ANSWERS TO QUESTIONS FOR REVISION

CLASS XII : MATHEMATICS

TOPICS: APPLICATION OF INTEGRATION, DIFFERENTIAL EQUATIONS

PREPARED BY
M SRINIVASAN, PGT(MATHS), KVS
1. The area of the region bounded by the curve 𝒚 = 𝒙 − 𝟐 , 𝒙 = 𝟏, 𝒙 = 𝟑 and 𝒙-axis is
a) 4 b) 2 c) 3 d) 1
𝟑 𝒙 − 𝟐 ,𝒙 − 𝟐 > 𝟎
Area = ‫𝒙𝒅 𝒚 𝟏׬‬ 𝒙 − 𝟐 = ቊ
− 𝒙 − 𝟐 ,𝒙 − 𝟐 < 𝟎

𝟑 𝒙 − 𝟐 ,𝒙 > 𝟐
𝒙 −𝟐 =ቊ
= ‫ 𝒙 𝟏׬‬− 𝟐 𝒅𝒙 − 𝒙 − 𝟐 ,𝒙 < 𝟐

𝟐 𝟑
= ‫ 𝟏׬‬−(𝒙 − 𝟐) 𝒅𝒙 + ‫ 𝒙( 𝟐׬‬− 𝟐) 𝒅𝒙
𝟐 𝟐 𝟐 𝟑 𝒙 −𝟐 𝟐 𝟐 𝒙 −𝟐 𝟐 𝟑
𝒙 −𝟐 𝒙 −𝟐
=− + =− +
𝟐 𝟐 𝟐 𝟏
𝟐 𝟐
𝟏 𝟐
𝟏 𝟏 𝟏 𝟏
=− 𝟎−𝟏 + 𝟏 −𝟎 = +
𝟐 𝟐 𝟐 𝟐

d) 1
 ‫𝟑‬
‫𝟏׬‬ ‫𝟏 = 𝒙𝒅 𝟐 ‪𝒙 −‬‬
2. The area enclosed in the first quadrant and bounded by the circle 𝒙𝟐 + 𝒚𝟐 = 𝟒 and
the lines 𝒙 = 𝟎 and 𝒙 = 𝟐 is
𝝅 𝝅 𝝅
a) 𝝅 b) c) d)
𝟐 𝟑 𝟒
𝟐
Area = ‫𝒙𝒅 𝒚 𝟎׬‬ 𝒙𝟐 + 𝒚𝟐 = 𝟒 ⟹ 𝒚𝟐 = 𝟒 − 𝒙𝟐 𝒚= 𝟒 − 𝒙𝟐
𝟐
Area = ‫𝟎׬‬ 𝟒 − 𝒙𝟐 𝒅𝒙
𝟐
𝒙 𝟐
𝟒 −𝟏
𝒙
= 𝟒 − 𝒙 + 𝒔𝒊𝒏
𝟐 𝟐 𝟐 𝟎

= 𝟎 + 𝟐𝒔𝒊𝒏−𝟏 (𝟏) − 𝟎 + 𝟐𝒔𝒊𝒏−𝟏 𝟎

𝝅
=𝟐 ×
𝟐
a) 𝝅
 𝟐
Area = ‫𝟎׬‬ 𝟒 − 𝒙𝟐 𝒅𝒙 = 𝝅
 𝟑
𝒅𝒚 𝟐 𝟐 𝒅𝟐 𝒚
3. The degree of the differential equation 𝟏 + = is
𝒅𝒙 𝒅𝒙𝟐
𝟑
a) 4 𝟑
b) c) 2 d) Not defined
𝟐
𝟐 𝟐
𝒅𝒚 𝒅𝟐 𝒚
𝟏+ =
𝒅𝒙 𝒅𝒙𝟐
𝟐 𝟑 𝟐 𝟐
𝒅𝒚 𝒅 𝒚
⟹ 𝟏+ =
𝒅𝒙 𝒅𝒙𝟐

𝒅𝟐 𝒚
Highest order derivative is and its exponent is 2
𝒅𝒙𝟐
Order of Differential Equation = 2 and Degree = 2

c) 𝟐
 𝒅𝒚
4. Integrating factor of 𝒙 − 𝒚 = 𝒙𝟒 − 𝟑𝒙 is
𝒅𝒙
𝟏
a) 𝒙 b) 𝒍𝒐𝒈 𝒙 c) d) – 𝒙
𝒙
𝒅𝒚
𝒙 − 𝒚 = 𝒙𝟒 − 𝟑𝒙
𝒅𝒙
𝒅𝒚 𝒚 𝒙𝟒 − 𝟑𝒙 𝒅𝒚
+ 𝑷𝒚 = 𝑸
− = 𝒅𝒙
𝒅𝒙 𝒙 𝒙
𝟏
𝑷=− 𝑸 = 𝒙𝟑 − 𝟑
𝒙 𝟏
‫ ׬‬−𝒙 𝒅𝒙
Integrating Factor = 𝒆‫= 𝒙𝒅𝑷 ׬‬ 𝒆 𝟏
−𝒍𝒐𝒈 𝒙 𝒍𝒐𝒈 𝒙 −𝟏 𝒍𝒐𝒈
=𝒆 =𝒆 = 𝒆 𝒙

𝟏
c)
𝒙
 𝒅𝒙 𝒅𝒚
5. The solution of Differential Equation + =𝟎
𝒙 𝒚
𝟏 𝟏
a) + = 𝑪 b) 𝒍𝒐𝒈 𝒙 – 𝒍𝒐𝒈 𝒚 = 𝑪 c) 𝒙𝒚 = 𝑪 d) 𝒙 + 𝒚 = 𝑪
𝒙 𝒚

𝒅𝒙 𝒅𝒚
+ =𝟎
𝒙 𝒚
𝒅𝒚 𝒅𝒙
=−
𝒚 𝒙
𝒅𝒚 𝒅𝒙
න
𝒚
= −න ⟹ 𝒍𝒐𝒈 𝒚 = − 𝒍𝒐𝒈 𝒙 + 𝒍𝒐𝒈 𝑪
𝒙
⟹ 𝒍𝒐𝒈 𝒙 + 𝒍𝒐𝒈 𝒚 = 𝒍𝒐𝒈 𝑪
⟹ 𝒍𝒐𝒈 𝒙𝒚 = 𝒍𝒐𝒈 𝑪 𝒆𝐥𝐨𝐠 𝒙𝒚 = 𝒆𝐥𝐨𝐠 𝑪

c) 𝒙𝒚 = 𝒄
6. Assertion : The area of the region enclosed by the curve 𝒚 = 𝒙𝟑 and the lines y = 0
and y = 8 is 12.
Reason : The area of the region enclosed by the curve 𝒚 = 𝒇(𝒙), x- axis, the lines
𝒃
𝒙 = 𝒂 and 𝒙 = 𝒃 is given by ‫𝒙𝒅 𝒚 𝒂׬‬
Assertion
𝟖 𝟖 𝟏
Area of the region enclosed by the curve 𝒚 = 𝒙 and
𝟑
= න 𝒙 𝒅𝒚 = න 𝒚𝟑 𝒅𝒚
the lines y = 0 and y = 8 𝟒 𝟖 𝟎 𝟎
𝒚𝟑
=
𝟒
𝟑 𝟎

𝟑 𝟒 𝟑
= 𝟖 𝟑 −𝟎 = 𝟏𝟔
𝟖 𝟒 𝟒
න 𝒙 𝒅𝒚 = 𝟏𝟐
𝟎
 𝟖
න 𝒙 𝒅𝒚 = 𝟏𝟐
𝟎
Area of the region enclosed by the curve 𝒚 = 𝒙𝟑 and the lines y = 0 and y = 8 is 12

Assertion is True
Reason
The area of the region enclosed by the curve 𝒚 = 𝒇(𝒙), x- axis, the lines 𝒙 = 𝒂
𝒃
and 𝒙 = 𝒃 is given by ‫𝒙𝒅 𝒚 𝒂׬‬

Reason is True but not used to prove Assertion

b) Both A and R are true but R is not the correct
explanation of A
 𝒅𝒙
7. Assertion : The integrating factor of differential equation 𝐜𝐨𝒕 𝒚 + 𝒙 = 𝒔𝒆𝒄𝟐 𝒚 is sec 𝒚
𝒅𝒚
𝒅𝒙
Reason: For the linear differential equation + 𝑹𝒙 = 𝑺, R and S are functions of y, the integrating
𝒅𝒚
factor is 𝒆‫𝒚𝒅𝑹 ׬‬
𝒅𝒙
Assertion 𝐜𝐨𝒕 𝒚 + 𝒙 = 𝒔𝒆𝒄𝟐 𝒚
𝒅𝒚
𝒅𝒙 𝒙 𝒔𝒆𝒄𝟐 𝒚 𝒅𝒙 𝒔𝒆𝒄𝟐 𝒚
 + =  + 𝒕𝒂𝒏𝒚 𝒙 =
𝒅𝒚 𝒄𝒐𝒕 𝒚 𝒕𝒂𝒏 𝒚 𝒅𝒚 𝒕𝒂𝒏 𝒚
𝒅𝒙
The differential equation is of the form + 𝑹𝒙 = 𝑺
𝒅𝒚

𝑹 = 𝐭𝐚𝐧 𝒚 𝒔𝒆𝒄𝟐 𝒚
𝑺=
𝒕𝒂𝒏 𝒚
‫𝒚𝒅𝑹 ׬‬
Integrating factor = 𝒆
 ‫𝒚𝒅𝑹 ׬‬
Integrating factor = 𝒆 𝑹 = 𝐭𝐚𝐧 𝒚

Integrating factor = 𝒆 ‫𝒚𝒅 𝒚 𝒏𝒂𝒕 ׬‬ = 𝒆𝒍𝒐𝒈 (𝒔𝒆𝒄 𝒚)

Integrating factor= 𝒔𝒆𝒄 𝒚 Assertion is True
𝒅𝒙
For the linear differential equation + 𝑹𝒙 = 𝑺, R and S are functions of y,
𝒅𝒚
Reason the integrating factor is 𝒆‫𝒚𝒅𝑹 ׬‬

Reason is True and used to prove Assertion

b) Both A and R are true and R is the correct

explanation of A
 𝒅𝒚
8. Solve the differential equation : = 𝟏 + 𝒙 + 𝒚 + 𝒙𝒚
𝒅𝒙
𝒅𝒚
= 𝟏 + 𝒙 + 𝒚 + 𝒙𝒚
𝒅𝒙
𝒅𝒚
= 𝟏 + 𝒙 + 𝒚(𝟏 + 𝒙)
𝒅𝒙
𝒅𝒚
= (𝟏 + 𝒙)(𝟏 + 𝒚)
𝒅𝒙
𝒅𝒚
= 𝟏 + 𝒙 𝒅𝒙
(𝟏 + 𝒚)
 𝒅𝒚
= 𝟏 + 𝒙 𝒅𝒙
(𝟏 + 𝒚)
Integrating both sides
𝒅𝒚
න = න 𝟏 + 𝒙 𝒅𝒙
(𝟏 + 𝒚)
𝟐
𝒙
𝒍𝐨𝐠 𝟏 + 𝒚 = 𝒙 + +𝑪
𝟐
9. Solve the differential equation 𝟏 + 𝒚𝟐 𝟏 + 𝒍𝒐𝒈 𝒙 𝒅𝒙 + 𝒙𝒅𝒚 = 𝟎
given that when 𝒙 = 𝟏, 𝒚 = 𝟏
𝟏 + 𝒚𝟐 𝟏 + 𝒍𝒐𝒈 𝒙 𝒅𝒙 + 𝒙𝒅𝒚 = 𝟎

𝒙𝒅𝒚 = − 𝟏 + 𝒚𝟐 𝟏 + 𝒍𝒐𝒈 𝒙 𝒅𝒙
𝒅𝒚 𝟏 + 𝒍𝒐𝒈 𝒙
𝟐
=− 𝒅𝒙
𝟏+𝒚 𝒙
Integrating both sides
𝒅𝒚 𝟏 + 𝒍𝒐𝒈 𝒙
න 𝟐
= −න 𝒅𝒙
𝟏+𝒚 𝒙
 𝒅𝒚 𝟏 + 𝒍𝒐𝒈 𝒙 Let 𝒕 = (𝟏 + 𝐥𝐨𝐠 𝒙)
න 𝟐
= −න 𝒅𝒙
𝟏+𝒚 𝒙 𝒅𝒙
𝒅𝒕 =
𝒅𝒚 𝒙
න 𝟐
= − න 𝒕 𝒅𝒕
𝟏+𝒚
𝟐
𝒕
𝒕𝒂𝒏−𝟏 𝒚 = − + 𝑪
𝟐
𝟐
−𝟏
𝟏 + 𝒍𝒐𝒈 𝒙
𝒕𝒂𝒏 𝒚 = − +𝑪
𝟐
𝟐
−𝟏
𝟏 + 𝒍𝒐𝒈 𝒙
𝒕𝒂𝒏 𝒚 + =𝑪
𝟐
 𝟐
−𝟏
𝟏 + 𝒍𝒐𝒈 𝒙
𝒕𝒂𝒏 𝒚 + =𝑪
𝟐
Given that when 𝒙 = 𝟏, 𝒚 = 𝟏
𝟐
𝟏 + 𝒍𝒐𝒈 (𝟏) 𝝅 𝟏
𝒕𝒂𝒏−𝟏 (𝟏) + =𝑪 + =𝑪
𝟐 𝟒 𝟐
The particular solution is
𝟐
−𝟏
𝟏 + 𝒍𝒐𝒈 𝒙 𝝅 𝟏
𝒕𝒂𝒏 𝒚 + = +
𝟐 𝟒 𝟐
𝝅 𝟏 𝟏 𝟐
𝒚 = 𝐭𝐚𝐧 + − 𝟏 + 𝐥𝐨𝐠 𝒙
𝟒 𝟐 𝟐
10. Find the equation of the curve through the point (𝟏 , 𝟎) if the slope of
𝒚−𝟏
the tangent to the curve at any point (𝒙 , 𝒚) is 𝟐
𝒙 +𝒙
𝒚−𝟏
The slope of the tangent to the curve at any point (𝒙, 𝒚) is
𝒙𝟐 +𝒙
𝒅𝒚 𝒚−𝟏
= 𝟐
𝒅𝒙 𝒙 + 𝒙
𝒅𝒚 𝒅𝒙
= 𝟐
𝒚−𝟏 𝒙 +𝒙
Integrating both sides
𝒅𝒚 𝒅𝒙
න =න 𝟐
𝒚−𝟏 𝒙 +𝒙
 𝒅𝒚 𝒅𝒙
න =න 𝟐
𝒚−𝟏 𝒙 +𝒙

𝒅𝒚 𝒅𝒙
න =න 𝟐
𝒚−𝟏 𝒙 +𝒙
𝒅𝒚 𝒅𝒙
න =න
𝒚−𝟏 𝒙(𝒙 + 𝟏)
𝒅𝒚 (𝒙 + 𝟏) − 𝒙
න =න 𝒅𝒙
𝒚−𝟏 𝒙(𝒙 + 𝟏)
𝒅𝒚 (𝒙 + 𝟏) 𝒙
න =න − 𝒅𝒙
𝒚−𝟏 𝒙(𝒙 + 𝟏) 𝒙(𝒙 + 𝟏)
 𝒅𝒚 (𝒙 + 𝟏) 𝒙
න =න − 𝒅𝒙
𝒚−𝟏 𝒙(𝒙 + 𝟏) 𝒙(𝒙 + 𝟏)
𝒅𝒚 𝟏 𝟏
න =න − 𝒅𝒙
𝒚−𝟏 𝒙 (𝒙 + 𝟏)

𝒍𝒐𝒈 𝒚 − 𝟏 = 𝐥𝐨𝐠 𝒙 − 𝐥𝐨𝐠 𝒙 + 𝟏 + 𝐥𝐨𝐠 𝑪

𝑪𝒙
𝒍𝒐𝒈 𝒚 − 𝟏 = 𝐥𝐨𝐠
𝒙+𝟏
𝑪𝒙
𝒚 −𝟏=
𝒙+𝟏
 𝑪𝒙
𝒚 −𝟏=
𝒙+𝟏
The curve passes through the point (𝟏 , 𝟎)
𝑪(𝟏)
𝟎 −𝟏=
𝟏+𝟏
𝑪 = −𝟐 −𝟐𝒙
The equation of the curve is:
𝒚 − 𝟏 =
𝒙+𝟏
𝟐𝒙
𝒚 =𝟏−
𝒙+𝟏
8. Find the area of the region bounded by 𝒚 = 𝟐𝒙 − 𝒙𝟐 and the
line 𝒚 = − 𝒙
𝟐
𝒚 = 𝟐𝒙 − 𝒙 Downward Parabola
𝒚 = −𝒙 Line though the origin
𝟐
⟹ −𝒙 = 𝟐𝒙 − 𝒙
𝟐
⟹ 𝒙 − 𝟑𝒙 = 𝟎
⟹ 𝒙(𝒙 − 𝟑) = 𝟎
⟹ 𝒙 = 𝟎, 𝟑
The curve and the line meet for 𝒙 = 𝟎, 𝟑
The curve and the line meet for 𝒙 = 𝟎, 𝟑
𝟑
Area = ‫ 𝒆𝒗𝒓𝒖𝒄 𝒆𝒑𝒑𝒖 𝒎𝒐𝒓𝒇 𝒚 𝟎׬‬− 𝒚 𝒇𝒓𝒐𝒎 𝒍𝒐𝒘𝒆𝒓 𝒄𝒖𝒓𝒗𝒆 𝒅𝒙
𝟑
Area = ‫𝟎׬‬ 𝒚 𝒇𝒓𝒐𝒎 𝒑𝒂𝒓𝒂𝒃𝒐𝒍𝒂 − 𝒚 𝒇𝒓𝒐𝒎 𝒍𝒊𝒏𝒆 𝒅𝒙
𝟑 𝟐
Area = ‫ 𝒙𝟐( 𝟎׬‬− 𝒙 ) − (−𝒙) 𝒅𝒙
𝟑 𝟐
Area = ‫𝟎׬‬ 𝟑𝒙 − 𝒙 𝒅𝒙
𝟐 𝟑 𝟑
𝟑𝒙 𝒙 𝟐𝟕 𝟐𝟕 𝟐𝟕 𝟗
Area = − = 𝟎 −
𝟐
−
𝟑
= 𝟎 −
𝟐
−𝟗 =
𝟐
𝟐 𝟑 𝟎
𝟗
The area of the region bounded by 𝒚 = 𝟐𝒙 − 𝒙𝟐 and the line 𝒚 = − 𝒙 is
𝟐
 𝟗
The area of the region bounded by 𝒚 = 𝟐𝒙 − 𝒙𝟐 and the line 𝒚 = − 𝒙 is
𝟐
9. Using integration, find the area of the region
𝒙, 𝒚 : 𝒙𝟐 + 𝒚𝟐 ≤ 𝟗, 𝒙 + 𝒚 ≥ 𝟑

𝒙𝟐 + 𝒚𝟐 ≤ 𝟗
Area inside the circle 𝒙𝟐 + 𝒚𝟐 = 𝟗
𝒙+𝒚≥𝟑
Area above the line 𝒙 + 𝒚 = 𝟑

Area of the region 𝒙, 𝒚 : 𝒙𝟐 + 𝒚𝟐 ≤ 𝟗, 𝒙 + 𝒚 ≥ 𝟑

 𝟐
Area = ‫𝟎׬‬ 𝒚 𝒇𝒓𝒐𝒎 𝒖𝒑𝒑𝒆 𝒄𝒖𝒓𝒗𝒆 − 𝒚 𝒇𝒓𝒐𝒎 𝒍𝒐𝒘𝒆𝒓 𝒄𝒖𝒓𝒗𝒆 𝒅𝒙
𝟐
Area = ‫𝟎׬‬ 𝒚 𝒇𝒓𝒐𝒎 𝒄𝒊𝒓𝒄𝒍𝒆 − 𝒚 𝒇𝒓𝒐𝒎 𝒍𝒊𝒏𝒆 𝒅𝒙

𝟑
Area = ‫𝟎׬‬ 𝟒 − 𝒙𝟐 − (𝟐 − 𝒙) 𝒅𝒙
𝟑 𝟑
Area = ‫𝟎׬‬ 𝟒− 𝒙𝟐 𝒅𝒙 − ‫𝟎׬‬ (𝟐 − 𝒙) 𝒅𝒙

𝟐 𝟐
𝒙 𝟒 −𝟏 𝒙 𝟐−𝒙 𝟐
Area = 𝟒 − 𝒙𝟐 + 𝒔𝒊𝒏 −
𝟐 𝟐 𝟎𝟐 𝟐(−𝟏)
𝟎
𝟐
𝒙 𝟒 𝒙 𝟏 𝟐 𝟐
Area = 𝟒 − 𝒙𝟐 + 𝒔𝒊𝒏−𝟏 + 𝟐−𝒙 𝟎
𝟐 𝟐 𝟐 𝟎 𝟐
 𝟐
𝒙 𝟐 𝟒 −𝟏 𝒙 𝟏 𝟐 𝟐
Area = 𝟒− 𝒙 + 𝒔𝒊𝒏 + 𝟐−𝒙 𝟎
𝟐 𝟐 𝟐 𝟎 𝟐
𝟐 𝟒 −𝟏 𝟐 −𝟏 𝟏
Area = 𝟒− 𝒙𝟐 + 𝒔𝒊𝒏 − 𝟎 + 𝟐𝒔𝒊𝒏 𝟎 + 𝟎−𝟒
𝟐 𝟐 𝟐 𝟐

−𝟏
Area = 𝟎 + 𝟐𝒔𝒊𝒏 𝟏 − 𝟎 −𝟐
𝝅
Area = 𝟐 × − 𝟐
𝟐

Area = 𝝅 − 𝟐

The area of the region 𝒙, 𝒚 : 𝒙𝟐 + 𝒚𝟐 ≤ 𝟗, 𝒙 + 𝒚 ≥ 𝟑 = 𝝅 − 𝟐

The area of the region 𝒙, 𝒚 : 𝒙𝟐 + 𝒚𝟐 ≤ 𝟗, 𝒙 + 𝒚 ≥ 𝟑 = 𝝅 − 𝟐
 𝒅𝒚 𝒚 𝒚
13. Solve the differential equation : 𝒙 𝒔𝒊𝒏 + 𝒙 − 𝒚 𝒔𝒊𝒏 = 𝟎,
𝒅𝒙 𝒙 𝒙
𝝅
𝒚 𝟏 =
𝟐
𝒅𝒚 𝒚 𝒚
𝒙 𝒔𝒊𝒏 + 𝒙 − 𝒚 𝒔𝒊𝒏 =𝟎
𝒅𝒙 𝒙 𝒙

𝒅𝒚 𝒚 𝒚 𝒚
⟹ 𝒔𝒊𝒏 + 𝟏 − 𝒔𝒊𝒏 =𝟎
𝒅𝒙 𝒙 𝒙 𝒙
𝒅𝒚 𝒚 𝒚 𝒚
⟹ 𝒔𝒊𝒏 = 𝒔𝒊𝒏 −𝟏
𝒅𝒙 𝒙 𝒙 𝒙
𝒚 𝒚
𝒅𝒚 𝒔𝒊𝒏 −𝟏 𝒅𝒚 𝒚 Homogenous
⟹ =− 𝒙 𝒙
𝒅𝒙 𝒚 =𝒇
𝒔𝒊𝒏 𝒅𝒙 𝒙 Differential Equation
𝒙
 𝒚 𝒚
𝒅𝒚 𝒔𝒊𝒏 −𝟏
= −𝒙 𝒙
𝒚
𝒅𝒙 𝒔𝒊𝒏
𝒙
𝒚 𝒅𝒚 𝒅𝒗
=𝒗 =𝒗+𝒙
𝒙 𝒅𝒙 𝒅𝒙
𝒅𝒗 𝒗 𝒔𝒊𝒏 𝒗 − 𝟏
𝒗+𝒙 =
𝒅𝒙 𝒔𝒊𝒏 𝒗
𝒅𝒗 𝒗 𝒔𝒊𝒏 𝒗 − 𝟏
𝒙 = −𝒗
𝒅𝒙 𝒔𝒊𝒏 𝒗
𝒅𝒗 𝒗 𝒔𝒊𝒏 𝒗 − 𝟏 − 𝒗 𝒔𝒊𝒏 𝒗
𝒙 =
𝒅𝒙 𝒔𝒊𝒏 𝒗
 𝒅𝒗 𝒗 𝒔𝒊𝒏 𝒗 − 𝟏 − 𝒗 𝒔𝒊𝒏 𝒗
𝒙 =
𝒅𝒙 𝒔𝒊𝒏 𝒗
𝒅𝒗 −𝟏 𝒅𝒙
𝒙 = 𝒔𝒊𝒏 𝒗 𝒅𝒗 = −
𝒅𝒙 𝒔𝒊𝒏 𝒗 𝒙

Integrating both sides

𝒅𝒙
න 𝒔𝒊𝒏 𝒗 𝒅𝒗 = − න
𝒙
− 𝐜𝐨𝐬 𝒗 = − 𝐥𝐨𝐠 𝒙 + 𝑪
𝒚
− 𝐜𝐨𝐬 = − 𝐥𝐨𝐠 𝒙 + 𝑪
𝒙
 𝒚
− 𝐜𝐨𝐬 = − 𝐥𝐨𝐠 𝒙 + 𝑪
𝒙
𝒚
𝒍𝐨𝐠 𝒙 − 𝒄𝒐𝒔 =𝑪
𝒙
𝝅 𝝅
Given 𝒚 𝟏 =  when 𝒙 = 𝟏, 𝒚 =
𝟐 𝟐
𝝅
𝒍𝐨𝐠 𝟏 − 𝒄𝒐𝒔 =𝑪 ⟹𝑪= 𝟎
𝟐
The particular solution is given by
𝒚 𝒚
𝒍𝐨𝐠 𝒙 − 𝒄𝒐𝒔 =𝟎 𝒄𝒐𝒔 = 𝒍𝒐𝒈 𝒙
𝒙 𝒙
 𝟐
𝒅𝒚
𝟏𝟒. 𝑺𝒐𝒍𝒗𝒆 ∶ 𝒄𝒐𝒔 𝒙 + 𝒚 = 𝒕𝒂𝒏 𝒙
𝒅𝒙
𝒅𝒚
𝒄𝒐𝒔𝟐 𝒙 + 𝒚 = 𝒕𝒂𝒏 𝒙
𝒅𝒙
𝒅𝒚 𝒚 𝒕𝒂𝒏𝒙
+ =
𝒅𝒙 𝒄𝒐𝒔𝟐 𝒙 𝒄𝒐𝒔𝟐 𝒙 Linear Differential Equation of the form
𝒅𝒚 𝒅𝒚
+ 𝒔𝒆𝒄𝟐 𝒙 𝒚 = 𝒕𝒂𝒏 𝒙 𝒔𝒆𝒄𝟐 𝒙 𝒅𝒙
+ 𝑷𝒚 = 𝑸
𝒅𝒙
𝑷 = 𝒔𝒆𝒄𝟐 𝒙 𝑸 = 𝒕𝒂𝒏 𝒙 𝒔𝒆𝒄𝟐 𝒙
‫𝒙𝒅𝑷 ׬‬ ‫𝒙𝒅𝒙 𝟐𝒄𝒆𝒔 ׬‬
Integrating factor = 𝒆 =𝒆
𝒕𝒂𝒏 𝒙
Integrating factor = 𝒆
Integrating factor = 𝒆𝒕𝒂𝒏 𝒙 𝑸 = 𝒕𝒂𝒏 𝒙 𝒔𝒆𝒄𝟐 𝒙
The general solution of the Differential Equation is
𝒚 𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒊𝒏𝒈 𝒇𝒂𝒄𝒕𝒐𝒓 = න 𝑸(𝒊𝒏𝒕𝒆𝒈𝒓𝒂𝒕𝒊𝒏𝒈 𝒇𝒂𝒄𝒕𝒐𝒓) 𝒅𝒙 + 𝑪
𝒚(𝒆𝒕𝒂𝒏 𝒙 ) = න 𝒕𝒂𝒏 𝒙 𝒔𝒆𝒄𝟐 𝒙(𝒆𝒕𝒂𝒏 𝒙 ) 𝒅𝒙 + 𝑪

Let 𝒕 = 𝐭𝐚𝐧 𝒙 𝒅𝒕 = 𝒔𝒆𝒄𝟐 𝒙 𝒅𝒙

𝒚(𝒆𝒕𝒂𝒏 𝒙 ) = න 𝒕 𝒆𝒕 𝒅𝒕 + 𝑪

𝒖=𝒕 𝒅𝒗 = 𝒆𝒕 𝒅𝒕
𝒚 𝒆𝒕𝒂𝒏 𝒙 = 𝒕 𝒆𝒕 − න 𝒆𝒕 𝒅𝒕 + 𝑪
𝒕
𝒅𝒖 = 𝒅𝒕 𝒗=𝒆
 𝒕𝒂𝒏 𝒙 𝒕 𝒕
𝒚 𝒆 = 𝒕 𝒆 − න 𝒆 𝒅𝒕 + 𝑪

𝒕𝒂𝒏 𝒙 𝒕 𝒕
𝒚 𝒆 =𝒕𝒆 −𝒆 +𝑪
𝒕𝒂𝒏 𝒙 𝒕𝒂𝒏 𝒙 𝒕𝒂𝒏 𝒙
𝒚 𝒆 = 𝒕𝒂𝒏 𝒙 𝒆 −𝒆 +𝑪

𝒚 𝒆 𝒕𝒂𝒏 𝒙 = 𝒆 𝒕𝒂𝒏 𝒙 𝒕𝒂𝒏 𝒙 − 𝟏 + 𝑪

15. At Gandhi Park, Chennai it is proposed to construct semi-
circular car parking at one of its corners
The circular edge of the floor of car parking is given by the
equation 𝒙𝟐 + 𝒚𝟐 = 𝟒𝟎𝟎 and the equation of the straight edge line
is given by 𝒚 = 𝟎
𝒂) Find the area of the floor of the car parking using integration
𝟐𝟎
Required Area= ‫׬‬−𝟐𝟎 𝒚 𝒅𝒙
𝟐𝟎
Required Area= 𝟐 ‫𝒙𝒅 𝒚 𝟎׬‬
𝟐𝟎
= 𝟐න 𝟒𝟎𝟎 − 𝒙𝟐 𝒅𝒙
𝟎
 𝟐𝟎
Required Area= 𝟐 ‫𝟎׬‬ 𝟒𝟎𝟎 − 𝒙𝟐 𝒅𝒙
𝟐𝟎
= 𝟐න 𝟐𝟎 𝟐 − 𝒙𝟐 𝒅𝒙
𝟎
𝟐𝟎
𝒙 𝟐
𝟒𝟎𝟎 −𝟏
𝒙
=𝟐 𝟒𝟎𝟎 − 𝒙 + 𝒔𝒊𝒏
𝟐 𝟐 𝟐𝟎 𝟎

= 𝟐 𝟎 + 𝟐𝟎𝟎 𝒔𝒊𝒏−𝟏 𝟏 − 𝟎 + 𝟐𝟎 𝒔𝒊𝒏−𝟏 𝟎

𝝅
= 𝟐 × 𝟐𝟎𝟎 ×
𝟐
Area of floor of the car parking = 𝟐𝟎𝟎 𝝅 Square units
 𝟐𝟎
Area = 𝟐 ‫𝟎׬‬ 𝟒𝟎𝟎 − 𝒙𝟐 𝒅𝒙 = 𝟐𝟎𝟎𝝅
b) Find the cost of flooring the car park at the rate of Rs.500/-
per square units. (Use  = 3.14)
Area of floor of the car parking = 𝟐𝟎𝟎 𝝅 Square units
Cost of flooring = 𝟐𝟎𝟎 𝝅 × 𝟓𝟎𝟎
= 𝟐𝟎𝟎 × 𝟑. 𝟏𝟒 × 𝟓𝟎𝟎
= 𝟑𝟏𝟒𝟎𝟎𝟎

The cost of flooring the car park = 𝑹𝒔. 𝟑𝟏𝟒𝟎𝟎𝟎