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