Uploaded by Lokesh Venkatesh

Question Bank

advertisement
QUESTION BANK FOR Calculus & Differential equations
Course CODE – 21MAT11
Course Outcomes
On completion of this course, students are able to:
CO1: Apply the knowledge of calculus to solve problems related to polar curves and its applications in
determining the bentness of a curve.
CO2: Lean the notion of partial differentiation to calculate rates of change of multivariate functions and
solve problems related to composite functions and Jacobians.
CO3: Solve first order linear/nonlinear differential equation analytically using standard methods.
CO4: Demonstrate various models through higher order differential equations and solve such linear
ordinary differential equations.
CO5: Make use of matrix theory for solving system of linear equations and compute eigen values and
eigenvectors required for matrix diagonalization process.
Prepared By
Dr. Lakshminarayanachari. K
Dr. Bhaskar. C
Prof. Naveena gn
Dr. Arunkumar. R
Dr. Madhura. K
Calculus and Differential Equations / 21MAT11
DEPARTMENT OF MATHEMATICS
MODULE – I: DIFFERENTIAL CALCULUS – I
1) With usual notation, prove that tan  ο€½ r
d
dr
[Aug 2021, June 2019, 2018, Dec 2018, 2019, 2017]
1
1
1
π‘‘π‘Ÿ 2
2) With usual notations prove that, the pedal equation of the form 𝑝2 = π‘Ÿ 2 + π‘Ÿ 4 (π‘‘πœƒ)
[Feb 2021, Aug 2020, Dec 2017]
3) Find the angle between the radius vector and the tangent for the following curves:
i) π‘Ÿ = π‘Ž(1 − cos πœƒ) at πœƒ = πœ‹⁄3
iii) π‘Ÿ π‘š = π‘Žπ‘š (cos π‘šπœƒ + sin π‘šπœƒ)
[Dec 2017]
ii) π‘Ÿπ‘ π‘’π‘ 2 (πœƒ⁄2) = 2π‘Ž
iv) π‘Ÿ 2 = π‘Ž2 sin 2πœƒ
4) Find the angle of intersection of the following curves:
π‘Žπœƒ
π‘Ž
i) π‘Ÿ = π‘Ž(1 + sin πœƒ) & π‘Ÿ = π‘Ž(1 − π‘ π‘–π‘›πœƒ) [June 2017]
ii) π‘Ÿ = 1+πœƒ & π‘Ÿ = 1+πœƒ2
[June 2016]
2
2
2
2
iii) π‘Ÿ = 2π‘ π‘–π‘›πœƒ & π‘Ÿ = 2(π‘ π‘–π‘›πœƒ + π‘π‘œπ‘ πœƒ) [June 2017]
iv) π‘Ÿ sin 2πœƒ = π‘Ž & π‘Ÿ cos 2πœƒ = 𝑏
v) π‘Ÿ(1 + π‘π‘œπ‘ πœƒ) = π‘Ž & π‘Ÿ(1 − π‘π‘œπ‘ πœƒ) = 𝑏 [Dec 2016]
vi) π‘Ÿ 𝑛 = π‘Žπ‘› (sin π‘›πœƒ + cos π‘›πœƒ) & π‘Ÿ 𝑛 = π‘Žπ‘› sin π‘›πœƒ
vii) π‘Ÿ 𝑛 = π‘Žπ‘› cos π‘›πœƒ & π‘Ÿ 𝑛 = 𝑏 𝑛 sin π‘›πœƒ [June 2018, Dec 2018]
viii) π‘Ÿ = π‘Žπ‘™π‘œπ‘”πœƒ & π‘Ÿ = π‘Ž/π‘™π‘œπ‘”πœƒ
[Aug 2021, Dec 2017, 2016, 2014, 2012]
ix) π‘Ÿ 2 sin 2πœƒ = 4 & π‘Ÿ 2 = 16𝑠𝑖𝑛2πœƒ
[Dec 2015, 2012, June 2015]
2
2
x) π‘Ÿ = π‘Ž(1 + π‘π‘œπ‘ πœƒ) & π‘Ÿ = π‘Ž cos 2πœƒ
[Dec 2016, June 2016]
xi) π‘Ÿ = π‘Ž(1 + sin πœƒ) & π‘Ÿ = π‘Ž(1 − cos πœƒ)
[Dec 2017, June 2016]
xii) π‘Ÿ = π‘Ž(1 + cos πœƒ) & π‘Ÿ = 𝑏(1 − cos πœƒ) [June 2019, 2018, 2016, 2015, 2009, Dec 2019, Dec 2017, 2015]
xiii) π‘Ÿ = sin πœƒ + cos πœƒ & π‘Ÿ = 2 sin πœƒ
[Dec 2019]
xiv) π‘Ÿ = 6 cos πœƒ & π‘Ÿ = 2(1 + cos πœƒ)
[Aug 2020]
xv) π‘Ÿ = 2 sin πœƒ & π‘Ÿ = 2 cos πœƒ
[Feb 2021]
5) Find the pedal equation for the following curves:
i) π‘Ÿ π‘š cos π‘šπœƒ = π‘Žπ‘š
[June 2016] ii) π‘Ÿ = 2π‘Ž⁄1 + cos πœƒ
[June 2018, 2011, Dec 2016]
𝑛
iii) π‘Ÿ = sec β„Ž π‘›πœƒ
[Dec 2014] iv) π‘Ÿ = π‘Ž + 𝑏 cos πœƒ
[June 2017]
𝑙
3
v) π‘Ÿ = π‘Ž sin 3πœƒ
vi) π‘Ÿ = 1 + 𝑒 cos πœƒ
vii) π‘Ÿ(1 − π‘π‘œπ‘ πœƒ) = 2π‘Ž
[Aug 2020, Dec 2012] viii) π‘Ÿ 𝑛 = π‘Ž(1 + π‘π‘œπ‘ π‘›πœƒ)
[Dec 2015]
2π‘Ž
2
2
ix) π‘Ÿ = 1 − sin πœƒ
x) π‘Ÿ = π‘Ž π‘π‘œπ‘ 2πœƒ
[Aug 2021, June 2014]
𝑛
𝑛
𝑛
𝑛
xi) π‘Ÿ sec π‘›πœƒ = π‘Ž or π‘Ÿ = π‘Ž cos π‘›πœƒ
[June 2019, Dec 2019, June 2017, Dec 2016, 2015, 2013]
πœƒπ‘π‘œπ‘‘π›Ό
xii) π‘Ÿ = π‘Žπ‘’
[Dec 2018]
xiii) π‘Ÿ = π‘Ž(1 + cos πœƒ)
[Aug 2021, June 2018, 2016, Dec 2017]
π‘š
π‘š (cos
xiv) π‘Ÿ = π‘Ž
π‘šπœƒ + sin π‘šπœƒ)
[Feb 2021, Dec 2016, 2011, June 2010]
6) Find the pedal equation of π‘Ÿ 𝑛 = π‘Žπ‘› sin π‘›πœƒ + 𝑏 𝑛 cos π‘›πœƒ in the form 𝑝2 (π‘Ž2𝑛 + 𝑏 2𝑛 ) = π‘Ÿ 2𝑛+2 .
[June 2015]
7) Derive Radius of Curvature in the Cartesian form.
2
8) Show that the radius of curvature of the curve 𝑦 = 𝑐 πΆπ‘œπ‘ β„Ž(π‘₯⁄𝑐) is y
, where c is a constant.
c
9) Find the radius of curvature of the parabola 𝑦 2 = 4π‘Žπ‘₯ at the point (a, a).
[Aug 2020, Dec 2016]
π‘Ž π‘Ž
10) Find the radius of curvature of the curve √π‘₯ + √𝑦 = √π‘Ž at (2 , 2) .
[Dec 2017]
πœ‹
11) Find the radius of curvature of the curve 𝑦 = 4𝑠𝑖𝑛π‘₯ − 𝑠𝑖𝑛2π‘₯ at π‘₯ = ⁄2 .
[Dec 2017]
π‘₯
π‘₯
12) Show that the radius of curvature of the curve 𝑦 = π‘Žπ‘™π‘œπ‘”π‘ π‘’π‘ (π‘Ž) is π‘Žπ‘ π‘’π‘ (π‘Ž)
[Dec 2018, 2019]
13) Find the radius of curvature for the curve π‘₯ 4 + 𝑦 4 = 2 π‘Žπ‘‘ (1,1).
[Dec 2016, June 2016]
3π‘Ž 3π‘Ž
3
3
14) Find the radius of curvature for the Folium De-Cartes π‘₯ + 𝑦 = 3π‘Žπ‘₯𝑦 at the point ( 2 , 2 ) on it.
[Aug 2021, June 2017, 2016, Dec 2015, 2014]
1
Calculus and Differential Equations / 21MAT11
DEPARTMENT OF MATHEMATICS
3 3
−3
15) Show that the radius of curvature of the curve π‘₯ 3 + 𝑦 3 = 3π‘₯𝑦 at the point (2 , 2) is 8√2 .
[Dec 2014]
2
2
2)
16) Find the radius of curvature for the curve π‘₯ 𝑦 = π‘Ž(π‘₯ + 𝑦 at the point (−2π‘Ž, 2π‘Ž).
[Dec 2017, 2012]
2
3
3
17) Find the radius of curvature for the curve π‘Ž 𝑦 = π‘₯ − π‘Ž at the point where the curve cuts π‘₯-axis.
[June 2019, Dec 2010]
18) Find the radius of curvature for the curve 𝑦 2 =
4π‘Ž2 (2π‘Ž−π‘₯)
π‘₯
π‘Ž2 (π‘Ž−π‘₯)
19) Find the radius of curvature of the curve 𝑦 2 =
π‘₯
3
where the curve meets the π‘₯-axis.
at the point (π‘Ž, 0).
[June 2018, Dec 2017, 2013]
20) Find the radius of curvature of the curve π‘₯𝑦 2 = π‘Ž − π‘₯ 3 at (π‘Ž, 0).
2
2𝜌 ⁄3
π‘Žπ‘₯
21) If 𝑦 = π‘Ž+π‘₯ , then show that ( π‘Ž )
𝑦 2
[June 2018]
π‘₯ 2
= (π‘₯ ) + (𝑦) where 𝜌 is the radius of curvature at (π‘₯, 𝑦)
[Feb 2021, June 2017, Dec 2016]
π‘₯
𝑦
22) Prove that the radius of curvature at any point (π‘₯, 𝑦) on the curve √π‘Ž + √𝑏 =1 is 𝜌 =
23) Find the radius of curvature for the curve π‘₯ = π‘Žπ‘™π‘œπ‘”(sec 𝑑 + tan 𝑑) , 𝑦 = π‘Ž sec 𝑑
πœ‹
2(π‘Žπ‘₯+𝑏𝑦)3/2
π‘Žπ‘
. [June 2011]
[July 2015]
24) Find the radius of curvature for the Astroid π‘₯ = π‘Žπ‘π‘œπ‘  πœƒ, 𝑦 = π‘Žπ‘ π‘–π‘› πœƒ π‘Žπ‘‘ πœƒ = 4
25) Find the radius of curvature of the curve π‘₯ = (cos 𝑑 + 𝑑 sin 𝑑), 𝑦 = π‘Ž(sin 𝑑 - 𝑑 cos 𝑑)
𝑑
26) Find the radius of curvature of the curve π‘₯ = π‘Ž [cos 𝑑 + log tan 2] 𝑦 = π‘Ž sin 𝑑
[Aug 2020, July 2015]
27) Show that the radius of curvature at any point πœƒ on the Cycloid π‘₯ = π‘Ž(πœƒ + sin πœƒ) , 𝑦 = π‘Ž(1 − cos πœƒ) is
πœƒ
4π‘Ž cos 2
[July 2012]
28) Derive Radius of Curvature in the Polar form.
29) Find the radius of curvature to the curve π‘Ÿ = π‘Ž sin π‘›πœƒ at the pole.
[June 2018]
𝑛
𝑛
𝑛−1
30) Show that the radius of curvature of the curve π‘Ÿ = π‘Ž cos π‘›πœƒ varies inversely as π‘Ÿ .
[Dec 2016]
31) Find the radius of curvature of the curve π‘Ÿ = π‘Ž(1 + cos πœƒ).
[Dec 2018, 2017]
32) Show that for the curve π‘Ÿ(1 − cos πœƒ) = 2π‘Ž, 𝜌2 varies as π‘Ÿ 3 .
[June 2019]
𝑛
𝑛
33) Find the radius of curvature of the curve π‘Ÿ = π‘Ž sin π‘›πœƒ.
[Dec 2017, 2015]
34) Find the radius of curvature of the polar curve π‘Ÿ 2 = π‘Ž2 cos 2πœƒ.
[June 2017]
35) Find the radius of curvature of the curve π‘Ÿ = π‘Ž(1 − cos πœƒ).
3
36) Find the radius of curvature for the Parabola
2π‘Ž
π‘Ÿ
3
= 1 + cos πœƒ
[Feb 2021]
πœƒ
37) Show that for the curve π‘Ÿπ‘π‘œπ‘  2 ( 2) = π‘Ž, 𝜌2 varies as π‘Ÿ 3 .
38) Show that the radius of curvature of the curve π‘π‘Ž2 = π‘Ÿ 3 is
2
π‘Ÿ3
𝜌2
39) Show that for the curve 𝑝 = 2π‘Ž , π‘Ÿ is a constant.
40) Show that for the curve 𝑝2 = π‘Žπ‘Ÿ, 𝜌2 varies as π‘Ÿ 3 .
41) Show that for the ellipse in pedal form,
1
𝑝2
1
1
π‘Ž2
3π‘Ÿ
π‘Ÿ2
= π‘Ž2 + 𝑏2 − π‘Ž2 𝑏2 , the radius of curvature at the point (p, r) is
π‘Ž2 𝑏2
𝑝3
.
Self-Study:
42) Find the coordinates of the center of curvature at any point of the parabola π‘₯ 2 = 4π‘Žπ‘¦. Hence find its evolute.
43) Show that the evolute of the parabola 𝑦 2 = 4π‘Žπ‘₯ is 27π‘Žπ‘¦ 2 = 4(π‘₯ − 2π‘Ž)3 .
[Aug 2021, 2020, Feb 2021, Dec 2019, June 2019, Dec 2018]
44) Show that the evolute of the cycloid π‘₯ = π‘Ž(πœƒ − π‘ π‘–π‘›πœƒ), 𝑦 = π‘Ž(1 − π‘π‘œπ‘ πœƒ) is another equal cycloid.
45) Find the evolute of the curve π‘₯ = π‘Žπ‘π‘œπ‘  3 πœƒ, 𝑦 = π‘Žπ‘ π‘–π‘›3 πœƒ, i,e π‘₯ 2/3 + 𝑦 2/3 = π‘Ž2/3 .
𝑑
π‘₯
46) Show that the evolute of the curve π‘₯ = π‘Ž [cos 𝑑 + π‘™π‘œπ‘” tan (2)] , 𝑦 = π‘Ž sin 𝑑 is 𝑦 = π‘Ž cosh (π‘Ž).
[Aug 2020]
2
47) Find the evolute of the curve π‘₯𝑦 = 𝑐 .
π‘₯2
𝑦2
48) Find the evolute of the ellipse π‘Ž2 + 𝑏2 = 1.
**************************************************
2
Calculus and Differential Equations / 21MAT11
DEPARTMENT OF MATHEMATICS
MODULE – II : DIFFERENTIAL CALCULUS – II
Taylor’s and Maclaurin’s series:
1.
2.
3.
4.
5.
6.
Obtain the Taylors series expansion of π‘™π‘œπ‘”(π‘π‘œπ‘ π‘₯) about the point at π‘₯ = πœ‹/3 up to the 4th degree term
Expand π‘‘π‘Žπ‘›π‘₯ about the point π‘₯ = πœ‹/4 upto the third degree term.
Expand tan−1 π‘₯ in powers of (π‘₯ − 1) up to the term containing fourth degree.
Find Maclaurin series expansion of 𝑠𝑒𝑐π‘₯ upto 4th degree term.
[Dec 2015]
5
Using Maclaurin’s series expand 𝑦 = π‘‘π‘Žπ‘›π‘₯ up to the term containing π‘₯ [Aug 2020, Dec 2019, June 2016, 2015]
Expand tan−1 (1 + π‘₯) in Maclaurin’s expansion upto first three non –vanishing terms.
7. Expand 𝑒 tan
−1 π‘₯
in Maclaurin’s expansion upto first three non –vanishing terms.
x 2 x3 x 4
ο€­ 
ο€­οŒ
2
6 24
[July 2021, Aug 2020, June 2018, Dec 2018, 2015, 2014]
8. Using Maclaurin’s theorem prove that 1  sin 2 x ο€½ 1  x ο€­
9. Using Maclaurin’s theorem prove that √1 + cos 2π‘₯ = √2 [1 −
π‘₯2
2
π‘₯4
+ 24 − β‹― ]
[Jan 2021]
10. Expand π‘™π‘œπ‘”(1 + π‘π‘œπ‘ π‘₯) by the Maclaurin’s series up to the term containing π‘₯ 4 .
[June 2019, 2017, Dec 2017]
4
11. Expand π‘™π‘œπ‘”(1 + 𝑠𝑖𝑛π‘₯) by the Maclaurin’s series up to the term containing π‘₯ .
[Dec 2012]
6
12. Expand π‘™π‘œπ‘”(𝑠𝑒𝑐π‘₯)by the Maclaurin’s series up to the term containing π‘₯
[Dec 2019, 2017, 2010, June 2017, 2016, 2012]
π‘₯
13. Expand π‘™π‘œπ‘”(1 + 𝑒 ) using Maclaurin’s series up to and including 3rd degree term. [June 2016, Dec 2016, 2013]
14. Expand π‘™π‘œπ‘”(1 + π‘₯) using Maclaurin’s series up to the term containing π‘₯ 4 .
[Dec 2017]
𝑠𝑖𝑛π‘₯
4
15. Expand 𝑒
by the Maclaurin’s series up to the term containing π‘₯ .
[Dec 2017]
π‘₯
4
16. Obtain Maclaurin series for 𝑒 π‘π‘œπ‘ π‘₯ upto the term containing π‘₯
[June 2018]
𝑒π‘₯
17. Expand 1+𝑒 π‘₯ using Maclaurin’s series up to and including 3rd degree term.
[Dec 2016]
18. Find the first four non zero terms in the expansion of 𝑓(π‘₯) = 𝑒 π‘₯ −1
[Dec 2014]
π‘₯
Evaluate the following Indeterminate forms:
19. limπ‘₯→πœ‹ (𝑠𝑖𝑛π‘₯)
π‘‘π‘Žπ‘›π‘₯
20.
2
π‘₯
21. limπ‘₯→0 (π‘Ž + π‘₯)
1
π‘₯
[Dec 2016]
1
π‘Žπ‘₯ +𝑏 π‘₯ π‘₯
23. limπ‘₯→0 (
)
2
1
[Aug 2020, Dec 2017]
π‘‘π‘Žπ‘›π‘₯ 2
limπ‘₯→0 ( π‘₯ )π‘₯
1
2π‘₯ +3π‘₯ +4π‘₯ π‘₯
22. limπ‘₯→0 (
)
3
24. limπ‘₯→0 (π‘π‘œπ‘ π‘₯)π‘π‘œπ‘‘
1
25.
[Dec 2016, June 2016]
2π‘₯
1 2 sin π‘₯
π‘Ž+π‘₯ π‘₯
limπ‘₯→0 (π‘Ž−π‘₯)
26. limπ‘₯→0 (π‘₯)
[Dec 2010]
[June 2017]
[Jan 2021]
πœ‹π‘₯
27. lim (2 −
π‘₯→π‘Ž
π‘₯ π‘‘π‘Žπ‘›( 2π‘Ž)
)
π‘Ž
[Dec 2017, June 2016] 28. limπ‘₯→πœ‹ (π‘‘π‘Žπ‘›π‘₯)π‘π‘œπ‘ π‘₯
2
1
29.
𝑠𝑖𝑛π‘₯ π‘₯
limπ‘₯→0 ( π‘₯ )
30.
[July 2021]
32. limπ‘₯→0 (
1
31. limπ‘₯→0 (π‘π‘œπ‘ π‘₯)π‘₯2
33. limπ‘₯→0 (
1
π‘Žπ‘₯ +𝑏 π‘₯ +𝑐 π‘₯ π‘₯
)
3
π‘₯
)
[June 2017]
[Aug 2020]
[Jan 2021, Aug 2020, Dec 2019, June 2019]
1
π‘Žπ‘₯ +𝑏 π‘₯ +𝑐 π‘₯ +𝑑π‘₯ π‘₯
34. limπ‘₯→0 (
1
𝑠𝑖𝑛π‘₯ 2
limπ‘₯→0 ( π‘₯ )π‘₯
1
tan π‘₯ ⁄π‘₯
[Dec 2014]
[Dec 2012]
4
)
[July 2021, June 2018, 2017, Dec 2018, 2015, 2012]
3
Calculus and Differential Equations / 21MAT11
DEPARTMENT OF MATHEMATICS
Partial Differentiation:
35. If 𝑒 = π‘₯ 𝑦 , then verify that
ο‚Ά 2u ο‚Ά 2u
ο€½
.
ο‚Άxο‚Άy ο‚Άyο‚Άx


ο€­1 / 2
ο‚Ά 2u ο‚Ά 2u ο‚Ά 2u
36. If u ο€½ x 2  y 2  z 2
, then prove that


ο€½ 0.
ο‚Άx 2
ο‚Άy 2
ο‚Άz 2
37. If 𝑒 = π‘₯ 2 𝑦 + 𝑦 2 𝑧 + 𝑧 2 π‘₯, then show that 𝑒π‘₯ + 𝑒𝑦 + 𝑒𝑧 = (π‘₯ + 𝑦 + 𝑧)2
πœ•π‘§
πœ•π‘§
38. If 𝑧 = 𝑒 π‘Žπ‘₯+𝑏𝑦 𝑓(π‘Žπ‘₯ − 𝑏𝑦) , show that 𝑏 πœ•π‘₯ + π‘Ž πœ•π‘¦ = 2π‘Žπ‘π‘§
Total derivative and Chain rule:
39. If 𝑒 = π‘₯ 3 𝑦 2 + π‘₯ 2 𝑦 3 where π‘₯ = π‘Žπ‘‘ 2 , 𝑦 = 2π‘Žπ‘‘, find
𝑑𝑒
𝑑𝑑
.
[Dec 2016]
π‘₯
𝑑𝑒
𝑦
𝑑𝑑
𝑑𝑒
40. If 𝑒 = π‘‘π‘Žπ‘›−1 (𝑦) where π‘₯ = 𝑒 𝑑 − 𝑒 −𝑑 , 𝑦 = 𝑒 𝑑 + 𝑒 −𝑑 , find
41. If 𝑒 = π‘‘π‘Žπ‘›−1 (π‘₯ ) where π‘₯ = 𝑒 𝑑 − 𝑒 −𝑑 , 𝑦 = 𝑒 𝑑 + 𝑒 −𝑑 , find
.
[June 2017]
.
[June 2018]
𝑑𝑑
2𝑑
42. If 𝑒 = π‘₯ 2 + 𝑦 2 + 𝑧 2 where π‘₯ = 𝑒 2𝑑 , 𝑦 = 𝑒 2𝑑 π‘π‘œπ‘ 2𝑑, 𝑧 = 𝑒 𝑠𝑖𝑛2𝑑, find
43. If 𝑒 = log(π‘₯ + 𝑦 + 𝑧) where π‘₯ = 𝑒 −𝑑 , 𝑦 = 𝑠𝑖𝑛𝑑, 𝑧 = π‘π‘œπ‘ π‘‘, find
𝑒
44. If 𝑧 = 𝑓(π‘₯, 𝑦) where π‘₯ = 𝑒 − 𝑒
−𝑣
,𝑦 = 𝑒
−𝑒
𝑑𝑒
𝑑𝑑
.
[Dec 2015]
𝑑𝑑
πœ•π‘§
𝑣
𝑑𝑒
πœ•π‘§
πœ•π‘§
πœ•π‘§
− 𝑒 prove that πœ•π‘’ − πœ•π‘£ = π‘₯ πœ•π‘₯ − 𝑦 πœ•π‘¦
2
1  ο‚Άz οƒΆ
 ο‚Άz οƒΆ  ο‚Άz οƒΆ
 ο‚Άz οƒΆ
45. If 𝑧 = 𝑓(π‘₯, 𝑦) where π‘₯ = π‘Ÿπ‘π‘œπ‘ πœƒ, 𝑦 = π‘Ÿπ‘ π‘–π‘›πœƒ prove that  οƒ·   οƒ·οƒ· ο€½  οƒ·  2  οƒ·
r   οƒΈ
 ο‚Άx οƒΈ  ο‚Άy οƒΈ
 ο‚Άr οƒΈ
2
2
[Dec 2015]
2
[June 2018, 2017, Dec 2017, 2014]
yο€­x zο€­xοƒΆ
ο‚Άu
ο‚Άu
ο‚Άu
οƒ·οƒ· , find the value of x 2
,
 y2
 z2
zx οƒΈ
ο‚Άx
ο‚Άy
ο‚Άz
 xy
46. If u ο€½ u
47. If 𝑒 = 𝑓 (𝑦 − 𝑧, 𝑧 − π‘₯, π‘₯ − 𝑦) then prove that
πœ•π‘’
πœ•π‘’
πœ•π‘₯
πœ•π‘’
πœ•π‘’
[Dec 2011]
πœ•π‘’
+ πœ•π‘¦ + πœ•π‘§ = 0
πœ•π‘’
48. If 𝑒 = 𝑓(π‘₯ − 𝑦, 𝑦 − 𝑧, 𝑧 − π‘₯), show that πœ•π‘₯ + πœ•π‘¦ + πœ•π‘§ = 0
π‘₯ 𝑦 𝑧
πœ•π‘’
πœ•π‘’
[Jan 2021, Dec 2018, 2017, June 2015]
[July 2021, Aug 2020, Dec 2019]
πœ•π‘’
49. If 𝑒 = 𝑓 (𝑦 , 𝑧 , π‘₯) then prove that π‘₯ πœ•π‘₯ + 𝑦 πœ•π‘¦ + 𝑧 πœ•π‘§ = 0 [June 2019, 2018, 2017, 2016, 2014, Dec 2019, 2016]
50. If 𝑒 = 𝑓 (2π‘₯ − 3𝑦, 3𝑦 − 4𝑧, 4𝑧 − 2π‘₯) then find
𝑦
πœ•π‘’
πœ•π‘’
1 πœ•π‘’
2 πœ•π‘₯
1 πœ•π‘’
1 πœ•π‘’
+ 3 πœ•π‘¦ + 4 πœ•π‘§
[June 2017, Dec 2013]
πœ•π‘’
51. If 𝑒 = 𝑓 (π‘₯𝑧, 𝑧 ) prove that π‘₯ πœ•π‘₯ − 𝑦 πœ•π‘¦ − 𝑧 πœ•π‘§ = 0
Maxima and Minima:
52. Find the extreme values of f ( x, y) ο€½ x3  3xy 2 ο€­ 15x 2 ο€­ 15 y 2  72 x
53. Find the extreme values of 𝑓(π‘₯, 𝑦) = π‘₯ 3 + 𝑦 3 − 3π‘₯ − 12𝑦 + 20
[Aug 2020, June 2017]
[June 2019, Dec 2019, June 2018]
54. Find the extreme values of 𝑓(π‘₯, 𝑦) = π‘₯ 3 𝑦 2 (1 − π‘₯ − 𝑦)
[Dec 2012]
55. Find the extreme values of 𝑓(π‘₯, 𝑦) = π‘₯ 3 + 𝑦 3 − 3π‘Žπ‘₯𝑦, π‘Ž > 0
[June 2011]
56. Find the extreme values of 𝑓(π‘₯, 𝑦) = π‘₯ 4 + 𝑦 4 − 2(π‘₯ − 𝑦)2
[June 2015]
57. Examine the function π‘₯𝑦(π‘Ž − π‘₯ − 𝑦) for extreme values.
[June 2016]
4
Calculus and Differential Equations / 21MAT11
DEPARTMENT OF MATHEMATICS
58. Examine the function 𝑓(π‘₯, 𝑦) = 1 + sin(π‘₯ 2 + 𝑦 2 ) for extreme values.
[Dec 2012]
59. Find the extreme values of 𝑓(π‘₯, 𝑦) = 𝑠𝑖𝑛π‘₯ + 𝑠𝑖𝑛𝑦 + sin(π‘₯ + 𝑦)
[June 2016]
60. Examine the function 𝑓(π‘₯, 𝑦) = 2 + 2π‘₯ + 2𝑦 − π‘₯ 2 − 𝑦 2 for its extreme values.
[Jan 2021]
𝒂 𝒂
61. Show that the function π‘₯𝑦(π‘Ž − π‘₯ − 𝑦) is maximum at (πŸ‘ , πŸ‘). Hence find maximum value if π‘Ž > 0.
[July 2021]
Jacobians:
ο‚Ά ( x, y )
ο‚Ά (u, v)
ο‚Ά (r , )
63. If π‘₯ = π‘Ÿπ‘π‘œπ‘ πœƒ and 𝑦 = π‘Ÿπ‘ π‘–π‘›πœƒ find
ο‚Ά ( x, y )
62. If π‘₯ = 𝑒 (1 − 𝑣), 𝑦 = 𝑒 𝑣 then evaluate
[Dec 2011]
64. If 𝑒 = π‘₯ 2 − 𝑦 2 , 𝑣 = 2π‘₯𝑦 and π‘₯ = π‘Ÿπ‘π‘œπ‘ πœƒ, 𝑦 = π‘Ÿπ‘ π‘–π‘›πœƒ determine the value of
65. If π‘₯ = π‘Ÿπ‘ π‘–π‘›πœƒπ‘π‘œπ‘ ∅, 𝑦 = π‘Ÿπ‘ π‘–π‘›πœƒπ‘ π‘–π‘›∅ and 𝑧 = π‘Ÿπ‘π‘œπ‘ πœƒ find
ο‚Ά (u, v)
ο‚Ά (r , )
ο‚Ά ( x, y , z )
ο‚Ά ( r , ,  )
[June 2018, 2016, Dec 2013]
ο‚Ά (u, v, w)
where 𝑒 = π‘₯ 2 + 𝑦 2 + 𝑧 2 , 𝑣 = π‘₯𝑦 + 𝑦𝑧 + 𝑧π‘₯, 𝑀 = π‘₯ + 𝑦 + 𝑧
ο‚Ά ( x, y , z )
[July 2021, Dec 2019, Dec 2018, 2017, 2016, 2014, June 2017]
 (u, v, w) οƒΆ
zx
xy
yz
οƒ·οƒ· ο€½ 4
67. If u ο€½
and v ο€½
and w ο€½
then show that J 
[Dec 2016, 2010, June 2016, 2015]
y
z
x
 ( x, y , z ) οƒΈ
66. Find
68. If 𝑒 = π‘₯ + 3𝑦 + 𝑧 3 , 𝑣 = π‘₯ 2 𝑦𝑧, 𝑀 = 2π‘₯ 2 – π‘₯𝑦, Evaluate
ο‚Ά (u, v, w)
at [1, −1, 0]
ο‚Ά ( x, y , z )
[June 2018]
ο‚Ά (u, v, w)
at [1, −1, 0]
ο‚Ά ( x, y , z )
[June 2019, Dec2018, Dec 2016, 2015]
ο‚Ά (u, v, w)
70. If 𝑒 = π‘₯ + 3𝑦 2 , 𝑣 = 4π‘₯ 2 𝑦𝑧, 𝑀 = 2𝑧 2 – π‘₯𝑦 Evaluate
at [1, −1, 0]
[Dec 2017]
ο‚Ά ( x, y , z )
69. If 𝑒 = π‘₯ + 3𝑦 2 − 𝑧 3 , 𝑣 = 4π‘₯ 2 𝑦𝑧, 𝑀 = 2𝑧 2 – π‘₯𝑦, Evaluate
71. If π‘₯ + 𝑦 + 𝑧 = 𝑒, 𝑦 + 𝑧 = 𝑒𝑣, 𝑧 = 𝑒𝑣𝑀, then evaluate
72. If π‘₯ + 𝑦 + 𝑧 = 𝑒, 𝑦 + 𝑧 = 𝑣, 𝑧 = 𝑒𝑣𝑀, find
ο‚Ά ( x, y, z)
ο‚Ά (u , v, w )
ο‚Ά ( x, y , z )
ο‚Ά (u, v, w)
[Dec 2017, 2011]
73. If 𝑒 = 3π‘₯ + 2𝑦 − 𝑧 ; 𝑣 = π‘₯ − 2𝑦 + 𝑧 ; 𝑀 = π‘₯ 2 + 2π‘₯𝑦 − π‘₯𝑧 π‘ β„Žπ‘œπ‘€ π‘‘β„Žπ‘Žπ‘‘
π‘₯
𝑦
𝑧
74. If 𝑒 = 𝑦−𝑧 , 𝑣 = 𝑧−π‘₯ , 𝑀 = π‘₯−𝑦 , find
[June 2018, 2015, Dec 2015]
πœ•(𝑒,𝑣,𝑀)
(π‘₯,𝑦,𝑧)
=0
ο‚Ά (u, v, w)
ο‚Ά ( x, y , z )
[June 2017]
75. Given π‘₯ = π‘Ž(𝑒 + 𝑣), 𝑦 = 𝑏(1 − 𝑣) and 𝑒 = π‘Ÿ 2 π‘π‘œπ‘ 2πœƒ, 𝑣 = π‘Ÿ 2 𝑠𝑖𝑛2πœƒ find
76. If 𝑒 = π‘₯ cos 𝑦 cos 𝑧 , 𝑣 = π‘₯ cos 𝑦 sin 𝑧 , 𝑀 = π‘₯ sin 𝑦 , π‘ β„Žπ‘œπ‘€ π‘‘β„Žπ‘Žπ‘‘
[Jan 2021]
πœ•(𝑒,𝑣,𝑀)
πœ•(π‘₯,𝑦,𝑧)
ο‚Ά ( x, y )
ο‚Ά (r , )
[Dec 2017]
= −π‘₯ 2 cos 𝑦.
[Aug 2020]
5
Calculus and Differential Equations / 21MAT11
DEPARTMENT OF MATHEMATICS
Self-Study:
Euler’s Theorem and Problems
77. If 𝑒 = tan−1 (
π‘₯ 3 +𝑦 3
π‘₯−𝑦
π‘₯ 3 +𝑦 3
78. If 𝑒 = sin−1 (
π‘₯−𝑦
π‘₯ 4 +𝑦4
79. If 𝑒 = log (
π‘₯+𝑦
πœ•π‘’
πœ•π‘’
πœ•π‘’
πœ•π‘’
) Prove that π‘₯ πœ•π‘₯ + 𝑦 πœ•π‘¦ = sin 2𝑒 .
) Prove that π‘₯ πœ•π‘₯ + 𝑦 πœ•π‘¦ = 2 tan 𝑒 .
) Prove that π‘₯𝑒π‘₯ + 𝑦𝑒𝑦 = 3 .
π‘₯2𝑦2
80. If 𝑒 = sin−1 ( π‘₯+𝑦 ) Prove that π‘₯𝑒π‘₯ + 𝑦𝑒𝑦 = 3 tan 𝑒 .
π‘₯3𝑦3
3
81. If 𝑒 = tan−1 (π‘₯ 3 +𝑦 3 ) Prove that π‘₯𝑒π‘₯ + 𝑦𝑒𝑦 = 2 sin 2𝑒 .
Lagrange Undetermined Multipliers with single constraint:
82. If 𝒙, π’š, 𝒛 are angles of a triangle, show that the maximum value of 𝑠𝑖𝑛π‘₯𝑠𝑖𝑛𝑦𝑠𝑖𝑛𝑧 is
πŸ‘√πŸ‘
πŸ–
83. Find the minimum value of π‘₯ 2 + 𝑦 2 + 𝑧 2 under the condition π‘₯ + 𝑦 + 𝑧 = 3π‘Ž.
[July 2021, June 2017]
[Dec 2018]
84. Find the maximum value of π’™π’š subject to the condition π‘₯ 2 + π‘₯𝑦 + 𝑦 2 = π‘Ž2 .
85. The temperature T at any point (π‘₯, 𝑦, 𝑧) in space in 𝑇 = 400π‘₯𝑦𝑧 2 . Find the highest temperature on the surface of
the unit sphere π‘₯ 2 + 𝑦 2 + 𝑧 2 = 1.
[Jan 2021]
86. A rectangular box open at top is to have a volume of 32 cubic feet. Find the dimensions of the box, if the total
surface area is minimum.
[June 2019, Dec 2018]
********************************************
6
Calculus and Differential Equations / 21MAT11
DEPARTMENT OF MATHEMATICS
MODULE – III : ORDINARY DIFFERENTIAL EQUATIONS OF FIRST ORDER
Solve the following Differential Equations:
dy
ο€½ xy 3 ο€­ xy
[June 2016]
2.  x 2  y 2  x  dx  xy dy ο€½ 0 [July 2021, June 2018, 2017]
dx
dy
dy
 y tan x ο€½ y 3 sec x [July 2021, Jan 2021, June 2019, Dec 2017] 4.
 y tan x ο€½ y 2 sec x [Dec 2018, 2014, June 2016]
3.
dx
dx
dy y cos x  sin y  y
dy

ο€½ 0 [Jan 2020, Dec 2018, 2017, 2015] 6. xy (1  xy 2 )
5.
ο€½ 1 [June 2018, 2017, 2015, Dec 2017]
dx sin x  x cos y  x
dx
1.
7. ye xy dx  ( xe xy  2 y)dy ο€½ 0
dy
9. x  y ο€½ x 3 y 6
dx
[Dec 2017]
8.
[Dec 2017]
dy
sin x cos 2 x
10.
ο€­ y tan x ο€½
dx
y2
[Dec 2015]
11. (2 x log x ο€­ xy )dy  2 ydx ο€½ 0
dy 2
y2
ο€­ yο€½ 3
dx x
x
[Dec 2016]
12. ( x 2  y 3  6 x)dx  y 2 xdy ο€½ 0 [June 2016, 2015]
[Dec 2016]
dy y
 ο€½ y2x
dx x
 dy οƒΆ
15. (4 xy  3 y 2 ο€­ x)dx  x( x  2 y)dy ο€½ 0 [Jan 2020, Dec 2015] 16. e y  1οƒ· ο€½ e x
 dx οƒΈ
13. (2 x 3 ο€­ xy 2 ο€­ 2 y  3)dx ο€­ ( x 2 y  2 x)dy ο€½ 0
[Dec 2017]
[Dec 2017]
14.
[Dec 2015]
17. (1  2 xy cos x 2 ο€­ 2 xy )dx  (sin x 2 ο€­ x 2 )dy ο€½ 0 [June 2015] 18. ( xy 3  y)dx  2( x 2 y 2  x  y 4 )dy ο€½ 0 [Dec 2014]
19. r sin  ο€­ cos
dr
ο€½ r2
d
[June 2018]
21. ( y 2 e xy  4 x 3 )dx  (2 xye xy ο€­ 3 y 2 )dy ο€½ 0 [June 2017]
2
23. (1  e
x
y
2
x
x
)dx  e y (1 ο€­ )dy ο€½ 0
y
π‘‘π‘Ÿ
25. π‘Ÿ sin πœƒ − cos πœƒ π‘‘πœƒ = π‘Ÿ 2
[June 2019]
20.
dy
 x sin 2 y ο€½ x 3 cos 2 y
dx



[June 2017]

22. y 4  2 y dx  xy 3  2 y 4 ο€­ 4 x dy ο€½ 0
𝑑𝑦
24. π‘₯ 3 𝑑π‘₯ − π‘₯ 2 𝑦 = −𝑦 4 cos π‘₯
[Aug 2020]
[Jan 2020]
26. [4π‘₯ 3 𝑦 2 + 𝑦 cos(π‘₯𝑦)]𝑑π‘₯ + [2π‘₯ 4 𝑦 + π‘₯ cos(π‘₯𝑦)]𝑑𝑦 = 0
1
27. Solve [𝑦 (π‘₯ + π‘₯) + π‘π‘œπ‘ π‘¦] 𝑑π‘₯ + [π‘₯ + π‘™π‘œπ‘”π‘₯ − π‘₯𝑠𝑖𝑛𝑦]𝑑𝑦
[Aug 2020]
[Jan 2021]
28. (cos x tan y  cos( x  y))dx  (sin x sec 2 y  cos( x  y)dy ο€½ 0
ORTHOGONAL TRAJECTORIES:
1.
2.
3.
4.
5.
Find the orthogonal trajectories of a family of circles π‘Ÿ = 2π‘Žπ‘π‘œπ‘ πœƒ, where ‘π‘Ž’ is a parameter.
[June 2017]
𝑛
𝑛
Find the orthogonal trajectories of the curve π‘Ÿ π‘π‘œπ‘  π‘›πœƒ = π‘Ž
[June 2019, Dec 2016]
Find the orthogonal trajectories of the curve π‘Ÿ 𝑛 = π‘Žπ‘› π‘π‘œπ‘  π‘›πœƒ
[Dec 2017, 2015 June 2016]
Find the orthogonal trajectories of the curve π‘Ÿ 𝑛 = π‘Žπ‘› 𝑠𝑖𝑛 π‘›πœƒ, where ‘π‘Ž’ is a parameter. [June 2017, Dec 2015]
Find the orthogonal trajectories of the curve π‘Ÿ 𝑛 𝑠𝑖𝑛 π‘›πœƒ = π‘Žπ‘› , where ‘π‘Ž’ is a parameter.
[June 2018]
7
Calculus and Differential Equations / 21MAT11
DEPARTMENT OF MATHEMATICS
πœƒ
6. Show that orthogonal trajectories of the families of cardioides π‘Ÿ = π‘Žπ‘π‘œπ‘  2 ( 2) is another family of cardioides
πœƒ
π‘Ÿ = 𝑏𝑠𝑖𝑛2 ( 2)
[Dec 2014]
2a
ο€½ 1 ο€­ cos 
r
8. Find the orthogonal trajectories of the curve r = 4 a sec θ tan θ
[Dec 2016, June 2015]
9. Find the orthogonal trajectories of the family of cardioides π‘Ÿ = π‘Ž(1 + π‘π‘œπ‘ πœƒ)
[Aug 2020]
2

k οƒΆ
10. Find the orthogonal trajectories of the family of curves  r  οƒ·οƒ· cos  =a, a being a parameter.
[July 2015]
r οƒΈ

7. Find the orthogonal trajectories of a family
11. Show that the family of parabolas 𝑦 2 = 4π‘Ž(π‘₯ + π‘Ž) is self-orthogonal.
[Jan 2021, Dec 2018, June 2018, 2016]
12. Find the orthogonal trajectories of the family of the curves 𝑦 = π‘₯ + 𝑐𝑒 −π‘₯ , where 𝑐 is the parameter. [Dec 2017]
13. Find the orthogonal trajectories of the curve 𝑦 = 𝑐π‘₯ 3
[Dec 2017]
2
2
x
y
 2
ο€½ 1 , λ is a parameter, is self-orthogonal.
14. Show that the family of the curves 2
a  b 
[June 2017, Dec 2015, Dec 2012]
2
2
x
y
ο€½ 1 , λ is a parameter
15. Find the orthogonal trajectories of the family of the curve 2  2
a
b 
[July 2021, June 2015]
16. Find the orthogonal trajectories of the family of the curve x 2 / 3  y 2 / 3 ο€½ a 2 / 3
[Dec 2016]
2
17. Find the orthogonal trajectories of the family of the curve 𝑦 = 4π‘Žπ‘₯
NEWTON’S LAW OF COOLING:
1. Water at temperature 10° 𝐢 takes 5 minutes to warm upto 20° 𝐢 in a room temperature 40° 𝐢. Find the
temperature after 20 minutes.
[June 2018, 2017]
2. A body originally at 80℃ cools down to 60℃ in 20 minutes. The temperature of the air being 40℃. What will
be the temperature of the body after 40 minutes from the original?
[Dec 2017, 2016, 2015, June 2017, 2015]
°
°
3. A body heated to 110 𝐢 and placed in air at 10 𝐢. After one hour its temperature becomes 60° 𝐢. How much
additional time is required for it to cool to 30° 𝐢?
[Dec 2017, 2015]
4. If the temperature of the air is 30℃ and the substance cools from 100℃ to 70℃ in 15 minutes. Find when the
temperature will be 40℃.
[June 2016, Dec 2014]
5. Water at temperature 100℃ cools down to 88℃ in 10 minutes in a room temperature of 25℃. Find the
temperature of the water after 20 minutes
[June 2018]
6. Suppose an object is heated to 300° 𝐹 and allowed to cool in a room whose temperature is 80° 𝐹. After 10
minutes the temperature of the object is 250° 𝐹. What will be its temperature after 20 minutes?
[June 2016]
°
7. A body in air at 25℃ cools from 100℃ to 75 𝐢 in 1 minute. Find the temperature of the body at the end of 3
minutes.
[July 2021, Jan 2021, Aug 2020, Dec 2018]
8. If the air is maintained at 30℃ and the temperature of the body cools from 80℃ to 60℃ in 12 minutes. Find the
temperature of the body after 24 minutes.
[June 2019]
NON LINEAR DIFFERENTIAL EQUATIONS OF FIRST ORDER:
1.
2.
3.
4.
5.
Solve 𝑃2 + 2π‘ƒπ‘¦π‘π‘œπ‘‘π‘₯ − 𝑦 2 = 0
Solve π‘₯𝑦(𝑃2 + 1) = (π‘₯ 2 + 𝑦 2 )𝑃
Solve 𝑃(𝑃 + 𝑦) = π‘₯(π‘₯ + 𝑦)
Solve yp 2  ( x ο€­ y) p ο€­ x ο€½ 0
Solve π‘₯𝑝2 − (2π‘₯ + 3𝑦)𝑝 + 6𝑦 = 0
[Jan 2020]
[Jan 2021, Dec 2018, June 2018]
[Dec 2011, June 2011]
[July 2021, June 2019]
[Dec 2015]
8
Calculus and Differential Equations / 21MAT11
6.
Solve π‘₯𝑦𝑃2 + 𝑃(3π‘₯ 2 − 2𝑦 2 ) − 6π‘₯𝑦 = 0
7.
Solve π‘₯ 2 (𝑑π‘₯ ) + π‘₯𝑦 𝑑π‘₯ − 6𝑦 2 = 0
𝑑𝑦 2
𝑑𝑦
8.
Solve π‘₯ 2 𝑝2 + 3π‘₯𝑦𝑝 + 2𝑦 2 = 0
9.
Solve
10.
11.
12.
13.
14.
Solve
Solve
Solve
Solve
Solve
𝑑𝑦
𝑑π‘₯
4
𝑑π‘₯
DEPARTMENT OF MATHEMATICS
π‘₯
𝑦
− 𝑑𝑦 = 𝑦 − π‘₯
𝑃 − (π‘₯ + 2𝑦 + 1)𝑃3 + (π‘₯ + 2𝑦 + 2π‘₯𝑦)𝑃2 − 2π‘₯𝑦𝑃 = 0
π‘₯𝑝3 − 𝑦𝑝2 + 1 = 0 by Clairaut’s equation
(𝑃π‘₯ − 𝑦)(𝑃𝑦 + π‘₯) = π‘Ž2 𝑃, use the substitution 𝑋 = π‘₯ 2 , π‘Œ = 𝑦 2
(𝑝π‘₯ − 𝑦)(𝑝𝑦 + π‘₯) = 2𝑝 by reducing into Clairaut’s form, taking 𝑋 = π‘₯ 2 , π‘Œ = 𝑦 2
(𝑝π‘₯ − 𝑦)(𝑝𝑦 + π‘₯) = 0 by reducing into Clairaut’s form, taking 𝑋 = π‘₯ 2 , π‘Œ = 𝑦 2
[Dec 2014, 2011]
[June 2011]
[June 2019]
[Dec 2018]
Self-Study:
L-R CIRCUITS:
1. A 12 volt battery is connected to a series circuit in which the inductance is ½ Henry and resistance is 10 ohms.
Determine the current if the initial current is zero.
[Dec 2017]
2. The R-L circuit differential equation acted by a electro motive force πΈπ‘ π‘–π‘›πœ”π‘‘, satisfies the differential equation,
di
L dt + Ri = Esinωt.if there is no current in the circuit initially, obtain the value of current at any time ‘𝑑’.
[Dec 2016]
3. A series circuit with resistance R, inductance L and electromotive force E is governed by the differential
𝑑𝑖
equation 𝐿 𝑑𝑑 = 𝑅𝑖 = 𝐸, where L and R are constants and initially the current i is zero. Find the current at any
time t.
[Jan 2021]
𝑑𝑖
4. Solve the equation 𝐿 + 𝑅𝑖 = 𝐸0 𝑠𝑖𝑛𝑀𝑑 Where L, R and 𝐸0 are constants and discuss the case when t
𝑑𝑑
increases indefinitely.
[Aug 2020, Jan 2020]
𝑑𝐼
5. Solve the L-R circuit 𝐿 𝐷𝑇 + 𝑅𝐼 = 𝐸 Initially I = 0 when t = 0
[July 2021]
6. When a switch is closed in a circuit containing a battery E, a resistance R and an inductance L. The current i
di
builds up at rate given by L dt + Ri = Esinωt. Find i as a function of t. How long will it be, before the current
has reached one half its final value if E = 6 volts, R = 100 ohms and L = 0.1 Henry.
Solvable for x and y :
1.
2.
3.
4.
Solve 𝑦 2 log 𝑦 = π‘₯𝑝𝑦 + 𝑝2
Solve 𝑝3 − 4π‘₯𝑦𝑝 + 8𝑦 2 = 0
Solve π‘₯𝑝4 − 2𝑦𝑝3 + 12π‘₯ 3 = 0
Solve 𝑦 = 2𝑝π‘₯ + 𝑝2 𝑦
**************************************
9
Calculus and Differential Equations / 21MAT11
DEPARTMENT OF MATHEMATICS
MODULE – IV: ORDINARY DIFFERENTIAL EQUATIONS OF HIGHER ORDER
Solve the following linear differential equations:
1.
3.
5.
6.
7.
8.
d3y
d2y
dy
ο€­
3
3 ο€­ y ο€½ 0
3
2
dx
dx
dx
d3y d2y
dy
 2  4  4y ο€½ 0
3
dx
dx
dx
4. (𝐷4 − π‘š4 )𝑦 = 0
[June 2018]
2.
(𝐷4 − 1)𝑦 = 0
[June 2010]
[June 2017]
[June 2017]
(𝐷 + 5𝐷 + 6𝐷 − 4𝐷 − 8)𝑦 = 0
[Dec 2016]
4
3
2
(4𝐷 − 4𝐷 − 23𝐷 + 12𝐷 + 36)𝑦 = 0
[Dec 2016, 2011, June 2015]
4
3
2
(4𝐷 − 8𝐷 − 7𝐷 + 11𝐷 + 6)𝑦 = 0 [Jan 2021, Jan 2020, Dec 2019, 2016, June 2018, 2016, 2015]
(𝐷4 + 4𝐷3 – 5𝐷2 – 36𝐷– 36) 𝑦 (π‘₯) = 0
[Dec 2011]
4
3
2
9. ( D 4 ο€­ 2D 3  2D 2 ο€­ 2D  1) y ο€½ 0
2
dx
(0) ο€½ 15
10. d 2x  5 dx  6 x ο€½ 0 given x(0) ο€½ 0,
dt
dt
dt
2
dy
13
ο€½ ο€­ for x ο€½ 0.
11. 9 d 2y  6 dy  y ο€½ 0 given y ο€½ 4 and
dx
3
dx
dx
[Dec 2017]
2
dy d y
d2y
dy
ο€½
for x ο€½ 0.
 4  5 y ο€½ 0 given y ο€½ 2 and
2
dx dx 2
dx
dx
d2y
dy
13.
[June 2018]
14.(𝐷2 + 4𝐷 + 3)𝑦 = 𝑒 −π‘₯ [Jan 2020, Dec 2019]
ο€­ 6  9 y ο€½ 6e 3 x
2
dx
dx
12.
d3y d2y
dy
 2 ο€­ 4 ο€­ 4 y ο€½ 3e x
3
dx
dx
dx
2
d y dy
17.

 y ο€½ (1 ο€­ e x ) 2
dx 2 dx
15.
19.
16.
[June 2018]
18. (𝐷2 + 4)𝑦 = π‘₯ 2 + π‘π‘œπ‘ 2π‘₯ + 2−π‘₯ [Dec 2017]
d3y d2y
dy
ο€­ 2  4 ο€­ 4 y ο€½ sinh(2 x  3) [Dec 2017]
3
dx
dx
dx
21. D 2 ο€­ 3D  2y ο€½ sin 2 x
[Dec 2017]
23. ( D ο€­ 4D  4) y ο€½ e  sin x
2
2x
[June 2019]
25. (𝐷2 – 4𝐷 + 4)𝑦 = 𝑒 2π‘₯ + π‘π‘œπ‘ 2π‘₯ + 4 [Dec 2014]
27.
d2y
 4 y ο€½ 2ο€­x
dx 2
[June 2018]
d3y
d 2 y dy
2 2 
ο€½ e ο€­ x  sin 2 x
3
dx
dx
dx
[Dec 2016, 2011]
[June 2015]
 xοƒΆ
20. y ο‚’ο‚’  2 y ο‚’  y ο€½ cosh οƒ·
2οƒΈ
22. (𝐷3 − 1)𝑦 = 3π‘π‘œπ‘ 2π‘₯ [Jan 2020, Dec 2019, 2010]
24. y '' ο€­ 4 y '  4 y ο€½ 8 cos 2 x
26. (𝐷2 – 4𝐷 + 3)𝑦 = 𝑠𝑖𝑛3π‘₯π‘π‘œπ‘ 2π‘₯
28. y ''  4 y ' ο€­ 12 y ο€½ e 2 x ο€­ 3sin 2 x
29. (𝐷3 − 𝐷)𝑦 = 2𝑒 π‘₯ + 4π‘π‘œπ‘ π‘₯
[June 2011]
30.
d2y
dy
 3  2y ο€½ x2
2
dx
dx
31. (𝐷2 + 4)𝑦 = π‘₯ 2 + 𝑒 −π‘₯
[June 2015]
32.
d2y
 4 y ο€½ x 2  2 ο€­x  log 2
dx 2
33. (𝐷2 + 3𝐷 + 2)𝑦 = 4π‘π‘œπ‘  2 π‘₯
[Aug 2020]
2
35.
d y
dy
ο€­ 6  25 y ο€½ e 2 x  sin x  x [June 2018]
2
dx
dx
37. ( D3  8) y ο€½ x 4  2 x  1
39.
d2y
dy
ο€­ 4  13 y ο€½ e3 x cosh 2 x  3x
2
dx
dx
d3y
d2y
dy

6
 11  6 y ο€½ e x  1
3
2
dx
dx
dx
2
d y
41.
ο€­ 4 y ο€½ cosh2 x ο€­ 1  3 x
2
dx
y
42. ''' ο€­ 6 y ''  11y ' ο€­ 6 y ο€½ 1  x  sin x
40.
[June 2018]
[June 2018]
[June 2018]
[Dec 2017]
34. 𝑦 ′′ + 2𝑦 ′ + 𝑦 = 2π‘₯ + π‘₯ 2
[Aug 2020]
36. ( D 2 ο€­ 3D  2) y ο€½ 2 x 2  sin 2 x
[June 2019]
38. (2𝐷2 + 2𝐷 + 3) 𝑦 = π‘₯ 2 – 2π‘₯– 1
[June 2018, 2017, 2016]
[June 2015, 2014]
[Jan 2021, June 2017, 2016, 2014]
[June 2017]
10
Calculus and Differential Equations / 21MAT11
𝑑3 𝑦
𝑑2 𝑦
DEPARTMENT OF MATHEMATICS
𝑑𝑦
43. 𝑑π‘₯ 3 + 𝑑π‘₯ 3 + 4 𝑑π‘₯ + 4𝑦 = π‘₯ 2 − 4π‘₯ − 6
[Jan 2021]
44. D2  3D  2y ο€½ 1  3x  x 2
[Dec 2016, 2011, June 2016]
45. (𝐷3 − 6𝐷2 + 11𝐷 − 6)𝑦 = 𝑒 2π‘₯ +2π‘₯ + π‘π‘œπ‘ 2π‘₯
[Dec 2016, June 2013]
46. (𝐷 − 2)2 𝑦 = 8(𝑒 2π‘₯ + 𝑠𝑖𝑛2π‘₯ + π‘₯ 2 )
[June 2017, 2016, 2012, Dec 2016, 2011]
dy
ο€½ 1 at π‘₯ = 0
47. y ''  4 y '  5 y ο€½ ο€­2 cosh x , find 𝑦 when y ο€½ 0 and
[June 2018, 2017]
dx
48. yο‚’ο‚’  4 y '  4 y ο€½ 8x 2 given 𝑦(0) = 1, 𝑦 ′ (0) = 2
[June 2016]
Solve the following linear differential equations by the method of variation of parameters:
1. yο‚’ο‚’  a 2 y ο€½ sec a x 
3. yο‚’ο‚’  y ο€½ sec x tan x
[Dec2016, 2014, 2010]
[Aug 2020,June 2018]
5. yο‚’ο‚’  y ο€½ cosec x
7. yο‚’ο‚’  y ο€½ tan x
[June 2016]
[Jan2021, June 2013]
9. yο‚’ο‚’ ο€­ 2yο‚’  2y ο€½ ex tanx
11. yο‚’ο‚’  y ο€½

1
1  Sinx

[June 2015]
[June 2018, 2015]
 
13. D2 ο€­ 3D  2 y ο€½ cos eο€­ x
2. yο‚’ο‚’  y ο€½ sec x 
4. yο‚’ο‚’  y ο€½ cosec x cot x
[June 2019, Dec 2017]
[Dec 2016]
-x
3
6. yο‚’ο‚’  2yο‚’  2y ο€½ e sec x
[June 2018]
8. yο‚’ο‚’  2 yο‚’  y ο€½ e log x
1
10. yο‚’ο‚’ ο€­ 3yο‚’  2y ο€½
1  e- x
ο€­x
12.
[Dec 2014]
[June 2010]
d2y
2
ο€­yο€½
2
dx
1 ex
[June 2016]
14. (𝐷2 − 2𝐷 + 1)𝑦 = 𝑒 π‘₯ π‘™π‘œπ‘”π‘₯
e 3x
x2
16. yο‚’ο‚’  4y ο€½ tan(2x)
15. yο‚’ο‚’ ο€­ 6yο‚’  9y ο€½
[Dec 2016, 2011, June 2011]
[Jan 2020, Dec 2019, 2017, 2011, June 2017, 2012]
Cauchy’s and Legendre’s linear differential equations:
Solve the following differential equations:
d2y
dy
ο€­ x  y ο€½ log x [Dec 2017, June 2017, 2010]
1. x
2
dx
dx
2
3. x 3
2
d3y
dy
2 d y

3
x
 x  y ο€½ x  log x
3
2
dx
dx
dx
5. x 2 yο‚’ο‚’  xyο‚’  9 y ο€½ 3x 2  sin(3 log x)
7. x 3
2
d3y
dy
2 d y

3
x
 x  8 y ο€½ 65 cos(log x)
3
2
dx
dx
dx
8. x 2
d2y
dy
ο€­ 3x  4 y ο€½ (1  x) 2
2
dx
dx
d2y
dy
 x  9 y ο€½ sin(3 log x)  3x 2
9. x
2
dx
dx
2
d2y
dy
ο€­ 4 x  6 y ο€½ cos(2 log x)
10. x
2
dx
dx
2
11. x yο‚’ο‚’ ο€­ 5xyο‚’  8 y ο€½ 2 log x
2
2
12. 3x  2 yο‚’ο‚’  53x  2yο‚’ ο€­ 3y ο€½ x 2  x  1
13. (2 x  5) 2
[Dec 2017]
2. x D 2 y ο€­ 2
y
1
ο€½ x 2
x
x
[June 2018]
4. x 2 yο‚’ο‚’  xyο‚’  y ο€½ 2 cos 2 (log x)
[June 2011]
d2y
dy
 5x  13 y ο€½ log x  x 2
2
dx
dx
[Dec 2016]
6. x 2
[Dec 2016]
[June 2016, Dec 2010]
[Aug 2020, Dec 2016, Dec 2011]
[June 2019]
[Jan 2020, Dec 2019]
[June 2018]
2
d y
dy
ο€­ 6(2 x  5)  8 y ο€½ 6 x
2
dx
dx
[Dec 2017]
11
Calculus and Differential Equations / 21MAT11
DEPARTMENT OF MATHEMATICS
14. (1  x) 2 yο‚’ο‚’  (1  x) yο‚’  y ο€½ 4 cos[log(1  x) ]
[June 2016]
d2y
dy
 6(2 x  3)  6 y ο€½ log( 2 x  3)
2
dx
dx
2
16. 3x  2 yο‚’ο‚’  33x  2yο‚’ ο€­ 36y ο€½ 8x 2  4x  1
15. (2 x  3) 2
[June 2015]
[June 2018, 2016, Dec 2014]
17. (1  x) 2 yο‚’ο‚’  (1  x) yο‚’  y ο€½ 2 sin[log(1  x)]
[June 2019, 2012, Dec 2017, 2016, 2011]
d2y
dy
ο€­ (2 x  3) ο€­ 12 y ο€½ 6 x
2
dx
dx
19. (1  x) 2 yο‚’ο‚’  (1  x) yο‚’  y ο€½ sin[log(1  x) 2 ]
18. (2 x  3) 2
[June 2018, 2017, Dec 2016]
[June 2017]
20. (2π‘₯ + 1)2 𝑦 ′′ − 6(2π‘₯ + 1)𝑦 ′ + 16𝑦 = 8(2π‘₯ + 1)2
[Aug 2020]
Self-Study:
Applications to oscillations of a spring and L-C-R circuits:
1. A condenser of capacity 𝐢 discharged through an inductance 𝐿 and resistance 𝑅 in series and charge q at
𝑑2 π‘ž
π‘‘π‘ž
π‘ž
time satisfying the equation 𝐿 𝑑𝑑 2 + 𝑅 𝑑𝑑 + 𝐢 = Esinpt. Given 𝐿 = 0.25𝐻, 𝑅 = 250 ohms 𝐢 = 2 × 10−6
and that when 𝑑 = 0 charge π‘ž = 0.02 and current
π‘‘π‘ž
= 0. Obtain π‘ž in time 𝑑
𝑑𝑑
𝑑2 π‘ž
π‘‘π‘ž
π‘ž
2. In an LCR circuit the charge π‘ž on a plate of condenser is given by 𝑑𝑑 2 + 𝑅 𝑑𝑑 + 𝐢 = Esinpt . The circuit is
1
tuned to resonance so that 𝑝2 = 𝐿𝐢. If initially the current 𝐼 and the charge be zero. Show that for all small
𝑅
𝐸𝑑
values of 𝐿 , the current in the circuit at a time t is given by 2𝐿 𝑠𝑖𝑛𝑝𝑑.
3. The current 𝑖 and the charge π‘ž in a series containing an inductance 𝐿, capacitance 𝐢, emf 𝐸, satisfy the
𝑑2 π‘ž
π‘ž
differential equation 𝐿 𝑑𝑑 2 + 𝐢 = E. Express π‘ž and 𝑖 in terms of ‘𝑑’ given that 𝐿, 𝐢, 𝐸 are constants and the
value of 𝑖 and π‘ž are both zero initially.
[June 2019]
2
4. The differential equation of a simple pendulum is
d x
2
 w0 x ο€½ F0 sin nt where w0 and F0 are constants.
2
dt
dx
ο€½ 0 . Solve it.
[Dec 2019]
dt
5. The differential equation of spring fixed at upper end and a weight at its lower end is given by
Also initially x ο€½ 0,
𝑑2 π‘₯
𝑑π‘₯
10 𝑑𝑑 2 + 𝑑𝑑 + 200π‘₯ = 0. The weight is pulled down 0.25cm below the equilibrium position and then
released. Find the expression for displacement of weight from its equilibrium positionat any time 𝑑 during
its first upward motion.
6. The differential equation of spring fixed at upper end and a weight at its lower end is given by
𝑑2 π‘₯
𝑑𝑑 2
πœ†π‘Ž
+ πœ†2 π‘₯ = λ2 π‘Žπ‘ π‘–π‘›π‘›π‘‘. Show that if ≠ πœ†, the displacement is given by π‘₯ = πœ†2 −𝑛2 (πœ†π‘ π‘–π‘›π‘›π‘‘ − π‘›π‘ π‘–π‘›πœ†π‘‘).
What happens when 𝑛 = πœ†?
**********************************
12
Calculus and Differential Equations / 21MAT11
DEPARTMENT OF MATHEMATICS
MODULE V: LINEAR ALGEBRA
Rank of a Matrix:
Find the rank of the following matrices by reducing it to echelon form:
5οƒΉ
0 1 ο€­ 3 ο€­ 1οƒΉ
2 1 3
4
οƒͺ1 0 1
οƒΊ
οƒͺ
οƒΊ
οƒͺ2
1οƒΊ
4 2 1
3οƒΊ
οƒͺ
οƒͺ
1.
[June 2018]
2.
[Dec 2017] 3. οƒͺ
οƒͺ3 1 0
οƒͺ8 4 7 13 οƒΊ
οƒͺ2
2οƒΊ
οƒͺ
οƒΊ
οƒͺ
οƒΊ
οƒͺ
1 1 ο€­ 2 0 
8 4 ο€­ 3 ο€­ 1
2
2 ο€­ 1 3
οƒͺ0 3 4
4. οƒͺ
οƒͺ2 3 7
οƒͺ
2 5 11
1
οƒͺ2
7. οƒͺ
οƒͺ3
οƒͺ
6
4οƒΉ
1οƒΊοƒΊ
[June 2016]
5οƒΊ
οƒΊ
6
2 3 0οƒΉ
4 3 2 οƒΊοƒΊ
2 1 3οƒΊ
οƒΊ
8 7 5
1 2 3 2 οƒΉ
10. οƒͺοƒͺ2 3 5 1οƒΊοƒΊ
οƒͺ1 3 4 5
1
οƒͺ2
12. οƒͺ
οƒͺ4
οƒͺ
1
3οƒΉ
8 οƒΊοƒΊ
8 12 16οƒΊ
οƒΊ
2 3 4
2
4
4
6
[Jan 2021]
1 2 3 4οƒΉ
5. οƒͺ5 6 7 8 οƒΊ
οƒͺ
οƒΊ
οƒͺ8 7 0 5
3 ο€­4 ο€­1
οƒͺ1 7 3
8. οƒͺ
οƒͺ5 ο€­2 5
οƒͺ
9 ο€­3 7
[Dec 2019, June 2014]
[Dec 2018, June 2011]
2οƒΉ
1 οƒΊοƒΊ
4οƒΊ
οƒΊ
7
[Dec 2015]
[Aug 2021]
0 2 1οƒΉ
1 3 4οƒΊοƒΊ
3 4 7οƒΊ
οƒΊ
3 1 4
[Dec 2017]
 1 2 0 ο€­ 1οƒΉ
6. οƒͺ 3 4 1 2 οƒΊ
οƒͺ
οƒΊ
οƒͺ 2 3 2 5 
[Dec 2013]
 2 ο€­1 ο€­3 ο€­1οƒΉ
οƒͺ1 2 3 ο€­1οƒΊ
οƒΊ
9. οƒͺ
οƒͺ1 0 1 1 οƒΊ
οƒͺ
οƒΊ
 0 1 1 ο€­1
[Aug 2020]
2 ο€­2 3 οƒΉ
1
οƒͺ2
5 ο€­ 4 6 οƒΊοƒΊ
οƒͺ
11.
[June 2019, Dec 2019, 2018, 2012]
οƒͺ ο€­ 1 ο€­ 3 2 ο€­ 2οƒΊ
οƒͺ
οƒΊ
4 ο€­1 6 
2
2 3 ο€­ 1 ο€­ 1 οƒΉ
οƒͺ1 ο€­ 1 ο€­ 2 ο€­ 4 οƒΊ
οƒΊ
13. οƒͺ
οƒͺ3 1
3 ο€­ 2οƒΊ
οƒͺ
οƒΊ
0 ο€­ 7
6 3
91
 2 ο€­ 1 ο€­ 3 ο€­ 1οƒΉ
οƒͺ92
οƒͺ1
οƒΊ
οƒͺ
2
3
ο€­
1
οƒΊ [Dec 2107, 2015, June 2015] 15. οƒͺ93
14. οƒͺ
οƒͺ1
0
1
1οƒΊ
οƒͺ
οƒͺ
οƒΊ
οƒͺ94
1 ο€­ 1 ο€­ 1
0
οƒͺ95
[June 2016, Dec 2016]
92 93 94 95οƒΉ
93 94 95 96οƒΊοƒΊ
94 95 96 97 οƒΊ
οƒΊ
95 96 97 98οƒΊ
96 97 98 99
[Dec 2010]
Consistency of System of Linear Equations:
1. For what values of  and  following simultaneous equations have i) no solution ii) a unique solution
iii) an infinite number of solutions? π‘₯ + 𝑦 + 𝑧 = 6, π‘₯ + 2𝑦 + 3𝑧 = 10,
π‘₯ + 2𝑦 +  𝑧 = 
[Aug 2021, Jan 2021, June 2019, 2016, 2014, Dec 2012, 2010]
2. Investigate the value of  and  so that the equations 2x + 3y + 5z = 9; 7x + 3y - 2z = 8; 2x + y + z = 
have i) no solution ii) a unique solution iii) an infinite number of solutions?
[June 2011]
3. Find the value of  and  such that the system x  2 y  3z ο€½ 6; x  3 y  5z ο€½ 9; 2 x  5 y  z ο€½  have
i) no solution ii) a unique solution iii) an infinite number of solutions?
[Jan 2019, Dec 2018]
13
Calculus and Differential Equations / 21MAT11
DEPARTMENT OF MATHEMATICS
4. Show that the following system is consistent and solve: x  y  z ο€½ 4; 2 x  y ο€­ z ο€½ 1; x ο€­ 2 y  2 z ο€½ 2
5. Test for consistency and solve: 5x + 3y + 7z = 4 ; 3x + 26y + 2z = 9; 7x + 2y +10z = 5
[June 2018, Dec 2013]
[Dec 2017, 2015]
6. Test for consistency and solve: 4π‘₯ − 2𝑦 + 6𝑧 = 8 , π‘₯ + 𝑦 − 3𝑧 = −1 , 15π‘₯ − 3𝑦 + 9𝑧 = 21
7. Test for consistency and solve: 5π‘₯ + 𝑦 + 3𝑧 = 20 ,2π‘₯ + 5𝑦 + 2𝑧 = 18 , 3π‘₯ + 2𝑦 + 𝑧 = 14
[Aug 2020]
8. Test for consistency and solve: x  2 y  2 z ο€½ 5; 2 x  y  3z ο€½ 6; 3x ο€­ y  2 z ο€½ 4; and x  y  z ο€½ ο€­1 [June 2015]
Gauss-Elimination Method:
Solve the following system of equations by Gauss Elimination method:
1. π‘₯ + 2𝑦 + 𝑧 = 3, 2π‘₯ + 3𝑦 + 2𝑧 = 5, 3π‘₯ − 5𝑦 + 5𝑧 = 2
[June 2017]
2. π‘₯ + 𝑦 + 𝑧 = 9, π‘₯ − 2𝑦 + 3𝑧 = 8, 2π‘₯ + 𝑦 − 𝑧 = 3
[Aug 2021, Dec 2017, 2012]
3. 2π‘₯ − 3𝑦 + 4𝑧 = 7, 5π‘₯ − 2𝑦 + 2𝑧 = 7, 6π‘₯ − 3𝑦 + 10𝑧 = 23
[June 2017]
4. 2π‘₯ + 𝑦 + 4𝑧 = 12, 4π‘₯ + 11𝑦 − 𝑧 = 33, 8π‘₯ − 3𝑦 + 2𝑧 = 20
[Dec 2016, 2015]
5. 4π‘₯ + 𝑦 + 𝑧 = 4, π‘₯ + 4𝑦 − 2𝑧 = 4, 3π‘₯ + 2𝑦 − 4𝑧 = 6
[June 2016]
6. 3π‘₯ − 𝑦 + 2𝑧 = 12, π‘₯ + 2𝑦 + 3𝑧 = 11, 2π‘₯ − 2𝑦 − 𝑧 = 2
[Dec 2014]
7. 2π‘₯ − 𝑦 + 3𝑧 = 1 , −3π‘₯ + 4𝑦 − 5𝑧 = 0 , π‘₯ + 3𝑦 + 6𝑧 = 0
[June 2014]
8. 2π‘₯ + 5𝑦 + 7𝑧 = 52 , 2π‘₯ + 𝑦 − 𝑧 = 0 , π‘₯ + 𝑦 + 𝑧 = 9
[Jan 2019, Dec 2018]
9. π‘₯ − 2𝑦 + 3𝑧 = 2 , 3π‘₯ − 𝑦 + 4𝑧 = 4 , 2π‘₯ + 𝑦 − 2𝑧 = 5
[Dec 2019]
10. 5π‘₯1 + π‘₯2 + π‘₯3 + π‘₯4 = 4 , π‘₯1 + 7π‘₯2 + π‘₯3 + π‘₯4 = 12 , π‘₯1 + π‘₯2 + 6π‘₯3 + π‘₯4 = −5 , π‘₯1 + π‘₯2 + π‘₯3 + 4π‘₯4 = −6
[June 2015]
Gauss-Jordan Method:
Solve the following system of equations by Gauss Jordan method:
1. π‘₯ + 𝑦 + 𝑧 = 9, π‘₯ − 2𝑦 + 3𝑧 = 8, 2π‘₯ + 𝑦 − 𝑧 = 3 [Jan 2021, June 2018, 2017, 2016, 2015, Dec 2017, 2010]
2. π‘₯ + 2𝑦 + 𝑧 = 3, 2π‘₯ + 3𝑦 + 3𝑧 = 10, 3π‘₯ − 𝑦 + 2𝑧 = 13
[Dec 2017]
3. 2π‘₯ + 𝑦 + 𝑧 = 10, 3π‘₯ + 2𝑦 + 3𝑧 = 18, π‘₯ + 4𝑦 + 9𝑧 = 16
[Dec 2016]
4. π‘₯ + 𝑦 + 𝑧 = 8, −π‘₯ − 𝑦 + 2𝑧 = −4, 3π‘₯ + 5𝑦 − 7𝑧 = 14
[Dec 2015]
5. 2π‘₯ − 3𝑦 + 𝑧 = 1, π‘₯ + 4𝑦 + 5𝑧 = 25, 3π‘₯ − 4𝑦 + 𝑧 = 2
[Dec 2017]
6. π‘₯ + 𝑦 + 𝑧 = 1, 4π‘₯ + 3𝑦 − 𝑧 = 6, 3π‘₯ + 5𝑦 + 3𝑧 = 4
[Dec 2016]
7. 2π‘₯ + 5𝑦 + 7𝑧 = 52 , 2π‘₯ + 𝑦 − 𝑧 = 0 , π‘₯ + 𝑦 + 𝑧 = 9
[Aug 2020, June 2019, Dec 2015, 2013, 2012]
8. 2π‘₯1 + π‘₯2 + 3π‘₯3 = 1, 4π‘₯1 + 4π‘₯2 + 7π‘₯3 = 1, 2π‘₯1 + 5π‘₯2 + 9π‘₯3 = 3
[Dec 2019]
9. π‘₯1 + π‘₯2 + π‘₯3 + π‘₯4 = 2 , 2π‘₯1 − π‘₯2 + 2π‘₯3 − π‘₯4 = −5 , 3π‘₯1 + 2π‘₯2 + 3π‘₯3 + 4π‘₯4 = 7 , π‘₯1 − 2π‘₯2 − 3π‘₯3 + 2π‘₯4 = 5
Gauss-Seidel Method:
Solve the following system of equations by Gauss Seidel Method performing 3 iterations:
1. 20π‘₯ + 𝑦 − 2𝑧 = 17 , 3π‘₯ + 20𝑦 − 𝑧 = −18 , 2π‘₯ − 3𝑦 + 20𝑧 = 25
[Aug 2021, June 2019, Dec 2019, 2018, 2017, Dec 2017, 2016, 2015, 2012]
2. 3π‘₯ + 8𝑦 + 29𝑧 = 71, 83π‘₯ + 11𝑦 − 4𝑧 = 95; 7π‘₯ + 52𝑦 + 13𝑧 = 104
[Jan 2021, June 2018]
3. 10π‘₯ + 2𝑦 + 𝑧 = 9 , π‘₯ + 10𝑦 − 𝑧 = −22 , −2π‘₯ + 3𝑦 + 10𝑧 = 22
[June 2017]
4. π‘₯ + 𝑦 + 54𝑧 = 110 , 27π‘₯ + 6𝑦 − 𝑧 = 85 , 6π‘₯ + 15𝑦 + 2𝑧 = 72
[Dec 2015, 2014]
9 4 6
5. 5π‘₯ − 𝑦 = 9, π‘₯ − 5𝑦 + 𝑧 = −4, 𝑦 − 5𝑧 = 6 taking (5 , 5 , 5) as first approximation.
6.
7.
8.
9.
[Dec 2011]
10π‘₯ + 𝑦 + 𝑧 = 12, π‘₯ + 10𝑦 + 𝑧 = 12, π‘₯ + 𝑦 + 10𝑧 = 12
[June 2016]
12π‘₯ + 𝑦 + 𝑧 = 31, 2π‘₯ + 8𝑦 − 𝑧 = 24, 3π‘₯ + 4𝑦 + 10𝑧 = 58
[Jan 2019, Dec 2018]
5π‘₯ + 2𝑦 + 𝑧 = 12 , π‘₯ + 4𝑦 + 2𝑧 = 15 , π‘₯ + 2𝑦 + 5𝑧 = 20
[Aug 2020]
5π‘₯ + 2𝑦 + 𝑧 = 12 , π‘₯ + 4𝑦 + 2𝑧 = 15 , π‘₯ + 2𝑦 + 5𝑧 = 20 taking initial approximation as (1,0,3)
14
Calculus and Differential Equations / 21MAT11
DEPARTMENT OF MATHEMATICS
Rayleigh Power Method:
Find the largest Eigen value and the corresponding Eigen vector of the following matrices using power method:
2 0 1 οƒΉ
1. 𝐴 = οƒͺ0 2 0οƒΊ . Take (1,0,0) as initial vector. Carry out six iterations.
οƒͺ
οƒΊ
οƒͺ1 0 2
[June 2019, Dec 2019, 2018, 2016, Dec 2017, 2016, 2015]
 6 ο€­2 2 οƒΉ
2. A ο€½ οƒͺο€­ 2 3 ο€­ 1οƒΊ . Take (1, 1, 1) as initial vector. Carry out five iterations.
οƒͺ
οƒΊ
οƒͺ 2 ο€­ 1 3 
[Aug 2021, Aug 2020, Jan 2019, Dec 2018, 2017, 2011, June 2016, 2015, 2012]
3.
4.
 2 ο€­1 0 οƒΉ
A ο€½ οƒͺοƒͺο€­ 1 2 ο€­ 1οƒΊοƒΊ . Take (1, 0, 0) as initial vector. Carry out four iterations.
οƒͺ 0 ο€­ 1 2 
[June 2018, 2012, Dec 2016, 2015, 2013]
1 6 1οƒΉ
[June 2018, 2017]
A ο€½ οƒͺοƒͺ1 2 0οƒΊοƒΊ . Take (1, 1, 0) as initial vector. Carry out six iterations.
οƒͺ0 0 3
5.
 1 3 ο€­ 1οƒΉ
A ο€½ οƒͺοƒͺ 3 2 4 οƒΊοƒΊ . Take (0, 0, 1) as initial vector. Carry out five iterations.
οƒͺ 1 4 10 
[Dec 2014]
6.
5 0 1 οƒΉ
A ο€½ οƒͺοƒͺ0 ο€­ 2 0οƒΊοƒΊ . Take (1, 0, 0) as initial vector. Carry out six iterations.
οƒͺ1 0 5
[Dec 2017]
7.
 4 1 ο€­ 1οƒΉ
A ο€½ οƒͺοƒͺ 2 3 ο€­ 1οƒΊοƒΊ . Take (1, 0, 0) as initial vector. Carry out seven iterations.
οƒͺ 2 1 5 
[Jan 2021, Dec 2014]
8.
25 1 2 οƒΉ
A ο€½ οƒͺοƒͺ 1 3 0 οƒΊοƒΊ . Take (1, 0, 0) as initial vector. Carry out seven iterations.
οƒͺ 2 0 ο€­ 4
[June 2013]
9.
10 2 1 οƒΉ
A ο€½ οƒͺοƒͺ 2 10 1 οƒΊοƒΊ . Take (1, 1, 0) as initial vector. Carry out six iterations.
οƒͺ 2 1 10
[June 2014]
Self-study:
Gauss-Jacobi iterative Method:
1.
2.
3.
4.
5.
28π‘₯ + 4𝑦 − 𝑧 = 32, 2π‘₯ + 17𝑦 + 4𝑧 = 35, π‘₯ + 3𝑦 + 10𝑧 = 24
10π‘₯ + 2𝑦 + 𝑧 = 9 , 2π‘₯ + 20𝑦 − 2𝑧 = −44 , −2π‘₯ + 3𝑦 + 10𝑧 = 22
6π‘₯ − 𝑦 − 𝑧 = 19 , 3π‘₯ + 4𝑦 + 𝑧 = 26 , π‘₯ + 2𝑦 + 6𝑧 = 22
3π‘₯ + 𝑦 + 𝑧 = 1 , π‘₯ + 3𝑦 − 𝑧 = 11 , π‘₯ − 2𝑦 + 4𝑧 = 21
23π‘₯ + 13𝑦 + 3𝑧 = 29 , 5π‘₯ + 23𝑦 + 7𝑧 = 37 , 11π‘₯ + 𝑦 + 23𝑧 = 43
15
Calculus and Differential Equations / 21MAT11
DEPARTMENT OF MATHEMATICS
Cayley-Hamilton Method:
1. By using Cayley-Hamilton theorem, find the inverse of the following matrices:
1 0 2οƒΉ
(i) A ο€½ οƒͺ 0 2 1 οƒΊ
οƒͺ
οƒΊ
οƒͺ 2 0 3 
 2 ο€­1 1 οƒΉ
(ii) A ο€½ οƒͺ ο€­1 2 ο€­1οƒΊ
οƒͺ
οƒΊ
οƒͺ 1 ο€­1 2 
1 0 2οƒΉ
(iii) A ο€½ οƒͺ 0 2 1 οƒΊ
οƒͺ
οƒΊ
οƒͺ 2 0 2 
*************************************
16
Download