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¦ ο½ r dο± dr [Aug 2021, June 2019, 2018, Dec 2018, 2019, 2017] 1 1 1 ππ 2 2) With usual notations prove that, the pedal equation of the form π2 = π 2 + π 4 (ππ) [Feb 2021, Aug 2020, Dec 2017] 3) Find the angle between the radius vector and the tangent for the following curves: i) π = π(1 − cos π) at π = π⁄3 iii) π π = ππ (cos ππ + sin ππ) [Dec 2017] ii) ππ ππ 2 (π⁄2) = 2π iv) π 2 = π2 sin 2π 4) Find the angle of intersection of the following curves: ππ π i) π = π(1 + sin π) & π = π(1 − π πππ) [June 2017] ii) π = 1+π & π = 1+π2 [June 2016] 2 2 2 2 iii) π = 2π πππ & π = 2(π πππ + πππ π) [June 2017] iv) π sin 2π = π & π cos 2π = π v) π(1 + πππ π) = π & π(1 − πππ π) = π [Dec 2016] vi) π π = ππ (sin ππ + cos ππ) & π π = ππ sin ππ vii) π π = ππ cos ππ & π π = π π sin ππ [June 2018, Dec 2018] viii) π = πππππ & π = π/ππππ [Aug 2021, Dec 2017, 2016, 2014, 2012] ix) π 2 sin 2π = 4 & π 2 = 16π ππ2π [Dec 2015, 2012, June 2015] 2 2 x) π = π(1 + πππ π) & π = π cos 2π [Dec 2016, June 2016] xi) π = π(1 + sin π) & π = π(1 − cos π) [Dec 2017, June 2016] xii) π = π(1 + cos π) & π = π(1 − cos π) [June 2019, 2018, 2016, 2015, 2009, Dec 2019, Dec 2017, 2015] xiii) π = sin π + cos π & π = 2 sin π [Dec 2019] xiv) π = 6 cos π & π = 2(1 + cos π) [Aug 2020] xv) π = 2 sin π & π = 2 cos π [Feb 2021] 5) Find the pedal equation for the following curves: i) π π cos ππ = ππ [June 2016] ii) π = 2π⁄1 + cos π [June 2018, 2011, Dec 2016] π iii) π = sec β ππ [Dec 2014] iv) π = π + π cos π [June 2017] π 3 v) π = π sin 3π vi) π = 1 + π cos π vii) π(1 − πππ π) = 2π [Aug 2020, Dec 2012] viii) π π = π(1 + πππ ππ) [Dec 2015] 2π 2 2 ix) π = 1 − sin π x) π = π πππ 2π [Aug 2021, June 2014] π π π π xi) π sec ππ = π or π = π cos ππ [June 2019, Dec 2019, June 2017, Dec 2016, 2015, 2013] ππππ‘πΌ xii) π = ππ [Dec 2018] xiii) π = π(1 + cos π) [Aug 2021, June 2018, 2016, Dec 2017] π π (cos xiv) π = π ππ + sin ππ) [Feb 2021, Dec 2016, 2011, June 2010] 6) Find the pedal equation of π π = ππ sin ππ + π π cos ππ in the form π2 (π2π + π 2π ) = π 2π+2 . [June 2015] 7) Derive Radius of Curvature in the Cartesian form. 2 8) Show that the radius of curvature of the curve π¦ = π πΆππ β(π₯⁄π) is y , where c is a constant. c 9) Find the radius of curvature of the parabola π¦ 2 = 4ππ₯ at the point (a, a). [Aug 2020, Dec 2016] π π 10) Find the radius of curvature of the curve √π₯ + √π¦ = √π at (2 , 2) . [Dec 2017] π 11) Find the radius of curvature of the curve π¦ = 4π πππ₯ − π ππ2π₯ at π₯ = ⁄2 . [Dec 2017] π₯ π₯ 12) Show that the radius of curvature of the curve π¦ = πππππ ππ (π) is ππ ππ (π) [Dec 2018, 2019] 13) Find the radius of curvature for the curve π₯ 4 + π¦ 4 = 2 ππ‘ (1,1). [Dec 2016, June 2016] 3π 3π 3 3 14) Find the radius of curvature for the Folium De-Cartes π₯ + π¦ = 3ππ₯π¦ 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 + π¦ 3 = 3π₯π¦ at the point (2 , 2) is 8√2 . [Dec 2014] 2 2 2) 16) Find the radius of curvature for the curve π₯ π¦ = π(π₯ + π¦ at the point (−2π, 2π). [Dec 2017, 2012] 2 3 3 17) Find the radius of curvature for the curve π π¦ = π₯ − π at the point where the curve cuts π₯-axis. [June 2019, Dec 2010] 18) Find the radius of curvature for the curve π¦ 2 = 4π2 (2π−π₯) π₯ π2 (π−π₯) 19) Find the radius of curvature of the curve π¦ 2 = π₯ 3 where the curve meets the π₯-axis. at the point (π, 0). [June 2018, Dec 2017, 2013] 20) Find the radius of curvature of the curve π₯π¦ 2 = π − π₯ 3 at (π, 0). 2 2π ⁄3 ππ₯ 21) If π¦ = π+π₯ , then show that ( π ) π¦ 2 [June 2018] π₯ 2 = (π₯ ) + (π¦) where π is the radius of curvature at (π₯, π¦) [Feb 2021, June 2017, Dec 2016] π₯ π¦ 22) Prove that the radius of curvature at any point (π₯, π¦) on the curve √π + √π =1 is π = 23) Find the radius of curvature for the curve π₯ = ππππ(sec π‘ + tan π‘) , π¦ = π sec π‘ π 2(ππ₯+ππ¦)3/2 ππ . [June 2011] [July 2015] 24) Find the radius of curvature for the Astroid π₯ = ππππ π, π¦ = ππ ππ π ππ‘ π = 4 25) Find the radius of curvature of the curve π₯ = (cos π‘ + π‘ sin π‘), π¦ = π(sin π‘ - π‘ cos π‘) π‘ 26) Find the radius of curvature of the curve π₯ = π [cos π‘ + log tan 2] π¦ = π sin π‘ [Aug 2020, July 2015] 27) Show that the radius of curvature at any point π on the Cycloid π₯ = π(π + sin π) , π¦ = π(1 − cos π) is π 4π cos 2 [July 2012] 28) Derive Radius of Curvature in the Polar form. 29) Find the radius of curvature to the curve π = π sin ππ at the pole. [June 2018] π π π−1 30) Show that the radius of curvature of the curve π = π cos ππ varies inversely as π . [Dec 2016] 31) Find the radius of curvature of the curve π = π(1 + cos π). [Dec 2018, 2017] 32) Show that for the curve π(1 − cos π) = 2π, π2 varies as π 3 . [June 2019] π π 33) Find the radius of curvature of the curve π = π sin ππ. [Dec 2017, 2015] 34) Find the radius of curvature of the polar curve π 2 = π2 cos 2π. [June 2017] 35) Find the radius of curvature of the curve π = π(1 − cos π). 3 36) Find the radius of curvature for the Parabola 2π π 3 = 1 + cos π [Feb 2021] π 37) Show that for the curve ππππ 2 ( 2) = π, π2 varies as π 3 . 38) Show that the radius of curvature of the curve ππ2 = π 3 is 2 π3 π2 39) Show that for the curve π = 2π , π is a constant. 40) Show that for the curve π2 = ππ, π2 varies as π 3 . 41) Show that for the ellipse in pedal form, 1 π2 1 1 π2 3π π2 = π2 + π2 − π2 π2 , the radius of curvature at the point (p, r) is π2 π2 π3 . Self-Study: 42) Find the coordinates of the center of curvature at any point of the parabola π₯ 2 = 4ππ¦. Hence find its evolute. 43) Show that the evolute of the parabola π¦ 2 = 4ππ₯ is 27ππ¦ 2 = 4(π₯ − 2π)3 . [Aug 2021, 2020, Feb 2021, Dec 2019, June 2019, Dec 2018] 44) Show that the evolute of the cycloid π₯ = π(π − π πππ), π¦ = π(1 − πππ π) is another equal cycloid. 45) Find the evolute of the curve π₯ = ππππ 3 π, π¦ = ππ ππ3 π, i,e π₯ 2/3 + π¦ 2/3 = π2/3 . π‘ π₯ 46) Show that the evolute of the curve π₯ = π [cos π‘ + πππ tan (2)] , π¦ = π sin π‘ is π¦ = π cosh (π). [Aug 2020] 2 47) Find the evolute of the curve π₯π¦ = π . π₯2 π¦2 48) Find the evolute of the ellipse π2 + π2 = 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 πππ(πππ π₯) about the point at π₯ = π/3 up to the 4th degree term Expand π‘πππ₯ about the point π₯ = π/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 π πππ₯ upto 4th degree term. [Dec 2015] 5 Using Maclaurin’s series expand π¦ = π‘πππ₯ 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 π tan −1 π₯ in Maclaurin’s expansion upto first three non –vanishing terms. x 2 x3 x 4 ο ο« οο 2 6 24 [July 2021, Aug 2020, June 2018, Dec 2018, 2015, 2014] 8. Using Maclaurin’s theorem prove that 1 ο« sin 2 x ο½ 1 ο« 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 πππ(1 + πππ π₯) by the Maclaurin’s series up to the term containing π₯ 4 . [June 2019, 2017, Dec 2017] 4 11. Expand πππ(1 + π πππ₯) by the Maclaurin’s series up to the term containing π₯ . [Dec 2012] 6 12. Expand πππ(π πππ₯)by the Maclaurin’s series up to the term containing π₯ [Dec 2019, 2017, 2010, June 2017, 2016, 2012] π₯ 13. Expand πππ(1 + π ) using Maclaurin’s series up to and including 3rd degree term. [June 2016, Dec 2016, 2013] 14. Expand πππ(1 + π₯) using Maclaurin’s series up to the term containing π₯ 4 . [Dec 2017] π πππ₯ 4 15. Expand π by the Maclaurin’s series up to the term containing π₯ . [Dec 2017] π₯ 4 16. Obtain Maclaurin series for π πππ π₯ upto the term containing π₯ [June 2018] ππ₯ 17. Expand 1+π 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 π(π₯) = π π₯ −1 [Dec 2014] π₯ Evaluate the following Indeterminate forms: 19. limπ₯→π (π πππ₯) π‘πππ₯ 20. 2 π₯ 21. limπ₯→0 (π + π₯) 1 π₯ [Dec 2016] 1 ππ₯ +π π₯ π₯ 23. limπ₯→0 ( ) 2 1 [Aug 2020, Dec 2017] π‘πππ₯ 2 limπ₯→0 ( π₯ )π₯ 1 2π₯ +3π₯ +4π₯ π₯ 22. limπ₯→0 ( ) 3 24. limπ₯→0 (πππ π₯)πππ‘ 1 25. [Dec 2016, June 2016] 2π₯ 1 2 sin π₯ π+π₯ π₯ limπ₯→0 (π−π₯) 26. limπ₯→0 (π₯) [Dec 2010] [June 2017] [Jan 2021] ππ₯ 27. lim (2 − π₯→π π₯ π‘ππ( 2π) ) π [Dec 2017, June 2016] 28. limπ₯→π (π‘πππ₯)πππ π₯ 2 1 29. π πππ₯ π₯ limπ₯→0 ( π₯ ) 30. [July 2021] 32. limπ₯→0 ( 1 31. limπ₯→0 (πππ π₯)π₯2 33. limπ₯→0 ( 1 ππ₯ +π π₯ +π π₯ π₯ ) 3 π₯ ) [June 2017] [Aug 2020] [Jan 2021, Aug 2020, Dec 2019, June 2019] 1 ππ₯ +π π₯ +π π₯ +ππ₯ π₯ 34. limπ₯→0 ( 1 π πππ₯ 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 π’ = π₯ π¦ , then verify that οΆ 2u οΆ 2u ο½ . οΆxοΆy οΆyοΆx ο© ο¨ ο1 / 2 οΆ 2u οΆ 2u οΆ 2u 36. If u ο½ x 2 ο« y 2 ο« z 2 , then prove that ο« ο« ο½ 0. οΆx 2 οΆy 2 οΆz 2 37. If π’ = π₯ 2 π¦ + π¦ 2 π§ + π§ 2 π₯, then show that π’π₯ + π’π¦ + π’π§ = (π₯ + π¦ + π§)2 ππ§ ππ§ 38. If π§ = π ππ₯+ππ¦ π(ππ₯ − ππ¦) , show that π ππ₯ + π ππ¦ = 2πππ§ Total derivative and Chain rule: 39. If π’ = π₯ 3 π¦ 2 + π₯ 2 π¦ 3 where π₯ = ππ‘ 2 , π¦ = 2ππ‘, find ππ’ ππ‘ . [Dec 2016] π₯ ππ’ π¦ ππ‘ ππ’ 40. If π’ = π‘ππ−1 (π¦) where π₯ = π π‘ − π −π‘ , π¦ = π π‘ + π −π‘ , find 41. If π’ = π‘ππ−1 (π₯ ) where π₯ = π π‘ − π −π‘ , π¦ = π π‘ + π −π‘ , find . [June 2017] . [June 2018] ππ‘ 2π‘ 42. If π’ = π₯ 2 + π¦ 2 + π§ 2 where π₯ = π 2π‘ , π¦ = π 2π‘ πππ 2π‘, π§ = π π ππ2π‘, find 43. If π’ = log(π₯ + π¦ + π§) where π₯ = π −π‘ , π¦ = π πππ‘, π§ = πππ π‘, find π’ 44. If π§ = π(π₯, π¦) where π₯ = π − π −π£ ,π¦ = π −π’ ππ’ ππ‘ . [Dec 2015] ππ‘ ππ§ π£ ππ’ ππ§ ππ§ ππ§ − π prove that ππ’ − ππ£ = π₯ ππ₯ − π¦ ππ¦ 2 1 ο¦ οΆz οΆ ο¦ οΆz οΆ ο¦ οΆz οΆ ο¦ οΆz οΆ 45. If π§ = π(π₯, π¦) where π₯ = ππππ π, π¦ = ππ πππ prove that ο§ ο· ο« ο§ο§ ο·ο· ο½ ο§ ο· ο« 2 ο§ ο· r ο¨ οΆο± οΈ ο¨ οΆx οΈ ο¨ οΆy οΈ ο¨ οΆr οΈ 2 2 [Dec 2015] 2 [June 2018, 2017, Dec 2017, 2014] ο¦yοx zοxοΆ οΆu οΆu οΆu ο·ο· , find the value of x 2 , ο« y2 ο« z2 zx οΈ οΆx οΆy οΆz ο¨ xy 46. If u ο½ uο§ο§ 47. If π’ = π (π¦ − π§, π§ − π₯, π₯ − π¦) then prove that ππ’ ππ’ ππ₯ ππ’ ππ’ [Dec 2011] ππ’ + ππ¦ + ππ§ = 0 ππ’ 48. If π’ = π(π₯ − π¦, π¦ − π§, π§ − π₯), show that ππ₯ + ππ¦ + ππ§ = 0 π₯ π¦ π§ ππ’ ππ’ [Jan 2021, Dec 2018, 2017, June 2015] [July 2021, Aug 2020, Dec 2019] ππ’ 49. If π’ = π (π¦ , π§ , π₯) then prove that π₯ ππ₯ + π¦ ππ¦ + π§ ππ§ = 0 [June 2019, 2018, 2017, 2016, 2014, Dec 2019, 2016] 50. If π’ = π (2π₯ − 3π¦, 3π¦ − 4π§, 4π§ − 2π₯) then find π¦ ππ’ ππ’ 1 ππ’ 2 ππ₯ 1 ππ’ 1 ππ’ + 3 ππ¦ + 4 ππ§ [June 2017, Dec 2013] ππ’ 51. If π’ = π (π₯π§, π§ ) prove that π₯ ππ₯ − π¦ ππ¦ − π§ ππ§ = 0 Maxima and Minima: 52. Find the extreme values of f ( x, y) ο½ x3 ο« 3xy 2 ο 15x 2 ο 15 y 2 ο« 72 x 53. Find the extreme values of π(π₯, π¦) = π₯ 3 + π¦ 3 − 3π₯ − 12π¦ + 20 [Aug 2020, June 2017] [June 2019, Dec 2019, June 2018] 54. Find the extreme values of π(π₯, π¦) = π₯ 3 π¦ 2 (1 − π₯ − π¦) [Dec 2012] 55. Find the extreme values of π(π₯, π¦) = π₯ 3 + π¦ 3 − 3ππ₯π¦, π > 0 [June 2011] 56. Find the extreme values of π(π₯, π¦) = π₯ 4 + π¦ 4 − 2(π₯ − π¦)2 [June 2015] 57. Examine the function π₯π¦(π − π₯ − π¦) for extreme values. [June 2016] 4 Calculus and Differential Equations / 21MAT11 DEPARTMENT OF MATHEMATICS 58. Examine the function π(π₯, π¦) = 1 + sin(π₯ 2 + π¦ 2 ) for extreme values. [Dec 2012] 59. Find the extreme values of π(π₯, π¦) = π πππ₯ + π πππ¦ + sin(π₯ + π¦) [June 2016] 60. Examine the function π(π₯, π¦) = 2 + 2π₯ + 2π¦ − π₯ 2 − π¦ 2 for its extreme values. [Jan 2021] π π 61. Show that the function π₯π¦(π − π₯ − π¦) is maximum at (π , π). Hence find maximum value if π > 0. [July 2021] Jacobians: οΆ ( x, y ) οΆ (u, v) οΆ (r ,ο± ) 63. If π₯ = ππππ π and π¦ = ππ πππ find οΆ ( x, y ) 62. If π₯ = π’ (1 − π£), π¦ = π’ π£ then evaluate [Dec 2011] 64. If π’ = π₯ 2 − π¦ 2 , π£ = 2π₯π¦ and π₯ = ππππ π, π¦ = ππ πππ determine the value of 65. If π₯ = ππ ππππππ ∅, π¦ = ππ ππππ ππ∅ and π§ = ππππ π find οΆ (u, v) οΆ (r ,ο± ) οΆ ( x, y , z ) οΆ ( r ,ο± , ο¦ ) [June 2018, 2016, Dec 2013] οΆ (u, v, w) where π’ = π₯ 2 + π¦ 2 + π§ 2 , π£ = π₯π¦ + π¦π§ + π§π₯, π€ = π₯ + π¦ + π§ οΆ ( x, y , z ) [July 2021, Dec 2019, Dec 2018, 2017, 2016, 2014, June 2017] ο¦ (u, v, w) οΆ zx xy yz ο·ο· ο½ 4 67. If u ο½ and v ο½ and w ο½ then show that J ο§ο§ [Dec 2016, 2010, June 2016, 2015] y z x ο¨ ( x, y , z ) οΈ 66. Find 68. If π’ = π₯ + 3π¦ + π§ 3 , π£ = π₯ 2 π¦π§, π€ = 2π₯ 2 – π₯π¦, Evaluate οΆ (u, v, w) at [1, −1, 0] οΆ ( x, y , z ) [June 2018] οΆ (u, v, w) at [1, −1, 0] οΆ ( x, y , z ) [June 2019, Dec2018, Dec 2016, 2015] οΆ (u, v, w) 70. If π’ = π₯ + 3π¦ 2 , π£ = 4π₯ 2 π¦π§, π€ = 2π§ 2 – π₯π¦ Evaluate at [1, −1, 0] [Dec 2017] οΆ ( x, y , z ) 69. If π’ = π₯ + 3π¦ 2 − π§ 3 , π£ = 4π₯ 2 π¦π§, π€ = 2π§ 2 – π₯π¦, Evaluate 71. If π₯ + π¦ + π§ = π’, π¦ + π§ = π’π£, π§ = π’π£π€, then evaluate 72. If π₯ + π¦ + π§ = π’, π¦ + π§ = π£, π§ = π’π£π€, find οΆ ( x, y, z) οΆ (u , v, w ) οΆ ( x, y , z ) οΆ (u, v, w) [Dec 2017, 2011] 73. If π’ = 3π₯ + 2π¦ − π§ ; π£ = π₯ − 2π¦ + π§ ; π€ = π₯ 2 + 2π₯π¦ − π₯π§ π βππ€ π‘βππ‘ π₯ π¦ π§ 74. If π’ = π¦−π§ , π£ = π§−π₯ , π€ = π₯−π¦ , find [June 2018, 2015, Dec 2015] π(π’,π£,π€) (π₯,π¦,π§) =0 οΆ (u, v, w) οΆ ( x, y , z ) [June 2017] 75. Given π₯ = π(π’ + π£), π¦ = π(1 − π£) and π’ = π 2 πππ 2π, π£ = π 2 π ππ2π find 76. If π’ = π₯ cos π¦ cos π§ , π£ = π₯ cos π¦ sin π§ , π€ = π₯ sin π¦ , π βππ€ π‘βππ‘ [Jan 2021] π(π’,π£,π€) π(π₯,π¦,π§) οΆ ( x, y ) οΆ (r ,ο± ) [Dec 2017] = −π₯ 2 cos π¦. [Aug 2020] 5 Calculus and Differential Equations / 21MAT11 DEPARTMENT OF MATHEMATICS Self-Study: Euler’s Theorem and Problems 77. If π’ = tan−1 ( π₯ 3 +π¦ 3 π₯−π¦ π₯ 3 +π¦ 3 78. If π’ = sin−1 ( π₯−π¦ π₯ 4 +π¦4 79. If π’ = log ( π₯+π¦ ππ’ ππ’ ππ’ ππ’ ) Prove that π₯ ππ₯ + π¦ ππ¦ = sin 2π’ . ) Prove that π₯ ππ₯ + π¦ ππ¦ = 2 tan π’ . ) Prove that π₯π’π₯ + π¦π’π¦ = 3 . π₯2π¦2 80. If π’ = sin−1 ( π₯+π¦ ) Prove that π₯π’π₯ + π¦π’π¦ = 3 tan π’ . π₯3π¦3 3 81. If π’ = tan−1 (π₯ 3 +π¦ 3 ) Prove that π₯π’π₯ + π¦π’π¦ = 2 sin 2π’ . Lagrange Undetermined Multipliers with single constraint: 82. If π, π, π are angles of a triangle, show that the maximum value of π πππ₯π πππ¦π πππ§ is π√π π 83. Find the minimum value of π₯ 2 + π¦ 2 + π§ 2 under the condition π₯ + π¦ + π§ = 3π. [July 2021, June 2017] [Dec 2018] 84. Find the maximum value of ππ subject to the condition π₯ 2 + π₯π¦ + π¦ 2 = π2 . 85. The temperature T at any point (π₯, π¦, π§) in space in π = 400π₯π¦π§ 2 . Find the highest temperature on the surface of the unit sphere π₯ 2 + π¦ 2 + π§ 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 ο½ xy 3 ο xy [June 2016] 2. ο¨ x 2 ο« y 2 ο« x ο© dx ο« xy dy ο½ 0 [July 2021, June 2018, 2017] dx dy dy ο« y tan x ο½ y 3 sec x [July 2021, Jan 2021, June 2019, Dec 2017] 4. ο« y tan x ο½ y 2 sec x [Dec 2018, 2014, June 2016] 3. dx dx dy y cos x ο« sin y ο« y dy ο« ο½ 0 [Jan 2020, Dec 2018, 2017, 2015] 6. xy (1 ο« xy 2 ) 5. ο½ 1 [June 2018, 2017, 2015, Dec 2017] dx sin x ο« x cos y ο« x dx 1. 7. ye xy dx ο« ( xe xy ο« 2 y)dy ο½ 0 dy 9. x ο« y ο½ x 3 y 6 dx [Dec 2017] 8. [Dec 2017] dy sin x cos 2 x 10. ο y tan x ο½ dx y2 [Dec 2015] 11. (2 x log x ο xy )dy ο« 2 ydx ο½ 0 dy 2 y2 ο yο½ 3 dx x x [Dec 2016] 12. ( x 2 ο« y 3 ο« 6 x)dx ο« y 2 xdy ο½ 0 [June 2016, 2015] [Dec 2016] dy y ο« ο½ y2x dx x ο¦ dy οΆ 15. (4 xy ο« 3 y 2 ο x)dx ο« x( x ο« 2 y)dy ο½ 0 [Jan 2020, Dec 2015] 16. e y ο§ ο«1ο· ο½ e x ο¨ dx οΈ 13. (2 x 3 ο xy 2 ο 2 y ο« 3)dx ο ( x 2 y ο« 2 x)dy ο½ 0 [Dec 2017] [Dec 2017] 14. [Dec 2015] 17. (1 ο« 2 xy cos x 2 ο 2 xy )dx ο« (sin x 2 ο x 2 )dy ο½ 0 [June 2015] 18. ( xy 3 ο« y)dx ο« 2( x 2 y 2 ο« x ο« y 4 )dy ο½ 0 [Dec 2014] 19. r sin ο± ο cosο± dr ο½ r2 dο± [June 2018] 21. ( y 2 e xy ο« 4 x 3 )dx ο« (2 xye xy ο 3 y 2 )dy ο½ 0 [June 2017] 2 23. (1 ο« e x y 2 x x )dx ο« e y (1 ο )dy ο½ 0 y ππ 25. π sin π − cos π ππ = π 2 [June 2019] 20. dy ο« x sin 2 y ο½ x 3 cos 2 y dx ο ο ο [June 2017] ο 22. y 4 ο« 2 y dx ο« xy 3 ο« 2 y 4 ο 4 x dy ο½ 0 ππ¦ 24. π₯ 3 ππ₯ − π₯ 2 π¦ = −π¦ 4 cos π₯ [Aug 2020] [Jan 2020] 26. [4π₯ 3 π¦ 2 + π¦ cos(π₯π¦)]ππ₯ + [2π₯ 4 π¦ + π₯ cos(π₯π¦)]ππ¦ = 0 1 27. Solve [π¦ (π₯ + π₯) + πππ π¦] ππ₯ + [π₯ + ππππ₯ − π₯π πππ¦]ππ¦ [Aug 2020] [Jan 2021] 28. (cos x tan y ο« cos( x ο« y))dx ο« (sin x sec 2 y ο« cos( x ο« y)dy ο½ 0 ORTHOGONAL TRAJECTORIES: 1. 2. 3. 4. 5. Find the orthogonal trajectories of a family of circles π = 2ππππ π, where ‘π’ is a parameter. [June 2017] π π Find the orthogonal trajectories of the curve π πππ ππ = π [June 2019, Dec 2016] Find the orthogonal trajectories of the curve π π = ππ πππ ππ [Dec 2017, 2015 June 2016] Find the orthogonal trajectories of the curve π π = ππ π ππ ππ, where ‘π’ is a parameter. [June 2017, Dec 2015] Find the orthogonal trajectories of the curve π π π ππ ππ = ππ , where ‘π’ is a parameter. [June 2018] 7 Calculus and Differential Equations / 21MAT11 DEPARTMENT OF MATHEMATICS π 6. Show that orthogonal trajectories of the families of cardioides π = ππππ 2 ( 2) is another family of cardioides π π = ππ ππ2 ( 2) [Dec 2014] 2a ο½ 1 ο cos ο± 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 = π(1 + πππ π) [Aug 2020] 2 ο¦ k οΆ 10. Find the orthogonal trajectories of the family of curves ο§ο§ r ο« ο·ο· cos ο± =a, a being a parameter. [July 2015] r οΈ ο¨ 7. Find the orthogonal trajectories of a family 11. Show that the family of parabolas π¦ 2 = 4π(π₯ + π) is self-orthogonal. [Jan 2021, Dec 2018, June 2018, 2016] 12. Find the orthogonal trajectories of the family of the curves π¦ = π₯ + ππ −π₯ , where π is the parameter. [Dec 2017] 13. Find the orthogonal trajectories of the curve π¦ = ππ₯ 3 [Dec 2017] 2 2 x y ο« 2 ο½ 1 , λ is a parameter, is self-orthogonal. 14. Show that the family of the curves 2 a ο«ο¬ b ο«ο¬ [June 2017, Dec 2015, Dec 2012] 2 2 x y ο½ 1 , λ is a parameter 15. Find the orthogonal trajectories of the family of the curve 2 ο« 2 a b ο«ο¬ [July 2021, June 2015] 16. Find the orthogonal trajectories of the family of the curve x 2 / 3 ο« y 2 / 3 ο½ a 2 / 3 [Dec 2016] 2 17. Find the orthogonal trajectories of the family of the curve π¦ = 4ππ₯ NEWTON’S LAW OF COOLING: 1. Water at temperature 10° πΆ takes 5 minutes to warm upto 20° πΆ in a room temperature 40° πΆ. Find the temperature after 20 minutes. [June 2018, 2017] 2. A body originally at 80β cools down to 60β in 20 minutes. The temperature of the air being 40β. 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 πΆ and placed in air at 10 πΆ. After one hour its temperature becomes 60° πΆ. How much additional time is required for it to cool to 30° πΆ? [Dec 2017, 2015] 4. If the temperature of the air is 30β and the substance cools from 100β to 70β in 15 minutes. Find when the temperature will be 40β. [June 2016, Dec 2014] 5. Water at temperature 100β cools down to 88β in 10 minutes in a room temperature of 25β. Find the temperature of the water after 20 minutes [June 2018] 6. Suppose an object is heated to 300° πΉ and allowed to cool in a room whose temperature is 80° πΉ. After 10 minutes the temperature of the object is 250° πΉ. What will be its temperature after 20 minutes? [June 2016] ° 7. A body in air at 25β cools from 100β to 75 πΆ 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 30β and the temperature of the body cools from 80β to 60β 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 π2 + 2ππ¦πππ‘π₯ − π¦ 2 = 0 Solve π₯π¦(π2 + 1) = (π₯ 2 + π¦ 2 )π Solve π(π + π¦) = π₯(π₯ + π¦) Solve yp 2 ο« ( x ο y) p ο x ο½ 0 Solve π₯π2 − (2π₯ + 3π¦)π + 6π¦ = 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₯π¦π2 + π(3π₯ 2 − 2π¦ 2 ) − 6π₯π¦ = 0 7. Solve π₯ 2 (ππ₯ ) + π₯π¦ ππ₯ − 6π¦ 2 = 0 ππ¦ 2 ππ¦ 8. Solve π₯ 2 π2 + 3π₯π¦π + 2π¦ 2 = 0 9. Solve 10. 11. 12. 13. 14. Solve Solve Solve Solve Solve ππ¦ ππ₯ 4 ππ₯ DEPARTMENT OF MATHEMATICS π₯ π¦ − ππ¦ = π¦ − π₯ π − (π₯ + 2π¦ + 1)π3 + (π₯ + 2π¦ + 2π₯π¦)π2 − 2π₯π¦π = 0 π₯π3 − π¦π2 + 1 = 0 by Clairaut’s equation (ππ₯ − π¦)(ππ¦ + π₯) = π2 π, use the substitution π = π₯ 2 , π = π¦ 2 (ππ₯ − π¦)(ππ¦ + π₯) = 2π by reducing into Clairaut’s form, taking π = π₯ 2 , π = π¦ 2 (ππ₯ − π¦)(ππ¦ + π₯) = 0 by reducing into Clairaut’s form, taking π = π₯ 2 , π = π¦ 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 πΈπ ππππ‘, 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 ‘π‘’. [Dec 2016] 3. A series circuit with resistance R, inductance L and electromotive force E is governed by the differential ππ equation πΏ ππ‘ = π π = πΈ, where L and R are constants and initially the current i is zero. Find the current at any time t. [Jan 2021] ππ 4. Solve the equation πΏ + π π = πΈ0 π πππ€π‘ Where L, R and πΈ0 are constants and discuss the case when t ππ‘ increases indefinitely. [Aug 2020, Jan 2020] ππΌ 5. Solve the L-R circuit πΏ π·π + π πΌ = πΈ 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 π¦ 2 log π¦ = π₯ππ¦ + π2 Solve π3 − 4π₯π¦π + 8π¦ 2 = 0 Solve π₯π4 − 2π¦π3 + 12π₯ 3 = 0 Solve π¦ = 2ππ₯ + π2 π¦ ************************************** 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 ο«3 ο y ο½ 0 3 2 dx dx dx d3y d2y dy ο« 2 ο« 4 ο« 4y ο½ 0 3 dx dx dx 4. (π·4 − π4 )π¦ = 0 [June 2018] 2. (π·4 − 1)π¦ = 0 [June 2010] [June 2017] [June 2017] (π· + 5π· + 6π· − 4π· − 8)π¦ = 0 [Dec 2016] 4 3 2 (4π· − 4π· − 23π· + 12π· + 36)π¦ = 0 [Dec 2016, 2011, June 2015] 4 3 2 (4π· − 8π· − 7π· + 11π· + 6)π¦ = 0 [Jan 2021, Jan 2020, Dec 2019, 2016, June 2018, 2016, 2015] (π·4 + 4π·3 – 5π·2 – 36π·– 36) π¦ (π₯) = 0 [Dec 2011] 4 3 2 9. ( D 4 ο 2D 3 ο« 2D 2 ο 2D ο« 1) y ο½ 0 2 dx (0) ο½ 15 10. d 2x ο« 5 dx ο« 6 x ο½ 0 given x(0) ο½ 0, dt dt dt 2 dy 13 ο½ ο for x ο½ 0. 11. 9 d 2y ο« 6 dy ο« y ο½ 0 given y ο½ 4 and dx 3 dx dx [Dec 2017] 2 dy d y d2y dy ο½ for x ο½ 0. ο« 4 ο« 5 y ο½ 0 given y ο½ 2 and 2 dx dx 2 dx dx d2y dy 13. [June 2018] 14.(π·2 + 4π· + 3)π¦ = π −π₯ [Jan 2020, Dec 2019] ο 6 ο« 9 y ο½ 6e 3 x 2 dx dx 12. d3y d2y dy ο« 2 ο 4 ο 4 y ο½ 3e x 3 dx dx dx 2 d y dy 17. ο« ο« y ο½ (1 ο e x ) 2 dx 2 dx 15. 19. 16. [June 2018] 18. (π·2 + 4)π¦ = π₯ 2 + πππ 2π₯ + 2−π₯ [Dec 2017] d3y d2y dy ο 2 ο« 4 ο 4 y ο½ sinh(2 x ο« 3) [Dec 2017] 3 dx dx dx 21. ο¨D 2 ο 3D ο« 2ο©y ο½ sin 2 x [Dec 2017] 23. ( D ο 4D ο« 4) y ο½ e ο« sin x 2 2x [June 2019] 25. (π·2 – 4π· + 4)π¦ = π 2π₯ + πππ 2π₯ + 4 [Dec 2014] 27. d2y ο« 4 y ο½ 2οx dx 2 [June 2018] d3y d 2 y dy ο«2 2 ο« ο½ e ο x ο« sin 2 x 3 dx dx dx [Dec 2016, 2011] [June 2015] ο¦ xοΆ 20. y ο’ο’ ο« 2 y ο’ ο« y ο½ coshο§ ο· ο¨2οΈ 22. (π·3 − 1)π¦ = 3πππ 2π₯ [Jan 2020, Dec 2019, 2010] 24. y '' ο 4 y ' ο« 4 y ο½ 8 cos 2 x 26. (π·2 – 4π· + 3)π¦ = π ππ3π₯πππ 2π₯ 28. y '' ο« 4 y ' ο 12 y ο½ e 2 x ο 3sin 2 x 29. (π·3 − π·)π¦ = 2π π₯ + 4πππ π₯ [June 2011] 30. d2y dy ο« 3 ο« 2y ο½ x2 2 dx dx 31. (π·2 + 4)π¦ = π₯ 2 + π −π₯ [June 2015] 32. d2y ο« 4 y ο½ x 2 ο« 2 οx ο« log 2 dx 2 33. (π·2 + 3π· + 2)π¦ = 4πππ 2 π₯ [Aug 2020] 2 35. d y dy ο 6 ο« 25 y ο½ e 2 x ο« sin x ο« x [June 2018] 2 dx dx 37. ( D3 ο« 8) y ο½ x 4 ο« 2 x ο« 1 39. d2y dy ο 4 ο« 13 y ο½ e3 x cosh 2 x ο« 3x 2 dx dx d3y d2y dy ο« 6 ο« 11 ο« 6 y ο½ e x ο« 1 3 2 dx dx dx 2 d y 41. ο 4 y ο½ coshο¨2 x ο 1ο© ο« 3 x 2 dx y 42. ''' ο 6 y '' ο« 11y ' ο 6 y ο½ 1 ο« x ο« sin x 40. [June 2018] [June 2018] [June 2018] [Dec 2017] 34. π¦ ′′ + 2π¦ ′ + π¦ = 2π₯ + π₯ 2 [Aug 2020] 36. ( D 2 ο 3D ο« 2) y ο½ 2 x 2 ο« sin 2 x [June 2019] 38. (2π·2 + 2π· + 3) π¦ = 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 π3 π¦ π2 π¦ DEPARTMENT OF MATHEMATICS ππ¦ 43. ππ₯ 3 + ππ₯ 3 + 4 ππ₯ + 4π¦ = π₯ 2 − 4π₯ − 6 [Jan 2021] 44. ο¨D2 ο« 3D ο« 2ο©y ο½ 1 ο« 3x ο« x 2 [Dec 2016, 2011, June 2016] 45. (π·3 − 6π·2 + 11π· − 6)π¦ = π 2π₯ +2π₯ + πππ 2π₯ [Dec 2016, June 2013] 46. (π· − 2)2 π¦ = 8(π 2π₯ + π ππ2π₯ + π₯ 2 ) [June 2017, 2016, 2012, Dec 2016, 2011] dy ο½ 1 at π₯ = 0 47. y '' ο« 4 y ' ο« 5 y ο½ ο2 cosh x , find π¦ when y ο½ 0 and [June 2018, 2017] dx 48. yο’ο’ ο« 4 y ' ο« 4 y ο½ 8x 2 given π¦(0) = 1, π¦ ′ (0) = 2 [June 2016] Solve the following linear differential equations by the method of variation of parameters: 1. yο’ο’ ο« a 2 y ο½ sec ο¨a x ο© 3. yο’ο’ ο« y ο½ sec x tan x [Dec2016, 2014, 2010] [Aug 2020,June 2018] 5. yο’ο’ ο« y ο½ cosec x 7. yο’ο’ ο« y ο½ tan x [June 2016] [Jan2021, June 2013] 9. yο’ο’ ο 2yο’ ο« 2y ο½ ex tanx 11. yο’ο’ ο« y ο½ ο¨ 1 1 ο« Sinx ο© [June 2015] [June 2018, 2015] ο¨ ο© 13. D2 ο 3D ο« 2 y ο½ cos eο x 2. yο’ο’ ο« y ο½ sec ο¨x ο© 4. yο’ο’ ο« y ο½ cosec x cot x [June 2019, Dec 2017] [Dec 2016] -x 3 6. yο’ο’ ο« 2yο’ ο« 2y ο½ e sec x [June 2018] 8. yο’ο’ ο« 2 yο’ ο« y ο½ e log x 1 10. yο’ο’ ο 3yο’ ο« 2y ο½ 1 ο« e- x οx 12. [Dec 2014] [June 2010] d2y 2 οyο½ 2 dx 1ο« ex [June 2016] 14. (π·2 − 2π· + 1)π¦ = π π₯ ππππ₯ e 3x x2 16. yο’ο’ ο« 4y ο½ tan(2x) 15. yο’ο’ ο 6yο’ ο« 9y ο½ [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 ο« y ο½ log x [Dec 2017, June 2017, 2010] 1. x 2 dx dx 2 3. x 3 2 d3y dy 2 d y ο« 3 x ο« x ο« y ο½ x ο« log x 3 2 dx dx dx 5. x 2 yο’ο’ ο« xyο’ ο« 9 y ο½ 3x 2 ο« sin(3 log x) 7. x 3 2 d3y dy 2 d y ο« 3 x ο« x ο« 8 y ο½ 65 cos(log x) 3 2 dx dx dx 8. x 2 d2y dy ο 3x ο« 4 y ο½ (1 ο« x) 2 2 dx dx d2y dy ο« x ο« 9 y ο½ sin(3 log x) ο« 3x 2 9. x 2 dx dx 2 d2y dy ο 4 x ο« 6 y ο½ cos(2 log x) 10. x 2 dx dx 2 11. x yο’ο’ ο 5xyο’ ο« 8 y ο½ 2 log x 2 2 12. ο¨3x ο« 2ο© yο’ο’ ο« 5ο¨3x ο« 2ο©yο’ ο 3y ο½ x 2 ο« x ο« 1 13. (2 x ο« 5) 2 [Dec 2017] 2. x D 2 y ο 2 y 1 ο½ xο« 2 x x [June 2018] 4. x 2 yο’ο’ ο« xyο’ ο« y ο½ 2 cos 2 (log x) [June 2011] d2y dy ο« 5x ο« 13 y ο½ log x ο« 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 ο« 5) ο« 8 y ο½ 6 x 2 dx dx [Dec 2017] 11 Calculus and Differential Equations / 21MAT11 DEPARTMENT OF MATHEMATICS 14. (1 ο« x) 2 yο’ο’ ο« (1 ο« x) yο’ ο« y ο½ 4 cos[log(1 ο« x) ] [June 2016] d2y dy ο« 6(2 x ο« 3) ο« 6 y ο½ log( 2 x ο« 3) 2 dx dx 2 16. ο¨3x ο« 2ο© yο’ο’ ο« 3ο¨3x ο« 2ο©yο’ ο 36y ο½ 8x 2 ο« 4x ο« 1 15. (2 x ο« 3) 2 [June 2015] [June 2018, 2016, Dec 2014] 17. (1 ο« x) 2 yο’ο’ ο« (1 ο« x) yο’ ο« y ο½ 2 sin[log(1 ο« x)] [June 2019, 2012, Dec 2017, 2016, 2011] d2y dy ο (2 x ο« 3) ο 12 y ο½ 6 x 2 dx dx 19. (1 ο« x) 2 yο’ο’ ο« (1 ο« x) yο’ ο« y ο½ sin[log(1 ο« x) 2 ] 18. (2 x ο« 3) 2 [June 2018, 2017, Dec 2016] [June 2017] 20. (2π₯ + 1)2 π¦ ′′ − 6(2π₯ + 1)π¦ ′ + 16π¦ = 8(2π₯ + 1)2 [Aug 2020] Self-Study: Applications to oscillations of a spring and L-C-R circuits: 1. A condenser of capacity πΆ discharged through an inductance πΏ and resistance π in series and charge q at π2 π ππ π time satisfying the equation πΏ ππ‘ 2 + π ππ‘ + πΆ = Esinpt. Given πΏ = 0.25π», π = 250 ohms πΆ = 2 × 10−6 and that when π‘ = 0 charge π = 0.02 and current ππ = 0. Obtain π in time π‘ ππ‘ π2 π ππ π 2. In an LCR circuit the charge π on a plate of condenser is given by ππ‘ 2 + π ππ‘ + πΆ = Esinpt . The circuit is 1 tuned to resonance so that π2 = πΏπΆ. If initially the current πΌ and the charge be zero. Show that for all small π πΈπ‘ values of πΏ , the current in the circuit at a time t is given by 2πΏ π ππππ‘. 3. The current π and the charge π in a series containing an inductance πΏ, capacitance πΆ, emf πΈ, satisfy the π2 π π differential equation πΏ ππ‘ 2 + πΆ = E. Express π and π in terms of ‘π‘’ given that πΏ, πΆ, πΈ are constants and the value of π and π are both zero initially. [June 2019] 2 4. The differential equation of a simple pendulum is d x 2 ο« w0 x ο½ F0 sin nt where w0 and F0 are constants. 2 dt dx ο½ 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 ο½ 0, π2 π₯ ππ₯ 10 ππ‘ 2 + ππ‘ + 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 π‘ 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 π2 π₯ ππ‘ 2 ππ + π2 π₯ = λ2 ππ ππππ‘. Show that if ≠ π, the displacement is given by π₯ = π2 −π2 (ππ ππππ‘ − ππ ππππ‘). What happens when π = π? ********************************** 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: 5οΉ ο©0 1 ο 3 ο 1οΉ ο©2 1 3 ο©4 οͺ1 0 1 οΊ οͺ οΊ οͺ2 1οΊ 4 2 1 3οΊ οͺ οͺ 1. [June 2018] 2. [Dec 2017] 3. οͺ οͺ3 1 0 οͺ8 4 7 13 οΊ οͺ2 2οΊ οͺ οΊ οͺ οΊ οͺ ο«1 1 ο 2 0 ο» ο«8 4 ο 3 ο 1ο» ο«2 ο©2 ο 1 3 οͺ0 3 4 4. οͺ οͺ2 3 7 οͺ ο«2 5 11 ο©1 οͺ2 7. οͺ οͺ3 οͺ ο«6 4οΉ 1οΊοΊ [June 2016] 5οΊ οΊ 6ο» 2 3 0οΉ 4 3 2 οΊοΊ 2 1 3οΊ οΊ 8 7 5ο» ο©1 2 3 2 οΉ 10. οͺοͺ2 3 5 1οΊοΊ οͺο«1 3 4 5οΊο» ο©1 οͺ2 12. οͺ οͺ4 οͺ ο«1 3οΉ 8 οΊοΊ 8 12 16οΊ οΊ 2 3 4ο» 2 4 4 6 [Jan 2021] ο©1 2 3 4οΉ 5. οͺ5 6 7 8 οΊ οͺ οΊ οͺο«8 7 0 5οΊο» ο©3 ο4 ο1 οͺ1 7 3 8. οͺ οͺ5 ο2 5 οͺ ο«9 ο3 7 [Dec 2019, June 2014] [Dec 2018, June 2011] 2οΉ 1 οΊοΊ 4οΊ οΊ 7ο» [Dec 2015] [Aug 2021] 0 2 1οΉ 1 3 4οΊοΊ 3 4 7οΊ οΊ 3 1 4ο» [Dec 2017] ο© 1 2 0 ο 1οΉ 6. οͺ 3 4 1 2 οΊ οͺ οΊ οͺο«ο 2 3 2 5 οΊο» [Dec 2013] ο© 2 ο1 ο3 ο1οΉ οͺ1 2 3 ο1οΊ οΊ 9. οͺ οͺ1 0 1 1 οΊ οͺ οΊ ο« 0 1 1 ο1ο» [Aug 2020] 2 ο2 3 οΉ ο©1 οͺ2 5 ο 4 6 οΊοΊ οͺ 11. [June 2019, Dec 2019, 2018, 2012] οͺ ο 1 ο 3 2 ο 2οΊ οͺ οΊ 4 ο1 6 ο» ο«2 ο©2 3 ο 1 ο 1 οΉ οͺ1 ο 1 ο 2 ο 4 οΊ οΊ 13. οͺ οͺ3 1 3 ο 2οΊ οͺ οΊ 0 ο 7ο» ο«6 3 ο©91 ο©ο 2 ο 1 ο 3 ο 1οΉ οͺ92 οͺ1 οΊ οͺ 2 3 ο 1 οΊ [Dec 2107, 2015, June 2015] 15. οͺ93 14. οͺ οͺ1 0 1 1οΊ οͺ οͺ οΊ οͺ94 1 ο 1 ο 1ο» ο«0 οͺο«95 [June 2016, Dec 2016] 92 93 94 95οΉ 93 94 95 96οΊοΊ 94 95 96 97 οΊ οΊ 95 96 97 98οΊ 96 97 98 99οΊο» [Dec 2010] Consistency of System of Linear Equations: 1. For what values of ο¬ and ο following simultaneous equations have i) no solution ii) a unique solution iii) an infinite number of solutions? π₯ + π¦ + π§ = 6, π₯ + 2π¦ + 3π§ = 10, π₯ + 2π¦ + ο¬ π§ = ο [Aug 2021, Jan 2021, June 2019, 2016, 2014, Dec 2012, 2010] 2. Investigate the value of ο¬ and ο so that the equations 2x + 3y + 5z = 9; 7x + 3y - 2z = 8; 2x + y + ο¬z = ο have i) no solution ii) a unique solution iii) an infinite number of solutions? [June 2011] 3. Find the value of ο¬ and ο such that the system x ο« 2 y ο« 3z ο½ 6; x ο« 3 y ο« 5z ο½ 9; 2 x ο« 5 y ο« ο¬z ο½ ο 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 ο« y ο« z ο½ 4; 2 x ο« y ο z ο½ 1; x ο 2 y ο« 2 z ο½ 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₯ − 2π¦ + 6π§ = 8 , π₯ + π¦ − 3π§ = −1 , 15π₯ − 3π¦ + 9π§ = 21 7. Test for consistency and solve: 5π₯ + π¦ + 3π§ = 20 ,2π₯ + 5π¦ + 2π§ = 18 , 3π₯ + 2π¦ + π§ = 14 [Aug 2020] 8. Test for consistency and solve: x ο« 2 y ο« 2 z ο½ 5; 2 x ο« y ο« 3z ο½ 6; 3x ο y ο« 2 z ο½ 4; and x ο« y ο« z ο½ ο1 [June 2015] Gauss-Elimination Method: Solve the following system of equations by Gauss Elimination method: 1. π₯ + 2π¦ + π§ = 3, 2π₯ + 3π¦ + 2π§ = 5, 3π₯ − 5π¦ + 5π§ = 2 [June 2017] 2. π₯ + π¦ + π§ = 9, π₯ − 2π¦ + 3π§ = 8, 2π₯ + π¦ − π§ = 3 [Aug 2021, Dec 2017, 2012] 3. 2π₯ − 3π¦ + 4π§ = 7, 5π₯ − 2π¦ + 2π§ = 7, 6π₯ − 3π¦ + 10π§ = 23 [June 2017] 4. 2π₯ + π¦ + 4π§ = 12, 4π₯ + 11π¦ − π§ = 33, 8π₯ − 3π¦ + 2π§ = 20 [Dec 2016, 2015] 5. 4π₯ + π¦ + π§ = 4, π₯ + 4π¦ − 2π§ = 4, 3π₯ + 2π¦ − 4π§ = 6 [June 2016] 6. 3π₯ − π¦ + 2π§ = 12, π₯ + 2π¦ + 3π§ = 11, 2π₯ − 2π¦ − π§ = 2 [Dec 2014] 7. 2π₯ − π¦ + 3π§ = 1 , −3π₯ + 4π¦ − 5π§ = 0 , π₯ + 3π¦ + 6π§ = 0 [June 2014] 8. 2π₯ + 5π¦ + 7π§ = 52 , 2π₯ + π¦ − π§ = 0 , π₯ + π¦ + π§ = 9 [Jan 2019, Dec 2018] 9. π₯ − 2π¦ + 3π§ = 2 , 3π₯ − π¦ + 4π§ = 4 , 2π₯ + π¦ − 2π§ = 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₯ + π¦ + π§ = 9, π₯ − 2π¦ + 3π§ = 8, 2π₯ + π¦ − π§ = 3 [Jan 2021, June 2018, 2017, 2016, 2015, Dec 2017, 2010] 2. π₯ + 2π¦ + π§ = 3, 2π₯ + 3π¦ + 3π§ = 10, 3π₯ − π¦ + 2π§ = 13 [Dec 2017] 3. 2π₯ + π¦ + π§ = 10, 3π₯ + 2π¦ + 3π§ = 18, π₯ + 4π¦ + 9π§ = 16 [Dec 2016] 4. π₯ + π¦ + π§ = 8, −π₯ − π¦ + 2π§ = −4, 3π₯ + 5π¦ − 7π§ = 14 [Dec 2015] 5. 2π₯ − 3π¦ + π§ = 1, π₯ + 4π¦ + 5π§ = 25, 3π₯ − 4π¦ + π§ = 2 [Dec 2017] 6. π₯ + π¦ + π§ = 1, 4π₯ + 3π¦ − π§ = 6, 3π₯ + 5π¦ + 3π§ = 4 [Dec 2016] 7. 2π₯ + 5π¦ + 7π§ = 52 , 2π₯ + π¦ − π§ = 0 , π₯ + π¦ + π§ = 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₯ + π¦ − 2π§ = 17 , 3π₯ + 20π¦ − π§ = −18 , 2π₯ − 3π¦ + 20π§ = 25 [Aug 2021, June 2019, Dec 2019, 2018, 2017, Dec 2017, 2016, 2015, 2012] 2. 3π₯ + 8π¦ + 29π§ = 71, 83π₯ + 11π¦ − 4π§ = 95; 7π₯ + 52π¦ + 13π§ = 104 [Jan 2021, June 2018] 3. 10π₯ + 2π¦ + π§ = 9 , π₯ + 10π¦ − π§ = −22 , −2π₯ + 3π¦ + 10π§ = 22 [June 2017] 4. π₯ + π¦ + 54π§ = 110 , 27π₯ + 6π¦ − π§ = 85 , 6π₯ + 15π¦ + 2π§ = 72 [Dec 2015, 2014] 9 4 6 5. 5π₯ − π¦ = 9, π₯ − 5π¦ + π§ = −4, π¦ − 5π§ = 6 taking (5 , 5 , 5) as first approximation. 6. 7. 8. 9. [Dec 2011] 10π₯ + π¦ + π§ = 12, π₯ + 10π¦ + π§ = 12, π₯ + π¦ + 10π§ = 12 [June 2016] 12π₯ + π¦ + π§ = 31, 2π₯ + 8π¦ − π§ = 24, 3π₯ + 4π¦ + 10π§ = 58 [Jan 2019, Dec 2018] 5π₯ + 2π¦ + π§ = 12 , π₯ + 4π¦ + 2π§ = 15 , π₯ + 2π¦ + 5π§ = 20 [Aug 2020] 5π₯ + 2π¦ + π§ = 12 , π₯ + 4π¦ + 2π§ = 15 , π₯ + 2π¦ + 5π§ = 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: ο©2 0 1 οΉ 1. π΄ = οͺ0 2 0οΊ . Take (1,0,0) as initial vector. Carry out six iterations. οͺ οΊ οͺο«1 0 2οΊο» [June 2019, Dec 2019, 2018, 2016, Dec 2017, 2016, 2015] ο© 6 ο2 2 οΉ 2. A ο½ οͺο 2 3 ο 1οΊ . Take (1, 1, 1) as initial vector. Carry out five iterations. οͺ οΊ οͺο« 2 ο 1 3 οΊο» [Aug 2021, Aug 2020, Jan 2019, Dec 2018, 2017, 2011, June 2016, 2015, 2012] 3. 4. ο© 2 ο1 0 οΉ A ο½ οͺοͺο 1 2 ο 1οΊοΊ . Take (1, 0, 0) as initial vector. Carry out four iterations. οͺο« 0 ο 1 2 οΊο» [June 2018, 2012, Dec 2016, 2015, 2013] ο©1 6 1οΉ [June 2018, 2017] A ο½ οͺοͺ1 2 0οΊοΊ . Take (1, 1, 0) as initial vector. Carry out six iterations. οͺο«0 0 3οΊο» 5. ο© 1 3 ο 1οΉ A ο½ οͺοͺ 3 2 4 οΊοΊ . Take (0, 0, 1) as initial vector. Carry out five iterations. οͺο«ο 1 4 10 οΊο» [Dec 2014] 6. ο©5 0 1 οΉ A ο½ οͺοͺ0 ο 2 0οΊοΊ . Take (1, 0, 0) as initial vector. Carry out six iterations. οͺο«1 0 5οΊο» [Dec 2017] 7. ο© 4 1 ο 1οΉ A ο½ οͺοͺ 2 3 ο 1οΊοΊ . Take (1, 0, 0) as initial vector. Carry out seven iterations. οͺο«ο 2 1 5 οΊο» [Jan 2021, Dec 2014] 8. ο©25 1 2 οΉ A ο½ οͺοͺ 1 3 0 οΊοΊ . Take (1, 0, 0) as initial vector. Carry out seven iterations. οͺο« 2 0 ο 4οΊο» [June 2013] 9. ο©10 2 1 οΉ A ο½ οͺοͺ 2 10 1 οΊοΊ . Take (1, 1, 0) as initial vector. Carry out six iterations. οͺο« 2 1 10οΊο» [June 2014] Self-study: Gauss-Jacobi iterative Method: 1. 2. 3. 4. 5. 28π₯ + 4π¦ − π§ = 32, 2π₯ + 17π¦ + 4π§ = 35, π₯ + 3π¦ + 10π§ = 24 10π₯ + 2π¦ + π§ = 9 , 2π₯ + 20π¦ − 2π§ = −44 , −2π₯ + 3π¦ + 10π§ = 22 6π₯ − π¦ − π§ = 19 , 3π₯ + 4π¦ + π§ = 26 , π₯ + 2π¦ + 6π§ = 22 3π₯ + π¦ + π§ = 1 , π₯ + 3π¦ − π§ = 11 , π₯ − 2π¦ + 4π§ = 21 23π₯ + 13π¦ + 3π§ = 29 , 5π₯ + 23π¦ + 7π§ = 37 , 11π₯ + π¦ + 23π§ = 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: ο©1 0 2οΉ (i) A ο½ οͺ 0 2 1 οΊ οͺ οΊ οͺο« 2 0 3 οΊο» ο© 2 ο1 1 οΉ (ii) A ο½ οͺ ο1 2 ο1οΊ οͺ οΊ οͺο« 1 ο1 2 οΊο» ο©1 0 2οΉ (iii) A ο½ οͺ 0 2 1 οΊ οͺ οΊ οͺο« 2 0 2 οΊο» ************************************* 16