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If y=(5/2)(e^(x/5)+e^(-x/5)), then prove that y"=y/25?
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differentiate'em twice ... and see what you get
\[y=5/2 (e^{x/5}+e^{-x/5})\]
\[y'=1/2e ^{-x/5}(e ^{x/5}-1)\]
you don't have to take out that e's .... leave it as it is.
\[y"=1/10e ^{-x/5}(e^{x/5}+1)\]
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