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