t = 0.
log y₀ = C
Substituting the value of C in equation (i), we get
log y = kt + log y₀
⇒ log y – log y₀ = kt
log \(\left(\frac{y}{y_{0}}\right)\) = kt
⇒ kt = log \(\left(\frac{y}{y_{0}}\right)\) ………… (ii)

Also, it is given that the number of bacteria increases by 10% in 2 hours.
⇒ y = \(\frac{110}{100}\) y₀
⇒ \(\frac{y}{y_{0}}=\frac{11}{10}\) ………….(iii)
Substituting the value in equation (ii), we get

Question 23.
The general solution of the differential equation \(\frac{d y}{d x}\) = e^{x + y} is
(A) e^x + e^{– y} = C
(B) e^x + e^y = C
(C) e^{– x} + e^y = C
(D) e^{– x} + e^{– y} = C
Solution.
\(\frac{d y}{d x}\) = e^{x + y}
= e^x . e^y
∫ \(\frac{d y}{e^{y}}\) = e^x dx
e^{– y} dy = e^x dx
Integrating both sides, we get
⇒ ∫ e^{– y} dy = ∫ e^x dx
⇒ – e^{– y} = e^x + k
⇒ e^x + e^{– y} = – k
⇒ e^x + e^{– y} = C (C = – k)
Hence, the correct answer is (A).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter Differential Equations Ex 9.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.1

Direction (1 – 10): Determine order and degree (if defined) of differential equation.

Question 1.
\(\frac{d^{4} y}{d x^{4}}\) + sin(y””) = 0.
Solution.
\(\frac{d^{4} y}{d x^{4}}\) + sin(y””) = 0
⇒ y”” + sin (y””) = 0
The highest order derivative present in the differential equation is y””, therefore, its order is 4.
The given differential equation is not a polynomial equation in its derivatives.
Hence, its degree is not defined.

Question 2.
y’ + 5y = 0.
Solution.
The given differential equation is y’ + 5y = 0
The highest order derivative present in the differential equation is y’.
Therefore, its order is 1.
It is a polynomial equation in y’. The highest power raised to y’ is 1.
Hence, its degree is 1.

Question 3.
\(\left(\frac{d s}{d t}\right)^{4}+3 \frac{d^{2} s}{d t^{2}}\) = 0
Solution.
\(\left(\frac{d s}{d t}\right)^{4}+3 \frac{d^{2} s}{d t^{2}}\) = 0
The highest order derivative present in the given differential equation is \(\frac{d^{2} s}{d t^{2}}\), therefore, its order is 2.
It is a polynomial equation in \(\frac{d^{2} s}{d t^{2}}\) and \(\frac{d s}{d t}\).
The highest power raised to \(\frac{d^{2} s}{d t^{2}}\) is 1.
Hence, its degree is 1.

Question 4.
\(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}\) + cos \(\left(\frac{d y}{d x}\right)\) = 0.
Solution.
\(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}\) + cos \(\left(\frac{d y}{d x}\right)\) = 0.
The highest order derivative present in the given differential equation is \(\frac{d^{2} y}{d x^{2}}\). Therefore, its order is 2.
The given differential equation is not a polynomial equation in its derivatives.
Hence, its degree is not defined.

Question 5.
\(\frac{d^{2} y}{d x^{2}}\) = cos 3x + sin 3x
Solution.
\(\frac{d^{2} y}{d x^{2}}\) = cos 3x + sin 3x

⇒ \(\frac{d^{2} y}{d x^{2}}\) – cos 3x – sin 3x = 0
The highest order derivative present in the differential equation is \(\frac{d^{2} y}{d x^{2}}\).
Therefore its order is 2.
It is a polynomial equation in \(\frac{d^{2} y}{d x^{2}}\) and the p[ower raised to \(\frac{d^{2} y}{d x^{2}}\) is 1.
Hence, its degree is 1.

Question 6.
(y”’)² + (y”)³ + (y’)⁴ + y⁵ = 0.
Solution.
(y”’)² + (y”)³ + (y’)⁴ + y⁵ = 0
The highest order derivative present in the differential equation is y”’.
Therefore, its order is 3.
The given differential equation is a polynomial equation in y”’, y”, and y’.
The highest power raised to y” is 2.
Hence, its degree is 2.

Question 7.
y”’ + 2y” + y’= 0.
Solution.
y”’ + 2y” + y’ = 0
The highest order derivative present in the differential equation is y”’.
Therefore, its order is 3.
It is polynomial equation in y”‘, y” and y’.
The highest power raised to y”‘ is 1.
Hence, its degree is 1.

Question 8.
y’ + y = e^x.
Solution.
y’ + y = e^x
⇒ y’ + y – e^x = 0
The highest order derivative present in the differential equation is y’.
Therefore, its order is 1.
The given differential equation is a polynomial equation in y’ and the highest power raised to y is 1.
Hence, its degree is 1.

Question 9.
y” + (y’)² + 2y = 0.
Solution.
y” + (y’)² + 2y = 0
The highest order derivative present in the differential equation is y”.
Therefore, its order is 2.
The given differential equation is a polynomial equation in y” and y’ and the highest power raised to y” is 1.
Hence, its degree is 1.

Question 10.
y” + 2y’ + sin y = 0.
Solution.
y” + 2y’ + sin y = 0
The highest order derivative present in the differential equation is y”.
Therefore, its order is 2.
This is a polynomial equation in y” and y’ and the highest power raised to y” is 1.
Hence, its degree is 1.

Direction (11 – 12): Choose the correct answer.

Question 11.
The degree of the differential equation
\(\left(\frac{d^{2} y}{d x^{2}}\right)^{3}+\left(\frac{d y}{d x}\right)^{2}\) + sin \(\left(\frac{d y}{d x}\right)\) + 1 = 0 is
(A) 3
(B) 2
(C) 1
(D) None of these
Solution.
\(\left(\frac{d^{2} y}{d x^{2}}\right)^{3}+\left(\frac{d y}{d x}\right)^{2}\) + sin \(\left(\frac{d y}{d x}\right)\) + 1 = 0
The given differential equation is not a polynomial equation in its derivatives. Therefore, its degree is not defined.
Hence, the correct answer is (D).

Question 12.
The order of the differential equation 2x² \(\frac{d^{2} y}{d x^{2}}\) – 3 \(\frac{d y}{d x}\) + y = 0
(A) 2
(B) 1
(C) zero
(D) None of these
Solution.
2x² \(\frac{d^{2} y}{d x^{2}}\) – 3 \(\frac{d y}{d x}\) + y = 0

The highest order derivative present in the given differential equation is \(\frac{d^{2} y}{d x^{2}}\)
Therefore, its order is 2.
Hence, the correct answer is (A).