If y = sin(log x), then show that x2 y2 + xy1 + y = 0.

1.	If y = sin(log x), then show that x2 y2 + xy1 + y = 0.
Answer» y = sin(log x) y₁ = cos(log x) \(\frac{d}{dx}\)(log x) y₁ = cos(log x).\(\frac{1}{x}\) ∴ xy₁ = cos(log x) Differentiating both sides with respect to x, we get xy₂ + y₁(1) = -sin(log x).\(\frac{1}{x}\) ⇒ x[xy₂ + y₁] = -sin(log x) ⇒ x² y₂ + xy₁ = -y ⇒ x² y₂ + xy₁ + y = 0

Answer»

y = sin(log x)

y₁ = cos(log x) \(\frac{d}{dx}\)(log x)

y₁ = cos(log x).\(\frac{1}{x}\)

∴ xy₁ = cos(log x)

Differentiating both sides with respect to x, we get

xy₂ + y₁(1) = -sin(log x).\(\frac{1}{x}\)

⇒ x[xy₂ + y₁] = -sin(log x)

⇒ x² y₂ + xy₁ = -y

⇒ x² y₂ + xy₁ + y = 0

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