Deriviative of function # 8

**skeske1234** · August 14th 2009, 12:31 PM

If y=5/2(e^(x/5)+e^(-x/5)), then prove that y''=y/25

Ok can someone just check with me.. I'm pretty close to the answer but i don't know what I did wrong ..

the final answer of y''=y/25.. I have

[e^(x/5)+e^(-x/5)]/[e^(x/5)/25+e^(-x/5)/25]

Don't know what happened to the 5/2 in y and.. the denominator is almost there but not quite. ^.^
Please explain your method to proving this. Thank you so much!

**Matt Westwood** · August 14th 2009, 01:01 PM

Presumably this is what you got (please learn latex):
$y = \frac 5 2 (e^{x/5} + e^{-x/5})$

Diffing $e^{ax}$ gives $ae^{ax}$ and doing it twice gives $a^2e^{ax}$ .

Diffing $e^{-ax}$ gives $-ae^{-ax}$ and doing it twice gives $(-a)^2e^{-ax} = a^2e^{ax}$ .

Adding them together gives $\frac {d^2} {dx^2} (e^{x/5} + e^{-x/5}) = \frac 1 {25} (e^{x/5} + e^{-x/5})$ .

Multiplying the lot by $\frac 5 2$ gives:
$\frac {d^2y} {dx^2} = \frac 5 2 \frac 1 {25} (e^{x/5} + e^{-x/5}) = \frac 1 {25} y$

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