logy=sin inversex show (1-x2)d2/dx2=xdy/dx+y
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Answered by
7
Given Log y = Sin⁻¹ x ---(1)
To show (1 - x²) d²y/dx² = x dy/dx + y
Differentiating given equation (1) wrt x.
1/y * y' = 1/√[1 - x²]
√[1 - x² ] * y' = y ---- (2)
Differentiate again wrt x
√[1 - x²] * y'' + y' * (1/2) * 1/√(1-x²) ] (-2x) = y'
√[1-x²] * y" = y' + x y'/√[1-x²]
(1-x²) y'' = √(1-x²) * y' + x y'
= y + x y' using (2).
Proved.
To show (1 - x²) d²y/dx² = x dy/dx + y
Differentiating given equation (1) wrt x.
1/y * y' = 1/√[1 - x²]
√[1 - x² ] * y' = y ---- (2)
Differentiate again wrt x
√[1 - x²] * y'' + y' * (1/2) * 1/√(1-x²) ] (-2x) = y'
√[1-x²] * y" = y' + x y'/√[1-x²]
(1-x²) y'' = √(1-x²) * y' + x y'
= y + x y' using (2).
Proved.
ABHAYSTAR:
Thanks you sir !
Answered by
3
ANSWER:-----
Log y = Sin⁻¹ x ---(1)
To show (1 - x²) d²y/dx² = x dy/dx + y
Differentiating given equation (1) wrt x.
1/y * y' = 1/√[1 - x²]
√[1 - x² ] * y' = y ---- (2)
Differentiate withx
√[1 - x²] * y'' + y' * (1/2) * 1/√(1-x²) ] (-2x) = y'
√[1-x²] * y" = y' + x y'/√[1-x²]
(1-x²) y'' = √(1-x²) * y' + x y'
= y + x y' using (2).
HENCE PROVED:----
hope it helps:----
Log y = Sin⁻¹ x ---(1)
To show (1 - x²) d²y/dx² = x dy/dx + y
Differentiating given equation (1) wrt x.
1/y * y' = 1/√[1 - x²]
√[1 - x² ] * y' = y ---- (2)
Differentiate withx
√[1 - x²] * y'' + y' * (1/2) * 1/√(1-x²) ] (-2x) = y'
√[1-x²] * y" = y' + x y'/√[1-x²]
(1-x²) y'' = √(1-x²) * y' + x y'
= y + x y' using (2).
HENCE PROVED:----
hope it helps:----
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