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If y = Aemx + Benx, show that y2 – (m + n)y1 + mny = 0. - Mathematics and Statistics

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प्रश्न

If y = Aemx + Benx, show that y2 – (m + n)y1 + mny = 0.

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उत्तर

y = Aemx + Benx 
Differentiating w.r.t. x, we get

`"dy"/"dx" = "A""d"/"dx"(e^(mx)) + "B""d"/"dx"(e^(nx))`

= `"Ae"^(mx)."d"/"dx"(mx) + "Be"^(nx)."d"/"dx"(nx)`

= Aemx . m + Benx . n
= y1 = mAemx + nBenx                    ...(1)
Differentiating again w.r.t. x, we get

y2 = `m"A""d"/"dx"(e^(mx)) + n"B""d"/"dx"(e^(nx))`

 = `m"Ae"^(mx)."d"/"dx"(mx) + n"Be"^(nx)."d"/"dx"(nx)`

= mAemx . m + nBenx . n

∴ y2 = m2Aemx + n2Benx                   ...(2)
∴ y2 – (m + n)y1 + mny = (m2Aemx + n2Benx) – (m + n)(mAemx + nBenx) + mn(Aemx + Benx)                           ...[By (1), (2)]

= m2Aemx + n2Benx – m2Aemx – mnBemx – n2Benx + mnAemx + mnBenx

= 0
∴ y2 – (m + n)y1 + mny = 0

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

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Differentiation
Miscellaneous Exercise 1 (II) | Q 7.5 | पृष्ठ ६४

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