Question

If $y=500e^{7x}+600e^{-7x}$,show that $\frac{d^2y}{dx^2}=49y$

Answer 1

It is given that,$y=500e^{7x}+600e^{-7x}$
Then,
$\frac{dy}{dx}=500.\frac{d}{dx}(e^{7x})+600.\frac{d}{dx}(e^{-7x})$
$=500.e^{7x}.\frac{d}{dx}(7x)+600.e^{-7x}.\frac{d}{dx}(-7x)$
$=3500e^{7x}-4200e^{-7x}$
$\frac{d^2y}{dx^2}=\frac{d}{dx}(3500e^{7x}-4200e^{-7x})$
$=3500.\frac{d}{dx}(e^{7x})-4200\frac{d}{dx}(e^{-7x})$
$=3500.e^{7x}.\frac{d}{dx}(7x)-4200.e^{-7x}.\frac{d}{dx}(-7x)$
$=7\times 3500\times e^{7x}+7\times 4200\times e^{-7x}$
$=49\times500e^{7x}+49\times600e^{-7x}$
$=49(500e^{7x}+600e^{-7x})$
$=49y$
Hence,proved