Question

if $y=500e^{7x}+600e^{-7x}$ . Then $\frac{d^{2}y}{d{x}^2}=49y$

• A. True
• B. False

Answer 1

The correct option is A True

$y=500e^{7x}+600e^{-7x}$

Differentiationg w.r.t. $x$

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

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

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

$\Rightarrow$ $\frac{dy}{dx}=500\times 7\times e^{7x}-600\times 7\times e^{-7x}$

Again differentiating w.r.t. $x$,

$\Rightarrow$ $\frac{d^{2}y}{d{x}^2}=500\times 7\times \frac{d\left(e^{7x}\right)}{dx}-600\times 7\times \frac{d\left(e^{-7x}\right)}{dx}$

$\Rightarrow$ $\frac{d^{2}y}{d{x}^2}=500\times 7\times e^{7x}\times 7-600\times 7\times (-7)\times e^{-7x}$

$\Rightarrow$ $\frac{d^{2}y}{d{x}^2}=500\times 7\times 7e^{7x}+600\times 7\times 7\times e^{-7x}$

$\Rightarrow$ $\frac{d^{2}y}{d{x}^2}=7\times 7(500e^{7x}+600e^{-7x})$

$\Rightarrow$ $\frac{d^{2}y}{d{x}^x}=49y$