Question

If y = sin(log x), then show that x²y₂ + xy₁ + y = 0.

Answer 1

y = sin(log x)

y₁ = cos(log x) `"d"/"dx"` (log x)

y₁ = cos(log x) . `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)

⇒ x²y₂ + xy₁ = -y

⇒ x²y₂ + xy₁ + y = 0