y=(x^2-1)^n prove that (x^2-1)yn+2 + 2nxyn+1 - n(n-1)yn =0
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43
Answer:
(x²-1)yₙ₊₂ + 2xyₙ₊₁ - n(n-1)yₙ = 0
Step-by-step explanation:
Question is
y = (x² - 1)ⁿ
Prove that
(x² - 1)yₙ₊₂ + 2xyₙ₊₁ - n(n-1)yₙ = 0
y = (x² - 1)ⁿ
taking log both sides
=> logy = n log(x² - 1)
Differentiating
=> (1/y) y₁ = n/(x²-1) * 2x
=> (x²-1)y₁ = 2nxy
Differentiating Again
=> (x²-1)y₂ + 2xy₁ = 2ny + 2nxy₁
=> (x²-1)y₂ + 2x(1 - n)y₁ = 2ny
Differentiating n times
=> (x²-1)yₙ₊₂ + ⁿC₁yₙ₊₁2x + 2.ⁿC₂yₙ + 2(1-n)(xyₙ₊₁ + ⁿC₁yₙ) = 2nyₙ
=> (x²-1)yₙ₊₂ + 2nxyₙ₊₁ + n(n-1)yₙ + 2xyₙ₊₁ + 2nyₙ -2nxyₙ₊₁ +2(1-n)nyₙ = 2nyₙ
=> (x²-1)yₙ₊₂ + 2nxyₙ₊₁ -2nxyₙ₊₁ + 2xyₙ₊₁ + n(n-1)yₙ -2(n-1)nyₙ = 2nyₙ - 2nyₙ
=> (x²-1)yₙ₊₂ + 2xyₙ₊₁ - n(n-1)yₙ = 0
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