Question 17
If y = cos (m cos ^{ -1 } β‘x)
Show that : (1 - x ^{ 2 } ) d ^{ 2 } y/dx ^{ 2 } β x dy/dx + m ^{ 2 } y = 0
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CBSE Class 12 Sample Paper for 2019 Boards
Last updated at March 23, 2023 by Teachoo
Question 17
If y = cos (m cos ^{ -1 } β‘x)
Show that : (1 - x ^{ 2 } ) d ^{ 2 } y/dx ^{ 2 } β x dy/dx + m ^{ 2 } y = 0
Get live Maths 1-on-1 Classs - Class 6 to 12
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