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If y = eax.sin(bx), show that y2 – 2ay1 + (a2 + b2)y = 0. - Mathematics and Statistics

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प्रश्न

If y = eax.sin(bx), show that y2 – 2ay1 + (a2 + b2)y = 0.

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उत्तर

y = eax.sin(bx)                                ...(1)

∴ `"dy"/"dx" = "d"/"dx"[e^(ax).sin(bx)]`

= `e^(ax)."d"/"dx"[sin(bx)] + sin(bx)."d"/"dx"(e^(ax))`

= `e^(ax).cos(bx)."d"/"dx"(bx) + sin(bx) xx e^(ax)."d"/"dx"(ax)`

= eax . cos (bx) x b + eax . sin (bx) x a
∴ y1 = eax[b cos (bx) + a sin (bx)]   ...(2)
Differentiating again w.r.t. x, we get

`"dy"_1/"dx" = "d"/'dx"[e^(ax){b  cos (bx) + a sin (bx)}]`

= `e^(ax)."d"/"dx"[b cos (bx) + a sin (bx)] + [b cos (bx) + a sin (bx)]."d"/"dx"(e^ax)`

= `e^(ax).[b{ - sin(bx)}."dy"/"dx"(bx) + a cos (bx)."dy"/"dx" (bx)] + [b cos(bx) + a sin (bx)] xx e^(ax)."d"/"dx"(ax)`

= eax [– b sin (bx) x b + a cos (bx) x b} + [b cos (bx) + a sin (bx)eax x a

= eax [–b2sin (bx) + ab cos (bx) + a2sin (bx)]
∴ y2 = eax [–b2sin (bx) + 2ab cos (bx) + a2sin (bx)]  ...(3)
∴ y2 – 2ay1 + (a2 + b2)y

= eax [– b2sin (bx) + 2ab cos (bx) + a2sin (bx)] – 2a.eax[b cos (bx) + a sin (bx)] + (a2 + b2)eax sin (bx)  ...[By (1), (2) and (3)]

= eax [– b2sin bx + 2ab cos (bx) + a2sin (bx) – 2ab cos (bx) – 2a2sin (bx) + a2sin (bx) + b2sin (bx)]

= eax x 0
∴ y2 – 2ay1 + (a2 + b2)y = 0.

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अध्याय 1: Differentiation - Exercise 1.5 [पृष्ठ ६०]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Differentiation
Exercise 1.5 | Q 3.05 | पृष्ठ ६०

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