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

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प्रश्न

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

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उत्तर

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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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Differentiation - Miscellaneous Exercise 1 (II) [पृष्ठ ६४]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Differentiation
Miscellaneous Exercise 1 (II) | Q 7.4 | पृष्ठ ६४

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