Class 12

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).

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

PSEB Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

Question 1.
Find the area under the given curves and given lines.
(i) y = x², x = 1, x = 2 and x – axis
(ii) y = x⁴, x = 1, x = 5 and x – axis
Solution.

(i) The required area is represented by the shaded area ADCBA as
Area of ADCBA = \(\int_{1}^{2}\) y dx

= \(\int_{1}^{2}\) x² dx

= \(\left[\frac{x^{3}}{3}\right]_{1}^{2}=\frac{8}{3}-\frac{1}{3}=\frac{7}{3}\) sq. unit.

(ii)

The required area is represented by y the shaded area ADCBA as

Area of ADCBA = \(\int_{1}^{5}\) x⁴ dx = \(\)
= \(\frac{(5)^{5}}{5}-\frac{1}{5}\)

= \((5)^{4}-\frac{1}{5}\)

= 625 – \(\frac{1}{5}\)

=624.8 sq.unit.

Question 2.
Find the area between the curves, y = x and y = x².
Solution.
The required area is represented by the shaded area OBADO as

The points of intersection of the curves, y = x and y = x² ,is A(1,1).
We draw AC perpendicular to x-axis.
∴ Area of OBAO = Area of AOCA – Area of OCABO ……….. (i)
= \(\int_{0}^{1}\) x dx – \(\int_{0}^{1}\) x² dx

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

= \(\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\) sq. unit.

Question 3.
Find the area of the region lying in the first quadrant and bounded by y = 4x², x = 0, y = 1 and y = 4.
Solution.
The area in the first quadrant bounded by y = 4x², x = 0, y = 1 and y = 4 is represented by the shaded area ABCDA as given in the figure.

∴ Area of ABCD = \(\int_{1}^{4}\) x dx
[∵ y = 4x²
⇒ x = \(\frac{\sqrt{y}}{2}\)]

= \(\int_{1}^{4} \frac{\sqrt{y}}{2} d x\)

= \(\frac{1}{2}\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{4}\)

= \(\frac{1}{3}\) [(4)^{\(\frac{3}{2}\) – 1]}

= \(\frac{1}{3}\) [8 – 1]

= \(\frac{7}{3}\) sq unit.

Question 4.
Sketch the graph of y = |x + 3| and evaluate \(\int_{-6}^{0}\) |x + 3| dx.
Solution.
The given equation is y = |x + 3|
The corresponding values of x and y are given in the following table.

On plotting these, points, we get the graph of y = |x + 3| as follows.

We know that, (x + 3) ≤ 0 for – 6 ≤ x – 3 and (x + 3) ≥ 0 for – 3 ≤ x ≤ 0

Question 5.
Find the area bounded by the curve y = sin x between x = 0 and x = 2π.
Solution.
The graph of y = sin x can be drawn as

∴ Required area = Area of OABO + Area of BCDB
= \(\int_{0}^{\pi}\) sin x dx + |\(\int_{0}^{2 \pi}\) sin x| dx

= \([-\cos x]_{0}^{\pi}+\left|[-\cos x]_{\pi}^{2 \pi}\right|\)

= [- cos π + cos 0] + |- cos 2π + cos π|
= 1 + 1 + |(- 1 – 1)|
= 2 + |- 2|
= 2 + 2 = 4 sq. unit

Question 6.
Find the area enclosed between the parabola y² = 4ax and the line y = mx.
Solution.
The area enclosed between the parabola, y² = 4ax and the line, y = mx, is represented by the shaded area OABO as given in the figure

The points of intersection of both the curves are (0, 0) and \(\left(\frac{4 a}{m^{2}}, \frac{4 a}{m}\right)\).

We draw AC perpendicular to x-axis.
∴ Area of OABO = Area of OCABO -Area of ∆OCA

Question 7.
Find the area enclosed by the parabola 4y = 3x² and the line 2y = 3x + 12.
Solution.
The area enclosed between the parabola, 4y = 3x² and the line, 2y = 3x +12 is represented by the shaded area OBAO as

The points of intersection of the given curves are A(- 2, 3) and (4, 12).
We draw AC and BD perpendicular to x-axis.
Area of OBAO = Area of CDBA – (Area of ODBO + Area of OACO).
= \(\int_{-2}^{4} \frac{1}{2}(3 x+12) d x-\int_{-2}^{4} \frac{3 x^{2}}{4} d x\)

= \(=\frac{1}{2}\left[\frac{3 x^{2}}{2}+12 x\right]_{-2}^{4}-\frac{3}{4}\left[\frac{x^{3}}{3}\right]_{-2}^{4}\)

= \(\frac{1}{2}\) [24 + 48 – 6] – \(\frac{1}{4}\) [64 + 8]

=\(\frac{1}{2}\) [90] – \(\frac{1}{4}\) [72]
= 45 – 18 = 27 sq. unit.

Question 8.
Find the area of the smaller region hounded by the ellipse \(\frac{x^{2}}{9}+\frac{y^{2}}{4}\) = 1 and the line \(\frac{x}{3}+\frac{y}{2}\) = 1.
Solution.
The area of the smaller region bounded by the ellipse, \(\frac{x^{2}}{9}+\frac{y^{2}}{4}\) = 1 and the line, \(\frac{x}{3}+\frac{y}{2}\) = 1, is represented by the shaded region BCAB as

∴ Area of BCAB = Area of OBCAO – Area of OBAO

= \(\int_{0}^{3} 2 \sqrt{1-\frac{x^{2}}{9}} d x-\int_{0}^{3} 2\left(1-\frac{x}{3}\right) d x\)

= \(\frac{2}{3}\left[\int_{0}^{3} \sqrt{9-x^{2}} d x\right]-\frac{2}{3} \int_{0}^{3}(3-x) d x\)

Question 9.
Find the area of the smaller region bounded by the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 and the line \(\frac{x}{a}+\frac{y}{b}\) = 1.
Solution.
The area of the smaller region bounded by the ellipse, \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1, and the line, \(\frac{x}{a}+\frac{y}{b}\) = 1, is represented by the shaded region BCAB as

∴ Area of BCAB = Area of OBCAO – Area of OBAO

Question 10.
Find the area of the region enclosed by the parabola x² = y, the line y = x + 2 and x- axis.
Solution.
We have x² = y.
It represents a parabola with vertex at (0, 0), axis along the positive direction of y-axis and it open upwards.
Also, y = x + 2 represents a straight line cutting x – axis at (- 2, 0)
Solving x² = y and y = x + 2, we get

x² = x + 2
=> x² – x – 2 = 0
=> (x – 2)(x + 1) = 0
x = 2, x = -1
When x = 2, y = (2)² = 4
So the two curves x² = y and y = x + 2 intersect at the points (2, 4) and (- 1, 1)
∴ Required area = Shaded region shown in the figure
= \(\int_{-1}^{2}\) (x + 2 – x²) dx

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

= \(\frac{1}{2}\) (4 – 1) + 2 (2 – 1) – \(\frac{1}{3}\) (8 + 1)

= \(\frac{3}{2}\) + 6 – 3

= \(\frac{3}{2}\) + 3 = \(\frac{9}{2}\) sq. unit.

Question 11.
Using the method of integration find the area bounded by the curve |x| + |y| = 1.
[Hint: The required region is bounded by lines x + y = 1, x – y = 1, – x + y = 1 and – x – y = 1.]
Solution.
The given curve is |x| + |y| = 1

In first quadrant, (x > 0, y > 0)
Then, the line AB is x + y = 1
In second quadrant, (x < 0, y > 0)
Then, the line BC is – x + y = 1
In third quadrant, (x < 0, y < 0) Then, the line CD is -x – y = 1 In fourth quadrant, (x > 0, y < 0)
Then, the line DA is x – y = 1
Since, ABCD is a square.
∴ Area of ADCB = 4 × Area of OBAO
= 4 \(\int_{0}^{1}\) (1 – x) dx
= 4 \(\left(x-\frac{x^{2}}{2}\right)_{0}^{1}\)
= 4 [1 – \(\frac{1}{2}\)]
= 4 \(\frac{1}{2}\) = 2 sq. unit.

Question 12.
Find the area bounded by curves {(x, y) : y ≥ x² and y = |x|}.
Solution.

The area bounded by the curves {(x, y): y > x² and y = |x|}, is represented by the shaded region as It can be observed that the required area is symmetrical about y-axis.
∴ Required area = 2 [Area of OCAO – Area of OCADO]
= 2 [\(\int_{0}^{1}\) x dx – \(\int_{0}^{1}\) x² dx]

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

= \(2\left[\frac{1}{2}-\frac{1}{3}\right]=2\left[\frac{1}{6}\right]=\frac{1}{3}\) sq unit.

Question 13.
Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A (2, 0), B (4, 5) and C (6, 3).
Solution.
The vertices of AABC are A (2, 0), B(4, 5) and C( 6, 3).

Equation of line segment AB is
y – 0 = \(\frac{5-0}{4-2}\) (x – 2)
⇒ 2y = 5x – 10
⇒ y = \(\frac{5}{2}\) (x – 2) …………..(i)
Equation of line segment BC is
y – 5 = \(\frac{3-5}{6-4}\) (x – 4)
⇒ 2y – 10 = – 2x + 8
⇒ 2y = – 2x + 18
y = – x + 9 ………………(ii)
Equation of line segment CA is
y – 3 = \(\frac{3-5}{6-4}\) (x – 6)
⇒ – 4y + 12 = – 3x + 18
4y = 3x – 6
⇒ y = \(\frac{3}{4}\) (x – 2) …………….(iii)
Area of ∆ABC = Area of ABLA + Area of BLMCB – Area of ACMA
= \(\int_{2}^{4} \frac{5}{2}(x-2) d x+\int_{4}^{6}(-x+9) d x-\int_{2}^{6} \frac{3}{4}(x-2) d x\)

= \(\frac{5}{2}\left[\frac{x^{2}}{2}-2 x\right]_{2}^{4}+\left[\frac{-x^{2}}{2}+9 x\right]_{4}^{6}-\frac{3}{4}\left[\frac{x^{2}}{2}-2 x\right]_{2}^{6}\)

= \(\frac{5}{2}\) [8 – 8 – 2 + 4] + [- 18 + 54 + 8 – 36] – \(\frac{3}{4}\) [18 – 12 – 2 + 4]
= 5 + 8 – \(\frac{3}{4}\) (8)
= 13 – 6 = 7 sq. unit.

Question 14.
Using the method of integration, find the area of the region bounded by lines:
2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0
Solution.
The given equations of lines are
2x + y = 4 ……………..(i)
3x – 2y = 6 …………….(ii)
and x-3y + 5 = 0 ………….(iii)

The area of the region bounded by the lines is the area of AABC. AL and CM are the perpendiculars on x-axis.
Area of ∆ABC = Area of ALMCA – Area of ALB – Area of CMB

Question 15.
Find the area of the region 4(x, y): y² ≤ 4x, 4x² + 4y² ≤ 9}.
Solution.
y² = 4x is a paraola whose vertex is the origin and 4x² + 4y² = 9 represents circle whose centre is (0, 0) 4x² + 4y² = 9 and radius = \(\frac{3}{2}\)
On solving y² = 4x and x² + y² = \(\frac{9}{4}\)

The points of intersection are A\(\left(\frac{1}{2}, \sqrt{2}\right)\) and C\(\left(\frac{1}{2}, -\sqrt{2}\right)\).
Both the curves are symmetrical about x-axis.
∴ Required area = area of the shaded region
= 2 (area of the region OBAO)
= 2 [(area of the region OMAO) + (area of the region MBAM)]

Direction (16 – 19): Choose the correct answer.

Question 16.
Area bounded by the curve y = x², the x-axis and the ordinates x = – 2 and x = 1 is
(A) – 9

(B) \(-\frac{15}{4}\)

(C) \(\frac{15}{4}\)

(D) \(\frac{17}{4}\)
Solution.

The curve is y = x³
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = 3x²
∴ Curve is an increasing curve
\(\frac{d y}{d x}\) = 0, x = 0 dx
∴ x-axis is the tangent at x = 0
f(- x) = – f(x)
⇒ (- x)³ = – x³
Curve is symmetrical in opposite quadrants.
Area bounded by the curve y = x³, the x- axis, x = Area of the region AQOBPOA
= Area of the region AQOA + Area of the region ABPO

Thus, the correct answer is (D).

Question 17.
The area bounded by the curve y = x |x|, x-axis and the ordinates x =1 and x = 1 is given by
(A) 0
(B) \(\frac{1}{3}\)

(C) \(\frac{2}{3}\)

(D) \(\frac{4}{3}\)
[Hint: y = x² if x > 0 and y = – x² if x < 0]
Solution.
Required area = \(\int_{-1}^{1}\) y dx
= \(\int_{-1}^{1}\) x |x| dx

= \(\int_{-1}^{0}\) x² dx + \(\int_{0}^{1}\) x² dx

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

= \(-\left(-\frac{1}{3}\right)+\frac{1}{3}=\frac{2}{3}\) sq. units
Thus, the correct answer is (C).

Question 18.
The area of the circle x² + y² = 16 exterior to the parabola y² = 6x is
(A) \(\frac{4}{3}\) (4π – √3)

(B) \(\frac{4}{3}\) (4π + √3)

(C) \(\frac{4}{3}\) (8π – √3)

(D) \(\frac{4}{3}\) (8π + √3)
Sol.
The given equations are
x² + y² = 16 …………..(i)
y² = 6x …………(ii)

Area ounded by the circle and parabola = 2 [Area of OADO + Area of ADBA]

= 2 \(\left[\int_{0}^{2} \sqrt{16} x d x+\int_{2}^{4} \sqrt{16-x^{2}} d x\right]\)

= \(2\left[\sqrt{6}\left\{\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right\}_{0}^{2}\right]+2\left[\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_{2}^{4}\)
Area of circle = π(r)²
= π(4)²
= 16π sq unit
Required area = 16π – \(\frac{4}{3}\) [4π + V3]
= \(\frac{4}{3}\) [4 × 3π – 4π – √3]
= \(\frac{4}{3}\) (8π – √3) sq. unit
Thus, the correct answer is (C).

Question 19.
The area bounded by the y – axis, y = cos x and y = sin x, when 0 ≤ x < \(\frac{\pi}{2}\) is
(A) 2 (√2 – 1)
(B) √2 – 1
(C) √2 + 1
(D) √2
Solution.
The two curves y = sin x and y = cos x meet where sin x = cos x, i.e., where tan x = 1.

∴ Required area (shown in shaded region) = \(\int_{0}^{\pi / 4}\) (cos x – sinx) dx
= \([\sin x+\cos x]_{0}^{\pi / 4}\)

= sin \(\frac{\pi}{4}\) + cos \(\frac{\pi}{4}\) – (0 + 1)

= \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\) – 1

= \(\frac{2}{\sqrt{2}}\) – 1
= √2 – 1
Thus, the correct answer is (B).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter Differential Equations Ex 9.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.3

Direction (1 – 5): In each question form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

Question 1.
\(\frac{x}{a}+\frac{y}{b}\) = 1
Solution.
Given, family is \(\frac{x}{a}+\frac{y}{b}\) = 1
Differentiating both sides w.r.t. x, we get
\(\frac{1}{a}+\frac{1}{b} \frac{d y}{d x}\) = 0

⇒ \(\frac{1}{a}+\frac{1}{b} y\) = 0
Again, differentiating both sides w.r.t. x, we get
0 + \(\frac{1}{4}\) y” = 0
⇒ \(\frac{4}{4}\) y” = 0
⇒ y” = 0
Hence, the required differential equation of the given curve is y” = 0.

Question 2.
y² = a(b² – x²)
Sol.
Given, family is y² = a(b2 —x2)
Differentiating both sides w.r.t. x, we get
2y \(\frac{d y}{d x}\) = a(- 2x)
2yy’ = – 2ax
yy’= – ax ……………(i)
Again, differentiating both sides w.r.t. x, we get
y’ . y’ + yy” = – a
= (y’)² + yy” = – a ……………(ii)
Dividing equation (ii) by equation (i), we get
\(\frac{\left(y^{\prime}\right)^{2}+y y^{\prime \prime}}{y y^{\prime}}=\frac{-a}{-a x}\)

⇒ xyy” + x (y’)² – yy” = 0
This is the required differential equation of the given curve.

Question 3.
y = ae^3x + be^{– 2x}
Solution.
Given, family is y = ae^3x + be^{– 2x} ………………(i)
Differentiating both sides w.r.t. x, we get
y’ = 3ae^3x – 2 be^{– 2x} ………….(ii)
Again, differentiating both sides w.r.t. x, we get
⇒ y” = 9ae^3x + 4 be^{– 2x} …………….(iii)
Multiplying equation (i) with equation (ii) and then adding it from equation (ii), we get
(2ae^3x + 2be^{– 2x}) + (3ae^3x – 2 be^{– 2x}) = 2y + y’
⇒ 5ae^3x = 2y + y’
⇒ ae^3x = \(\frac{2 y+y^{\prime}}{5}\)
Now, multiplying equation (i) with equation (iii) and subtracting equation (ii) for, it, we get
(3ae^3x + 3be^{– 2x}) – (3ae^3x – 2be^{– 2x}) = 3y – y’
⇒ 5be^{– 2x} = 3y – y’
⇒ be^{– 2x} = \(\frac{3 y-y^{\prime}}{5}\)
Substituting the values of ae^3x and be^{– 2x} in equation (iii), we get
y” = \(9 \cdot \frac{\left(2 y+y^{\prime}\right)}{5}+4 \frac{\left(3 y-y^{\prime}\right)}{5}\)

⇒ y” = \(\frac{18 y+9 y^{\prime}}{5}+\frac{12 y-4 y^{\prime}}{5}\)

⇒ y” = \(\frac{30 y+5 y^{\prime}}{5}\)

⇒ y” = 6y + y’
⇒ y” – y’ – 6y = 0
This is the required differential equation of the given curve.

Question 4.
y = e^2x (a + bx)
Solution.
Given, y = e^2x (a + bx) ………….(i)
Differentiating both sides w.r.t. x, we get
y’ = 2e^2x (a + bx) + e^2x . b
⇒ y’ = e^2x (2a + 2 bx + b) ……………(ii)
Multiplying equation (i) with equation (ii) and then subtracting it from equation (ii), we get
y’ – 2y = e^2x (2a + 2bx + b) – e^2xx (2a + 2bx)
⇒ y’ – 2 = be^2x ……………..(iii)
Differentiating both sides w.r.t. x, we get
y”k – 2y’ = 2be^2x …………..(iv)
Dividing equation (iv) by equation (iii), we get
\(\frac{y^{\prime \prime}-2 y^{\prime}}{y^{\prime}-2 y}\) = 2
⇒ y” – 2y’ = 2y’ – 4y
⇒ y” – 4y’ + 4y = 0
This is the required differential equation of the given curve.

Question 5.
y = e^x (a cos x + b sin x)
Solution.
Given, y = e^x (a cos x + b sin x) …………..(i)
Differentiating both sides w.r.t. x, we get
y’ = e^x (a cos x + b sin x) + e^x (- a sin x + b cos x)
=> y’ = e^x [(a + b) cos x – (a – b) sin x] ……………(ii)
Again, differentiating both sides w.r.t. x, We get
y” = e^x [(a + b) cos x) – (a – b) sin x] + e^x [- (a + b) sin x – (a – b) cos x]
y” = e^x (2 b cos x – 2a sin x)
⇒ y” = 2 e^x (b cos x – a sin x)
Adding equations (i) and (iii), we get
y + \(\frac{y^{\prime \prime}}{2}\) = e^x [(a + b) cos x – (a – b) sin x]
⇒ y + \(\frac{y^{\prime \prime}}{2}\) = y’
⇒ 2y + y” = 2y
⇒ y” – 2y’ + 2y = 0
This is the required differential equation of the given curve.

Question 6.
Form the differential equation of the family of circles touching the y-axis at the origin.
Solution.
The equation of the circle with centre (a, 0) and radius a, which touches y-axis at origin
(x – a)² + y² = a²
x² + y² = 2ax ………….(i)

Differentiating equation w.r.t x
2x + 2y y’ = 2a
or x + y y’ = a
Put value of a in (i)
x² + y² = 2x (x + y y’)
= 2x² + 2xy y’
∴ Required differential equation is
2xy \(\frac{d y}{d x}\) + x² – y² = 0

Question 7.
Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
Solution.

The equation of the parabola having the vertex at origin and the axis along the positive y-axis is
x² = 4ay ………..(i)
Differentiating equation (i) w.r.t. x, we get 2x = 4ay’ ………….(ii)
Dividing equation (ii) by equation (i), wex
\(\frac{2 x}{x^{2}}=\frac{4 a y^{\prime}}{4 a y}\)

⇒ \(\frac{2}{x}=\frac{y^{\prime}}{y}\)

⇒ xy’ = 2y
⇒ xy’ – 2y = 0
This is the required differential equation.

Question 8.
Form the differential equation of the family of ellipse having foci on y-is and centre at origin.
Solution.

The equation of the family of ellipse having foci on they-axis and the centre at origin is as follows:
\(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}\) = 1 …………….(i)
Differentiating equation (i) w.r.t. x, we get
\(\frac{2 x}{b^{2}}+\frac{2 y y^{\prime}}{a^{2}}\) = 0

⇒ \(\frac{x}{b^{2}}+\frac{y y^{\prime}}{a^{2}}\) = 0 …………..(ii)

Again, differentiating w.r.t. x, we get
\(\frac{1}{b^{2}}+\frac{y^{\prime} \cdot y^{\prime}+y \cdot y^{\prime \prime}}{a^{2}}\) = 0

⇒ \(\frac{1}{b^{2}}+\frac{1}{a^{2}}\left(y^{\prime 2}+y y^{\prime \prime}\right)\) = 0

⇒ \(\frac{1}{b^{2}}=-\frac{1}{a^{2}}\left(y^{2}+y y^{\prime \prime}\right)\)

x \(\left[-\frac{1}{a^{2}}\left(\left(y^{\prime}\right)^{2}+y y^{\prime \prime}\right)\right]+\frac{y y^{\prime}}{a^{2}}\) = 0

⇒ – x (y’)² – xyy” + yy’ = 0
⇒ xyy’ + x(y’)² – yy’ = 0
This is the required differential equation.

Question 9.
Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.
Solution.

The equation of the family of hyperbolas with the centre at origin and foci along the x-axis is
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 ………………(i)

Differentiating both sides of equation (j) w.r.t. x, we get
\(\frac{2 x}{a^{2}}-\frac{2 y y^{\prime}}{b^{2}}\) = 0

\(\frac{x}{a^{2}}-\frac{y y^{\prime}}{b^{2}}\) = 0 …………….(ii)

Again, differentiating w.r.t. x, we get

\(\frac{1}{a^{2}}-\frac{y^{\prime} \cdot y^{\prime}+y y^{\prime \prime}}{b^{2}}\) = 0

\(\frac{1}{a^{2}}=\frac{1}{b^{2}}\left(\left(y^{\prime}\right)^{2}+y y^{\prime \prime}\right)\)
Substituting the value of in equation (ii), we get
\(\left.\frac{x}{b^{2}}\left(y^{\prime}\right)^{2}+y y^{\prime \prime}\right)-\frac{y y^{\prime}}{b^{2}}\) = 0

⇒ x (y’)² + xyy” – yy’ = 0
⇒ xyy” + x(y’)² – yy’ = 0
This is the required differential equation.

Question 10.
Form the differential equation of the family of circles having centre on y-axis and radius 3 units.
Solution.

Let the centre of the circle on y-axis be (0, b).
The differential equation of the family of Circles with centre at (0, b) and radius 3 is as follows
x² + (y – b)² = 3²
⇒ x² + (y – b)² = 9 ……………(i)
Differentiating equation (i) w.r.t. x,
2x + 2(y – b) . y’ = 0
⇒ (y – b) . y’ = – x
⇒ y – b = \(\)
Substituting the value of (y – b) in equation (i), we get
x² + \(\left(\frac{-x}{y^{\prime}}\right)^{2}\) = 9

⇒ x² \(\left[1+\frac{1}{\left(y^{\prime}\right)^{2}}\right]\) = 9

⇒ x² ((y’)² + 1) = 9 (y’)x²

⇒ (x² – 9) (y’)² + x² = 0

This is the required differential equation.

Question 11.
Which of the following differential equations has y = c₁ e^x + c₂ e^x as the general solution?
(A) \(\frac{d^{2} y}{d x^{2}}\) + y = 0

(B) \(\frac{d^{2} y}{d x^{2}}\) – y = 0

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

(D) \(\frac{d^{2} y}{d x^{2}}\) – 1 = 0
Solution.
Given general solution is y = c₁ e^x + c₂ e^-x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = c₁ e^x – c₂ e^-x
Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = c₁ e^x + c₂ e^-x
⇒ \(\frac{d^{2} y}{d x^{2}}\) = y
⇒ \(\frac{d^{2} y}{d x^{2}}\) – y = 0
This is the required differential equation of the given equation of curve.
Hence, the correct answer is (B).

Question 12.
Which of the following differential equations has y = z as one of its particular solution?
(A) \(\frac{d^{2} y}{d x^{2}}-x^{2} \frac{d y}{d x}\) + xy = x

(B) \(\frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}\) + xy = x

(C) \(\frac{d^{2} y}{d x^{2}}-x^{2} \frac{d y}{d x}\) + xy = 0

(D) \(\frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}\) + xy = 0
Solution.
Given solution is y = x
Differentiating w.r.t. x, we get

\(\frac{d y}{d x}\) = 1 ……………….(i)

Again, differentiating w.r.t x, we get

\(\frac{d^{2} y}{d x^{2}\) = 0 ………………..(ii)

Now, on substituting the value of y, \(\frac{d^{2} y}{d x^{2}}\) and \(\frac{d y}{d x}\) from equation (i) and (ii) in each of the given alternatives, we find that only the differential
equation given in alternative C is correct.
\(\frac{d^{2} y}{d x^{2}}\) – x² \(\frac{d y}{d x}\) + xy = 0 – x² . 1 + x . x
= – x² + x² = 0
Hence, the correct answer is (C).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter Differential Equations Ex 9.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.2

Direction (1 – 10): In each of the questions verify that the given function (explicit or implicit) is a solution of the corresponding differential equation.

Question 1.
y = e^x + 1 ; y” – y’ = 0 ,
Solution.
Given, y = e^x + 1
Differentiating both sides of this equation w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (e^x + 1)
⇒ y’ = e^x
Now, differentiating equation (i) w.r.t. x, we get
\(\frac{d}{d x}\) (y’) = \(\frac{d}{d x}\) (e^x)
⇒ y” = e^x
Substituting the values of y’ and y” in the given differential equation, we get the L.H.S. as
y” – y’ = e^x – e^x = 0 = R.H.S.
Thus, the given function in the solution of the corresponding differential equation.

Question 2.
y = x² + 2x + C : y’ – 2x – 2 = 0
Solution.
Given, y = x² + 2x + C
Differentiating both sides of this equation w.r.t x, we get
y’ = \(\frac{d}{d x}\) (x² + 2x + C)
⇒ y’ = 2x+2 ax
Substituting the value of y’ in the given differential equation, we get
L.H.S. = y’ – 2x – 2 = 2x + 2 – 2x – 2 = 0 = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.

Question 3.
y = cos x + C: y’ + sin x – 0
Solution.
Given, y = cos x + C
Differentiating both sides of this equation w.r.t. x, we get
y’ = \(\frac{d}{d x}\) (cos x + C)
=> y’ = – sin x dx
Substituting the value of y in the given differential equation, we get
L.H.S. y’ + sin x = – sin x + sin x = 0 = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.

Question 4.
y = \(\sqrt{1-x^{2}}\); y’ = \(\frac{x y}{1+x^{2}}\)
Solution.
Given, y = \(\sqrt{1-x^{2}}\)
Differentiating both sides of this equation w.r.t. x, we get

∴ L.H.S. = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.

Question 5.
y = Ax: xy’ = y, (x ≠ 0)
Solution.
Given, y = Ax
Differentiating both sides of this equation w.r.t. x, we get
y = -^-G4x) =^y = A
ax
Substituting the value of y’ in the given differential equation, we get
L.H.S. = xy’ = x . A = Ax = y= R.H.S.
Hence, the given function is the solution of the corresponding differential equation.

Question 6.
y = x sin x : xy’ = y + x \(\sqrt{x^{2}-y^{2}}\) (x ≠ 0 and x > y or x< – y)
Solution.
Given, y = x sin x
Differentiating both sides of this equation w.r.t. x, we get
y’ = \(\frac{d}{d x}\) (x sin x)
y’ = sin x . \(\frac{d}{d x}\) (x) + x . \(\frac{d}{d x}\) (sin x)
⇒ y = sin x + x cos x
Substituting the values of y in the given differential equation, we get
L.H.S. = xy = x (sin x + x cos x)
= x sin x + x² cos x
= y + x² . \(\sqrt{1-\sin ^{2} x}\)
= y + x² \(\sqrt{1-\left(\frac{y}{x}\right)^{2}}\)
= y + x \(\sqrt{y^{2}-x^{2}}\) = R.H.S
Hence, the given function is the solution of the corresponding differential equation.

Question 7.
xy = log y + C : y’ = \(\frac{y^{2}}{1-x y}\) (xy ≠ 1)
Solution.
Given, xy = log y + C
Differentiating both sides of this equation w.r.t. x, we get
\(\frac{d}{d x}\) (xy) = \(\frac{d}{d x}\) (log y)
⇒ y . \(\frac{d}{d x}\) (x) + x . \(\frac{d y}{d x}\) = \(\frac{1}{y}\) \(\frac{d y}{d x}\)
⇒ y + xy’ = \(\frac{1}{y}\) y’
⇒ y² + xy y’ = y’
⇒ (xy – 1) y’ = – y²
⇒ y’ = \(\frac{y^{2}}{1-x y}\)
.-. L.H.S. = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.

Question 8.
y – cos y = x : (y sin y + cos y + x) y’ = y
Solution.
Given, y – cos y = x …………(i)
Differentiating both sides of this equation w.r.t. x, we get
\(\frac{d y}{d x}\) – \(\frac{d}{d x}\) (cos y) = \( \frac{d y}{d x} (x)\)
⇒ y’ + sin y . y’ = 1
⇒ y’ (1 + sin y) = 1
⇒ y’ = \(\frac{1}{1+\sin y}\)
Substituting the value of y in equation, we get
L.H.S = (y sin y + cos y + x) y’
= (y sin y + cos y + y – cos y) × \(\frac{1}{1+\sin y}\)
= y (1 + sin y) \(\frac{1}{1+\sin y}\)
= y = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.

Question 9.
x + y = tan^-1 y : y² y’ + y² + 1 = 0
Solution.
Given, x + y = tan^-1 y
Differentiating both sides of this equation w.r.t. x, we get

Substituting the values of y’ in the given differential equation, we get
L.H.S = y² y’ + y² + 1
= y² \(\left[\frac{-\left(1+y^{2}\right)}{y^{2}}\right]\) + y² + 1
= – 1 – y² + y² + 1 = 0
= R.H.S
Hence, the given function is the solution of the corresponding differential equation.

Question 10.
y = \(\sqrt{a^{2}-x^{2}}\), x ∈ (- a, a): x + y \(\frac{d y}{d x}\) = 0, (y ≠ 0)
Solution.
Given, y = \(\sqrt{a^{2}-x^{2}}\)
Differentiating both sides of this equation w.r.t. x, we get

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

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

Question 11.
The number of arbitrary constants in the general solution of a differential equation of fourth order is
(A) Zero
(B) 2
(C) 3
(D) 4
Solution.
We know that the number of constants in the general solution of a differential equation of order n is equal to its order. Therefore, the number of constants in the general equation of fourth order differential equation is four.
Hence, the correct answer is (D).

Question 12.
The number of arbitrary constants in the particular solution of a differential equation of third order is
(A) 3
(B) 2
(C) 1
(D) zero
Solution.
In a particular solution of a differential equation, there are no arbitrary constants.
Hence, the correct answer is (D).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 8 Application of Integrals Ex 8.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.1

Question 1.
Find the area of the region bounded by the curve y 2 = x and the lines x = 1, x = 4 and the x-axis in the first quadrant.
Solution.
The area of the region bounded by the curve, y² = x, the lines, x = 1 and x = 4 and the x-axis is the area ABCD.

Question 2.
Find the area of the region bounded by y² = 9x, x = 2, x = 4 and the x-axis in the first quadrant.
Solution.
The area of the region bounded by the curve, y² = 9x, x = 2 and x = 4 and the x-axis is the area ABCD.

Question 3.
Find the area of the region bounded by x² – 4y, y – 2, y = 4 and the y-axis in the first quadrant.
Solution.
The area of the region bounded by the curve, x² = 4y, y = 2 and y = 4 and the y-axis is the area ABCD.

Question 4.
Find the area of the region bounded by the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{9}\) = 1.
Solution.
The given equation of the ellipse is \(\frac{x^{2}}{16}+\frac{y^{2}}{9}\) = 1

The given ellipse is symmetrical about both axes as it contains only even power of y and x.
Now, \(\frac{x^{2}}{16}+\frac{y^{2}}{9}\) = 1

\(\frac{y^{2}}{9}=1-\frac{x^{2}}{16}\)

y² = \(9\left(1-\frac{x^{2}}{16}\right)\)

y = ± 3 \(\sqrt{\left(1-\frac{x^{2}}{16}\right)}\)

Area bounded by ellipse = 4 × Area of OAB

Therefore, area bounded by the ellipse = 4 × 3π = 12π sq. unit.

Question 5.
Find the area of the region bounded by the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}\) = 1
Solution.
The given equation of the ellipse can be represented as \(\frac{x^{2}}{4}+\frac{y^{2}}{9}\) = 1

It can be observed that the ellipse is symmetrical about x-axis and y-axis.
∴ Area bounded by ellipse = 4 × Area of OAB
∴ Area of OAB = \(\int_{0}^{2}\) y dx

= \(\int_{0}^{2} 3 \sqrt{1-\frac{x^{2}}{4}}\) [Using EQuestion (i)]

= \(\frac{3}{2} \int_{0}^{2} \sqrt{4-x^{2}} d x=\frac{3}{2}\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{0}^{2}\)

= \(\frac{3}{2}\left[\frac{2 \pi}{2}\right]=\frac{3 \pi}{2}\)
Therefore, area bounded by the ellipse = 4 × \(\frac{3 \pi}{2}\) = 6π sq. unit

Question 6.
Find the area of the region in the first quadrant enclosed by x-axis, line x = √3y and the circle x² + y² = 4.
Solution.

The area of the region bounded by the circle, x² + y² = 4, x = √3y, and the x-axis is the area OAB.
The point of intersection of the line and the circle in the first quadrant is (√3, 1).
Area of OAB = Area of ∆OCA + Area of ACB
Area of OAC = \(\frac{1}{2}\) × OC × AC
= \(\frac{1}{2}\) × √3 × 1
= \(\frac{\sqrt{3}}{2}\) …………..(i)

Therefore, are enclosed by x-axis, the line x = √3y, and the circle x² + y² = 4 in the first quadrant = \(\frac{\sqrt{3}}{2}+\frac{\pi}{3}-\frac{\sqrt{3}}{2}=\frac{\pi}{3}\) sq. unit

Question 7.
Find the area of the smaller part of the circle x² + y² = a² cut off by the line x = \(\frac{a}{\sqrt{2}}\).
Solution.
The area of the smaller part of the circle x² + y² = a², cut off the line, x = \(\frac{a}{\sqrt{2}}\), is the area ABCD.
It can be observed that the area of ABCD is symmetrical about x-axis.
∴ Area of ABCD = 2 × Area of ABC
Area of ABC = \(\int_{\frac{a}{\sqrt{2}}}^{a} y d x=\int_{\frac{a}{\sqrt{2}}}^{a} \sqrt{a^{2}-x^{2}} d x\)

Therefore, the area of smaller part of the circle, x² + y² = a²,, cut off by the line, x = \(\frac{a}{\sqrt{2}}\), is \(\frac{a^{2}}{2}\left(\frac{\pi}{2}-1\right)\) sq. unit.

Question 8.
The area between x = y² and x = 4 is divided into two equal parts by the line x = a, find the value of a.
Solution.

The line, x = a, divides the area bounded by the parabola and x = 4 into two equal parts.
∴ Area of OAD = Area of ABCD
It can be observed that the given area is symmetrical about x-axis.
⇒ Area of OED = Area of EFCD.
Area of OED = \(\int_{0}^{a}\) y dx = \(\int_{0}^{a}\) √x dx

⇒ 2 . \((a)^{\frac{3}{2}}\) = 8

⇒ \((a)^{\frac{3}{2}}\) = 4

⇒ a = \((4)^{\frac{3}{2}}\)

Therefore, the value of a is \((4)^{\frac{3}{2}}\) units.

Question 9.
Find the area of the region bounded by the parabola y = x and y = |x|.
Solution.

The area bounded by the parabola, x² = y and the line, y = |x| can be represented as in the given figure.
The given area is symmetrical about y-axis.
∴ Area of OACO = Area of ODBO
The point of intersection of (O O) parabola, x² = y, and line, y = x, is A(1, 1).
Area of OACO = Area of ∆OAB – Area of OBACO
∴ Area of OAB = \(\frac{1}{2}\) x OB x AB
= \(\frac{1}{2}\) × 1 × 1 = \(\frac{1}{2}\)
Area of OBACO = \(\int_{0}^{1}\) y dx = \(\int_{0}^{1}\) x² dx
= \(\left[\frac{x^{3}}{3}\right]_{0}^{1}=\frac{1}{3}\)
⇒ Area of OACO = Area of ∆OAB – Area of OBACO
= \(\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)
Therefore, the required area = 2 \(\left[\frac{1}{6}\right]=\frac{1}{3}\) sq. units.

Question 10.
Find the area hounded by the curve xx² = 4y and the line x = 4y – 2.
Solution.
The area bounded by the curve, xx² = 4y and line, x = 4y – 2, is represented by the shaded area OBAQ.
Let A and B be the points of intersection of the line and parabola.
Coordinates of point A are (- 1, \(\frac{1}{4}\))
Coordinates of point B are (2, 1).
We draw AL and BM perpendicular to x-axis.
It can be observed that,
Area of OBAO = Area of OBCD + Area of OACO ………….(i)
Then, Area of OBCO = Area of OMBC – Area of OMBO
= \(\int_{0}^{2} \frac{x+2}{4} d x-\int_{0}^{2} \frac{x^{2}}{4} d x\)

= \(\frac{1}{4}\left[\frac{x^{2}}{2}+2 x\right]_{0}^{2}-\frac{1}{4}\left[\frac{x^{3}}{3}\right]_{0}^{2}\)

Therefore, required area = \(\left(\frac{5}{6}+\frac{7}{24}\right)=\frac{9}{8}\) sq. unit.

Question 11.
Find the area of the region bounded by the curve y² = 4x and the line x – 3.
Solution.
The region bounded by the parabola, y² = 4x and the line, x = 3, is the area OACO.

The area OACO is symmetrical about x-axis.
∴ Area of OACO = 2 (Area of OAB)
Area of OACO = 2[\(\int_{0}^{3}\) y dx]
[∵ y² = 4x
⇒ y = 2√x]
= 2 \(\int_{0}^{3}\) 2√x dx

= 4 \(\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{3}\)

= \(\frac{8}{3}\left[(3)^{\frac{3}{2}}\right]\)

= 8√3

Direction (12 – 13): Choose the correct answer:

Question 12.
Area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2 is.
(A) π
(B) \(\frac{\pi}{2}\)

(C) \(\frac{\pi}{3}\)

(D) \(\frac{\pi}{4}\)
Solution.
The area bounded by the circle and the lines, x = 0 and x = 2, in the first quadrant is represented as given in the figure.

Question 13.
Area of the region bounded by the curve y² = 4x, y-axis and the line y = 3 is
(A) 2

(B) \(\frac{9}{4}\)

(C) \(\frac{9}{3}\)

(D) \(\frac{9}{2}\)
Solution.
The area bounded by the curve y² = 4x, y-axis and y = 3 is represented as

Area of OAB = \(\int_{0}^{3}\) x dy
[∵ y² = 4x
⇒ x = \(\frac{y^{2}}{4}\)]

= \(\int_{0}^{3} \frac{y^{2}}{4} d y=\frac{1}{4}\left[\frac{y^{3}}{3}\right]_{0}^{3}\)

= \(\frac{1}{12}(27)=\frac{9}{4}\) sq unit.

Thus, the correct answer is (B).

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

PSEB Solutions for Class 12 Maths Chapter 7 Integrals Miscellaneous Exercise

Question 1.
\(\frac{1}{x-x^{3}}\)
Solution.
\(\frac{1}{x-x^{3}}=\frac{1}{x\left(1-x^{2}\right)}=\frac{1}{x(1-x)(1+x)}\)

Let \(\frac{1}{x(1-x)(1+x)}=\frac{A}{x}+\frac{B}{(1-x)}+\frac{C}{1+x}\) ………………..(i)

Equating the coefficients of x², x, and constant term, we get
– A + B – C = 0
B + C = 0, A = 1
On solving these equations, we get
A = 1, B = \(\frac{1}{2}\) and C = – \(\frac{1}{2}\)
From equation (i), we get

Question 2.
\(\frac{1}{\sqrt{x+a}+\sqrt{(x+b)}}\)
Solution.

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

= \(\frac{2}{3(a-b)}\left[(x+a)^{\frac{3}{2}}-(x+b)^{\frac{3}{2}}\right]\) + C

Question 3.
\(\frac{1}{x \sqrt{a x-x^{2}}}\) [Hint: Put x = \(\frac{a}{t}\)]
Solution.

Question 4.
\(\frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}}\)
Solution.

Question 5.
Evaluate: \(\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}\) [Hint: \(\), put x = t⁶]
Solution.

Question 6.
\(\frac{5 x}{(x+1)\left(x^{2}+9\right)}\)
Solution.
Let \(\frac{5 x}{(x+1)\left(x^{2}+9\right)}=\frac{A}{(x+1)}+\frac{B x+C}{\left(x^{2}+9\right)}\) ………..(i)
⇒ 5x = A(x² + 9) + (Bx + C) (x + 1)
⇒ 5x = Ax² + 9A + Bx² +Bx + Cx + C
Equating the coefficients of x², x, and constant term, we get
A + B = 0;
B + C = 5;
9A + C = 0
On solving these equations, we get
A = – \(\frac{1}{2}\); B = \(\frac{1}{2}\) and C = \(\frac{9}{2}\)
From equation (i), we get

Question 7.
\(\frac{\sin x}{\sin (x-a)}\)
Solution.
\(\frac{\sin x}{\sin (x-a)}\)
Let x – a = t
⇒ dx = dt
∫ \(\frac{\sin x}{\sin (x-a)}\) dx = ∫ \(\frac{\sin (t+a)}{\sin t}\) dt

= ∫ \(\frac{\sin t \cos a+\cos t \sin a}{\sin t}\) dt

= ∫ (cos a + cos t sin a) dt
= t cos a + sin a log |sin t| + C₁
= (x – a) cos a + sin a log sin (x – a) + C₁
= x cos a + sin a log sin (x – a) – a cos a + C₁
= sin a log sin (x – a) + x cos a + C

Question 8.
\(\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}\)
Solution.

Question 9.
\(\frac{\cos x}{\sqrt{4-\sin ^{2} x}}\)
Solution.

Question 10.
\(\frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}\)
Solution.

Question 11.
\(\frac{1}{\cos (x+a) \cos (x+b)}\)
Solution.

Question 12.
\(\frac{x^{3}}{1-x^{8}}\)
Solution.
\(\frac{x^{3}}{1-x^{8}}\)
Let x⁴ = t
⇒ 4x³ dx = dt
⇒ \(\int \frac{x^{3}}{\sqrt{1-x^{8}}} d x=\frac{1}{4} \int \frac{d t}{\sqrt{1-t^{2}}}\)

= \(\frac{1}{4}\) sin^-1 t + C

= \(\frac{1}{4}\) sin^-1 (x⁴) + C
[∵ ∫ \(\frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1}\left(\frac{x}{a}\right)\)]

Question 13.
\(\frac{e^{x}}{\left(1+e^{x}\right)\left(2+e^{x}\right)}\)
Solution.

Question 14.
\(\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}\)
Solution.
\(\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C x+D}{\left(x^{2}+4\right)}\) …………..(i)

⇒ 1 = (Ax + B) (x² + 4) + (Cx + D) (x² + 1)
⇒ 1 = Ax³ + 4Ax + Bx² + 4B + Cx³ + Cx + Dx² + D
Equating the coefficients of x³, x², x, and constant term, we get
A + C = 0;
B + D = 0;
4A + C = 0;
4B + D = 1
On solving these equations, we get
A = 0, B = 1, C = 0 and D = – \(\frac{1}{3}\)

From equation (i), we get

Question 15.
cos³ x e^{log sin x}
Solution.
cos³ x e^{log sin x} = cos³ × x sin x
Let cos x = t
⇒ – sin x dx = dt
⇒ ∫ cos³ x e^{log sin x} dx = ∫ cos³ x sin x dx
= – ∫ t³ dt
= \(\frac{t^{4}}{4}\) + C

= – \(\frac{\cos ^{4} x}{4}\) + C

Question 16.
e^{3 log x} (x⁴ + 1)^-1
Solution.

Question 17.
f'(ax + b) [f(ax + b)]ⁿ
Solution.
f'(ax + b) [f(ax + b)]ⁿ
Let f (ax + b) = t
⇒ af’(ax + b) dx = dt
⇒ f’(ax + b)[f(ax + b)]ⁿ dx = \(\frac{1}{a}\) ∫ tⁿ dt

= \(\frac{1}{a}\left[\frac{t^{n+1}}{n+1}\right]\)

= \(\frac{1}{a(n+1)}\) f(ax + b)ⁿ⁺¹ + C

Question 18.
\(\frac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}}\)
Solution.
\(\frac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}}=\frac{1}{\sqrt{\sin ^{3} x(\sin x \cos \alpha+\cos x \sin \alpha)}}\)

= \(\frac{1}{\sqrt{\sin ^{4} x \cos \alpha+\sin ^{3} x \cos x \sin \alpha}}\)

Question 19.
\(\frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}}\), x ∈ [0, 1]
Solution.

Question 20.
\(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\)
Solution.

Question 21.
\(\frac{2+\sin 2 x}{1+\cos 2 x}\) e^x
Solution.
Let I = ∫ (\(\frac{2+\sin 2 x}{1+\cos 2 x}\)) e^x dx

= ∫ \(\left(\frac{2+2 \sin x \cos x}{2 \cos ^{2} x}\right)\) e^x dx

= ∫ \(\left(\frac{1+\sin x \cos x}{\cos ^{2} x}\right)\) e^x dx

= ∫ (sec² x + tan x) e^x dx
Let f(x) = tan x
⇒ f’(x) = sec² x
I = ∫ [f(x) + f'(x)] e^x dx
= e^x f(x) + C
= e^x tan x + C

Question 22.
\(\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}\)
Solution.
\(\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^{2}}+\frac{C}{(x+2)}\) …………..(i)

⇒ x² + x + 1 = A (x + 1) (x + 2) + B (x + 2) + C(x² +2x+1)
⇒ x² + x + 1 = A(x² + 3x + 2) + B (x + 2) + C(x² + 2x + 1)
⇒ x² + x + 1 = (A + C) x² + (3A +B + 2C) x + (2A + 2B + C)
Equating the coefficients of x2, x, and constant term, we get
A + C = 1;
3A + B + 2C = 1;
2A + 2B + C = 1
On solving these equations, we get
A = – 2, B = 1 andC = 3
From equation (i), we get

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

∫ \(\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}\) dx = \(-2 \int \frac{1}{x+1} d x+3 \int \frac{1}{(x+2)} d x+\int \frac{1}{(x+1)^{2}} d x\)

= – 2 log |x + 1| + 3 log |x + 2| – \(\frac{1}{(x+1)}\) + C.

Question 23.
tan^-1 \(\sqrt{\frac{1-x}{1+x}}\)
Solution.
Let I = ∫ tan^-1 \(\sqrt{\frac{1-x}{1+x}}\) dx
Let x = cos θ
⇒ dx = – sin θ dθ
I = ∫ tan^-1 \(\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}\) (- sin θ dθ)

= – ∫ tan^-1 \(\sqrt{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}}\) sin θ dθ

= – ∫ tan^-1 tan \(\frac{\theta}{2}\) . sin θ dθ

= – \(\frac{1}{2}\) ∫ θ . sin θ dθ

= – \(\frac{1}{2}\) [θ – cos θ] – ∫ 1 . (- cos θ) dθ

= – \(\frac{1}{2}\) [- θ cos θ + sin θ]

= + \(\frac{1}{2}\) θ cos θ – \(\frac{1}{2}\) sin θ

= \(\frac{1}{2}\) cos^-1 x . x – \(\frac{1}{2}\) \(\sqrt{1-x^{2}}\) + C

= \(\frac{x}{2}\) cos^-1 x – \(\frac{1}{2}\) \(\sqrt{1-x^{2}}\) + C

= \(\frac{1}{2}\) [x cos^-1 x – \(\sqrt{1-x^{2}}\)) + C

Question 24.
\(\frac{\sqrt{x^{2}+1\left[\log \left(x^{2}+1\right)-2 \log x\right]}}{x^{4}}\)
Solution.

Direction (25 – 33): Evaluate the definite integral.

Question 25.
\(\cdot \int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right)\) dx
Solution.

Question 26.
\(\int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x}\) dx
Solution.
Let I = \(\int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x}\) dx

= ∫ \(\int_{0}^{\frac{\pi}{4}} \frac{\cos ^{4} x}{\frac{\cos ^{4} x+\sin ^{4} x}{\cos ^{4} x}}\) dx

= ∫ \(\int_{0}^{\frac{\pi}{4}} \frac{\tan x \sec ^{2} x}{1+\tan ^{4} x}\) dx

Let tan² x = t
⇒ 2 tan x sec² x dx = dt
When x = 0, t = 0 and when x = \(\frac{\pi}{4}\), t = 1
∴ I = \(\frac{1}{2}\left[\tan ^{-1} t\right]_{0}^{1}\)

= \(\frac{1}{2}\) [tan^-1 1 – tan^-21 0]

= \(\frac{1}{2}\left[\frac{\pi}{4}\right]=\frac{\pi}{8}\)

Question 27.
\(\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x d x}{\cos ^{2} x+4 \sin ^{2} x}\) dx
Solution.

Let 2 tan x = t
⇒ 2 sec² x dx = dt
When x = 0, t = 0 and when x = \(\frac{\pi}{2}\), t = ∞
⇒ I = \(\int_{0}^{\frac{\pi}{2}} \frac{2 \sec ^{2} x}{1+4 \tan ^{2} x} d x=\int_{0}^{\infty} \frac{d t}{1+t^{2}}\)

= \(\left[\tan ^{-1} t\right]_{0}^{\infty}\)

= [tan (∞) – tan (o)] = \(\frac{\pi}{2}\)
Therefore, from Eq. Ci), we get
I = \(-\frac{\pi}{6}+\frac{2}{3}\left[\frac{\pi}{2}\right]=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}\)

Question 28.
\(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}}\)
Solution.
Let I = \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}}\) dx

= \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x+\cos x)}{\sqrt{-(-\sin 2 x)}}\) dx

Question 29.
\(\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{x}}\)
Solution.

Question 30.
\(\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x}\) dx
Solution.
Let I = \(\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x}\) dx
Also, let sin x – cos x = t
⇒ (cos x + sin x) dx = dt
When x = 0, t= – 1 and when x =\(\frac{\pi}{4}\), t = 0
⇒ (sin x – cosx)² = t²
⇒ sin² x + cos² x – 2 sin x cos x = t²
⇒ 1 – sin 2x = t²
⇒ sin 2x = 1 – t²

Question 31.
\(\int_{0}^{\frac{\pi}{2}}\) sin 2x tan^-1 (sin x) dx
Solution.
Let I = \(\int_{0}^{\frac{\pi}{2}}\) sin 2x tan^-1 (sin x) dx

= 2 \(\int_{0}^{\frac{\pi}{2}}\) sin x cos x tan^-1 (sin x) dx

Let sin x = t
cos dx = dt
when x = 0 , t = 0 and when x = \(\frac{\pi}{2}\), then t = 1

Question 32.
\(\int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x}\) dx
Solution.

Question 33.
\(\int_{1}^{4}\) [|x – 1| + |x – 2| + |x – 3|] dx
Solution.

Direction (34 – 39): Prove the following:

Question 34.
\(\int_{1}^{3} \frac{d x}{x^{2}(x+1)}=\frac{2}{3}+\log \frac{2}{3}\)
Solution.
Let I = \(\int_{1}^{3} \frac{d x}{x^{2}(x+1)}\)

Also, let \(\frac{1}{x^{2}(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}\)

1 = Ax (x + 1) + B(x + 1) + C (x²)
1 = Ax²) + Ax + Bx + B + Cx²)
Equating the coefficients of x²), x and constant term, we get
A + C = 0;
A + B = 0;
B = 1
On solving these equations, we get
A = – 1; C = 1 and B = 1

Hence, the given result is proved.

Question 35.
\(\int_{0}^{1}\) x e^x dx = 1
Solution.
Let I = \(\int_{0}^{1}\) x e^x dx dx
Integrating by parts, we get
I = \(x \int_{0}^{1} e^{x} d x-\int_{0}^{1}\left\{\left(\frac{d}{d x}(x)\right) \int e^{x} d x\right\}\) dx

= \(\left[x e^{x}\right]_{0}^{1}\) – \(\int_{0}^{1}\) e^x dx

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

= e – e + 1 = 1

Hence, the given result is proved.

Question 36.
\(\int_{-1}^{1}\) x¹⁷ cos⁴ x dx
Solution.
Let I = \(\int_{-1}^{1}\) x¹⁷ cos⁴ x dx
Also, f(x) = (- x)¹⁷ cos⁴ (- x)
= – x¹⁷ cos⁴ x = – f(x)
Therefore, f(x) is an odd function.
We know that if f(x) is an odd function, then \(\int_{-a}^{a}\) f(x) dx = o
∴ I = \(\int_{-1}^{1}\) x¹⁷ cos⁴ x dx = o
Hence, the given result is proved.

Question 37.
\(\int_{0}^{\frac{\pi}{2}}\) sin² x dx = \(\frac{2}{3}\)
Solution.
Let I = \(\int_{0}^{\frac{\pi}{2}}\) sin² x dx

= \(\int_{0}^{\frac{\pi}{2}}\) sin³ x sin x dx

= \(\int_{0}^{\frac{\pi}{2}}\) (1 – cos² x) sin x dx

= \(\int_{0}^{\frac{\pi}{2}}\) sin x dx – \(\int_{0}^{\frac{\pi}{2}}\) cos² x sin x dx

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

= \(1+\frac{1}{3}[-1]=1-\frac{1}{3}=\frac{2}{3}\)
Hence, the given result is proved.

Question 38.
\(\int_{0}^{\frac{\pi}{4}}\) 2 tan³ x dx = 1 – log 2
Solution.
Let I = \(\int_{0}^{\frac{\pi}{4}}\) 2 tan³ x dx

= 2 \(\int_{0}^{\frac{\pi}{4}}\) tan² x tan x dx

= 2 \(\int_{0}^{\frac{\pi}{4}}\) (sec² x – 1) tan x dx

= 2 \(\int_{0}^{\frac{\pi}{4}}\) sec² x tan x dx – 2 \(\int_{0}^{\frac{\pi}{4}}\) tan x dx
= \(2\left[\frac{\tan ^{2} x}{2}\right]_{0}^{\frac{\pi}{4}}+2[\log \cos x]_{0}^{\frac{\pi}{4}}\)

=1 + 2 [log cos \(\frac{\pi}{4}\) – log cos 0]
= 1 + 2 [log \(\frac{1}{\sqrt{2}}\) – log 1]
= 1 – log 2 – log 1
= 1 – log 2
Hence, the given result is proved.

Question 39.
\(\int_{0}^{1} \sin ^{-1} x d x=\frac{\pi}{2}-1\)
Solution.
Let I = \(\int_{0}^{1}\) sin^-1 x dx
= \(\int_{0}^{1}\) sin^-1 x . 1. dx

Hence, the given result is proved.

Question 40.
Eva1uate \(\int_{0}^{1}\) e^{2 – 3x} dx as a limit of a sum.
Solution.
Let I = \(\int_{0}^{1}\) e^{2 – 3x} dx
We know that,

Direction (41 – 44): Choose the correct answer:

Question 41.
\(\int \frac{d x}{e^{x}+e^{-x}}\) is equal to
(A) tan^-1 (e^x) + C
(B) tan^-1 (e^x) + C
(C) log (e^x – e^-x) + C
(D) log (e^x + e^-x) + C
Solution.
Let I = \(\int \frac{d x}{e^{x}+e^{-x}}\)
= ∫ \(\frac{e^{x}}{e^{2 x}+1}\) dx
Also, let e^x = t
⇒ e^x dx = dt
∴ I = ∫ \(\frac{d t}{1+t^{2}}\) tan^-1 t + C
= tan^-1 (e^x) + C
Hence, the correct answer is (A).

Question 42.
∫ \(\frac{\cos 2 x}{(\sin x+\cos x)^{2}}\) dx is equal to

(A) \(\frac{-1}{\sin x+\cos x}\)

(B) log |sin x + cos x| + C

(C) log |sin x – cos x| + C

(D) \(\frac{1}{(\sin x+\cos x)^{2}}\)
Solution.

Let cos x + sin x = t
⇒ (cos x – sin x) dx = dt
∴ I = \(\int \frac{d t}{t}\)
= log |t| + C
= log |cos x+ sin x| + C
Hence, the correct answer is (B).

Question 43.
If f(a + b – x) = f(x), then \(\int_{a}^{b}\) x f(x) dx is equal to
(A) \(\frac{a+b}{2} \int_{a}^{b} f(b-x)\) dx

(B) \(\frac{a+b}{2} \int_{a}^{b} f(b+x)\) dx

(C) \(\frac{b-a}{2} \int_{a}^{b} f(x)\) dx

(D) \(\frac{b+a}{2} \int_{a}^{b} f(x)\) dx
Sol.
Let I = \(\int_{a}^{b}\) x f(x) dx …………….(i)
= \(\int_{a}^{b}\) (a + b – x) f (a + b – x) dx
[∵ \(\int_{a}^{b}\) f(x) dx = \(\int_{a}^{b}\) f(a + b – x) dx]
= \(\int_{a}^{b}\) (a + b – x) f(x) dx
⇒ I = (a + b) \(\int_{a}^{b}\) f(x) dx – I [Using Eq.(i)]
⇒ I = (a + b) \(\int_{a}^{b}\) f(x) dx
⇒ 2I = (a + b) \(\int_{a}^{b}\) f(x) dx
= \(\left(\frac{a+b}{2}\right) \int_{a}^{b} f(x) d x\)
Hence, the correct answer is (D).

Question 44.
The value of \(\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right)\) dx is
(A) 1
(B) 0
(C) – 1
(D) \(\frac{\pi}{4}\)
Solution.

Adding Eqs. (i) and (ii), we get
2I = \(\int_{0}^{1}\) (tan^-1 x + tan^-1 (1 – x) – tan^-1 (1 – x) – tan^-1 x dx
⇒ 2I = 0
⇒ I = 0
Hence, the correct answer is (B).