Question

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

Answer 1

y = 500e^7x + 600e^-7x

On differentiating with respect to x,

${\displaystyle \frac{dy}{dx}=\frac{d}{dx}(500e^{7x}+600e^{-7x})}$

${\displaystyle =500\frac{d}{dx} e^{7x}+600 \frac{d}{dx} e^{-7x}}$

${\displaystyle =500 e^{7x}\frac{d}{dx}\left(7x\right)+600 e^{-7x}\frac{d}{dx}(-7x)}$

= 500 e^7x . 7 + 600 e^-7x. (-7)

= 3500 e^7x - 4200 e^-7x

Differentiating again with respect to x,

${\displaystyle \frac{d^{2}y}{{dx}^2}=3500 \frac{d}{dx} e^{7x}-4200 \frac{d}{dx} e^{-7x}}$

${\displaystyle =3500\times e^{7x}\cdot 7-4200 e^{-7x}(-7)}$

= 500 × 49 e^7x + 600 × 49 e^-7x

= 49(500 e^7x + 600 e^-7x)

= 49 y

∴ ${\displaystyle \frac{d^{2}y}{{dx}^2}=49y.}$