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