Question 17
If y = cos (m cos ^{ -1 } x)
Show that : (1 - x ^{ 2 } ) d ^{ 2 } y/dx ^{ 2 } – x dy/dx + m ^{ 2 } y = 0
CBSE Class 12 Sample Paper for 2019 Boards
Question 1 Important
Question 2
Question 3
Question 4 (Or 1st) Important
Question 4 (Or 2nd)
Question 5
Question 6
Question 7 Important
Question 8 (Or 1st) Important
Question 8 (Or 2nd)
Question 9
Question 10 (Or 1st) Important
Question 10 (Or 2nd)
Question 11 Important
Question 12 (Or 1st)
Question 12 (Or 2nd)
Question 13 (Or 1st) Important
Question 13 (Or 2nd)
Question 14 Important
Question 15
Question 16 (Or 1st)
Question 16 (Or 2nd) Important
Question 17 You are here
Question 18
Question 19 Important
Question 20 Important
Question 21 (Or 1st)
Question 21 (Or 2nd) Important
Question 22
Question 23 Important
Question 24 (Or 1st)
Question 24 (Or 2nd) Important
Question 25
Question 26 (Or 1st) Important
Question 26 (Or 2nd)
Question 27 (Or 1st) Important
Question 27 (Or 2nd) Important
Question 28
Question 29 Important
CBSE Class 12 Sample Paper for 2019 Boards
Last updated at April 16, 2024 by Teachoo
Question 17
If y = cos (m cos ^{ -1 } x)
Show that : (1 - x ^{ 2 } ) d ^{ 2 } y/dx ^{ 2 } – x dy/dx + m ^{ 2 } y = 0
Question 17 If y = cosβ‘γ(π cos^(β1)β‘π₯)γ Show that : (1βπ₯^2) (π^2 π¦)/(ππ₯^2 ) β π₯ ππ¦/ππ₯ + π^2 π¦ = 0 y = cosβ‘γ(π cos^(β1)β‘π₯)γ Differentiating w.r.t. x ππ¦/ππ₯ = π(cosβ‘(π cos^(β1)β‘π₯ ) )/ππ₯ ππ¦/ππ₯ = sinβ‘(π cos^(β1)β‘π₯ ) π(π cos^(β1)β‘π₯ )/ππ₯ ππ¦/ππ₯ = sinβ‘(π cos^(β1)β‘π₯ )Γπ π(cos^(β1)β‘π₯ )/ππ₯ ππ¦/ππ₯ = sinβ‘(π cos^(β1)β‘π₯ )ΓπΓ(β1)/β(1 β π₯^2 ) ππ¦/ππ₯ = βm sinβ‘(π cos^(β1)β‘π₯ )Γ1/β(1 β π₯^2 ) π¦β² = βm sinβ‘(π cos^(β1)β‘π₯ )Γ1/β(1 β π₯^2 ) Now, finding π¦β²β² will be complicated So, we multiply β(π β π^π ) to the left side β(1βπ₯^2 ) π¦β² = βm sinβ‘(π cos^(β1)β‘π₯ ) Squaring both sides (β(1βπ₯^2 ) )^2 (π¦^β² )^2 = (βm)^2 sin^2β‘(π cos^(β1)β‘π₯ ) ππ¦/ππ₯ = sinβ‘(π cos^(β1)β‘π₯ )ΓπΓ(β1)/β(1 β π₯^2 ) ππ¦/ππ₯ = βm sinβ‘(π cos^(β1)β‘π₯ )Γ1/β(1 β π₯^2 ) π¦β² = βm sinβ‘(π cos^(β1)β‘π₯ )Γ1/β(1 β π₯^2 ) Now, finding π¦β²β² will be complicated So, we multiply β(π β π^π ) to the left side β(1βπ₯^2 ) π¦β² = βm sinβ‘(π cos^(β1)β‘π₯ ) Squaring both sides (β(1βπ₯^2 ) )^2 (π¦^β² )^2 = (βm)^2 sin^2β‘(π cos^(β1)β‘π₯ ) ππ¦/ππ₯ = sinβ‘(π cos^(β1)β‘π₯ )ΓπΓ(β1)/β(1 β π₯^2 ) ππ¦/ππ₯ = βm sinβ‘(π cos^(β1)β‘π₯ )Γ1/β(1 β π₯^2 ) π¦β² = βm sinβ‘(π cos^(β1)β‘π₯ )Γ1/β(1 β π₯^2 ) Now, finding π¦β²β² will be complicated So, we multiply β(π β π^π ) to the left side β(1βπ₯^2 ) π¦β² = βm sinβ‘(π cos^(β1)β‘π₯ ) Squaring both sides (β(1βπ₯^2 ) )^2 (π¦^β² )^2 = (βm)^2 sin^2β‘(π cos^(β1)β‘π₯ ) (1βx^2 ) (π¦^β² )^2 = π^2 sin^2β‘(π cos^(β1)β‘π₯ ) We know that sin2 x = 1 β cos2 x (1βx^2 ) (π¦^β² )^2 = π^2 (1βcos^2β‘(π cos^(β1)β‘π₯ ) ) Putting y = cosβ‘γ(π cos^(β1)β‘π₯)γ (1βx^2 ) (π¦^β² )^2 = π^2 (1βπ¦^2 ) Differentiating again w.r.t. x π(1 β x^2 )/ππ₯ (π¦^β² )^2 + (1βx^2 ) (π(π¦^β² )^2)/ππ₯ = π^2 π(1 β π¦^2 )/ππ₯ β2x (π¦^β² )^2 + (1βx^2 )2π¦^β² π¦β²β² = π^2Γβ2π¦π¦β² 2π¦^β² (βπ₯π¦^β² "+ " (1βx^2 )π¦β²β²) = β2y^β²Γπ^2 π¦ βπ₯π¦^β² "+ " (1βx^2 )π¦β²β² = βπ^2 π¦ (1βx^2 )π¦β²β² β π₯π¦^β² + π^2 π¦ = 0 (1βπ₯^2) (π^2 π¦)/(ππ₯^2 ) β π₯ ππ¦/ππ₯ + π^2 π¦ = 0 Hence proved