Question 17 - CBSE Class 12 Sample Paper for 2019 Boards
Last updated at Oct. 1, 2019 by Teachoo

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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

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

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