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Q.If y=aenx+be−nxy = ae^{nx} + be^{-nx}, then prove that y′′=n2yy'' = n^2 y.

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2020Subjective· 2mImportance★★★★★
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Differentiate yy twice with respect to xx and factor out n2n^2 to recover the original expression for yy.

Given y=aenx+be−nxy = ae^{nx} + be^{-nx}.

First derivative: y′=anenx−bne−nxy' = an e^{nx} - bn e^{-nx}

Second derivative: y′′=an2enx+bn2e−nx=n2(aenx+be−nx)y'' = an^2 e^{nx} + bn^2 e^{-nx} = n^2\left(ae^{nx} + be^{-nx}\right)

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