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Question
If y = A cos (log x) + B sin (log x), show that x2y2 + xy1 + y = 0.
Solution
y = A cos (log x) + B sin (log x) ...(1)
Differentiating both sides w.r.t. x, we get
`"dy"/"dx" = "A""d"/"dx"[cos(logx)] + "B""d"/"dx"[sin(log x)]`
= `"A"[-sin (logx)]."d"/"dx"(logx) + "B"cos(logx)."d"/"dx"(logx)`
= `"A"sin(logx) xx (1)/x "B"cos(logx) xx(1)/x`
∴ `x"d"/"dx"(dy/dx) + "dy"/"dx"."d"/"dx"(x) = -"A""d"/"dx"[sin(logx)] +"B""d"/"dx"[cos(logx)]`
∴ `x(d^2y)/(dx2) + "dy"/"dx" xx 1 = -"A"cos(logx)."d"/"dx"(logx) + "B"[-sin(logx)]."d"/"dx"(logx)`
∴ xy2 + y1 = `-"A"cos(logx) xx(1)/x - "B"sin(logx) xx (1)/x`
∴ x2y2 + xy1 = – [A cos (log x) + B sin (log x)] ...[By (1)]
∴ x2y2 + xy1 + y = 0.
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