**
Question 17
**

If y = cos (m cos
^{
-1
}
β‘x)

Show that : (1 - x
^{
2
}
) d
^{
2
}
y/dx
^{
2
}
– x dy/dx + m
^{
2
}
y = 0

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Last updated at Oct. 1, 2019 by Teachoo

**
Question 17
**

If y = cos (m cos
^{
-1
}
β‘x)

Show that : (1 - x
^{
2
}
) d
^{
2
}
y/dx
^{
2
}
– x dy/dx + m
^{
2
}
y = 0

Subscribe to our Youtube Channel - https://you.tube/teachoo

Transcript

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

CBSE Class 12 Sample Paper for 2019 Boards

Paper Summary

Question 1

Question 2

Question 3

Question 4 (Or 1st)

Question 4 (Or 2nd)

Question 5

Question 6

Question 7

Question 8 (Or 1st)

Question 8 (Or 2nd)

Question 9

Question 10 (Or 1st)

Question 10 (Or 2nd)

Question 11

Question 12 (Or 1st)

Question 12 (Or 2nd)

Question 13 (Or 1st)

Question 13 (Or 2nd)

Question 14

Question 15

Question 16 (Or 1st)

Question 16 (Or 2nd)

Question 17 You are here

Question 18

Question 19

Question 20

Question 21 (Or 1st)

Question 21 (Or 2nd)

Question 22

Question 23

Question 24 (Or 1st)

Question 24 (Or 2nd)

Question 25

Question 26 (Or 1st)

Question 26 (Or 2nd)

Question 27 (Or 1st)

Question 27 (Or 2nd)

Question 28

Question 29

Class 12

Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.