Question

Which of the following differential equation $y=c_1{e}^x+c_2{e}^{-x}$ as a general solution?

• A. $\frac{d^{2}y}{d{x}^2}+y=0$
• B. $\frac{d^{2}y}{d{x}^2}-y=0$
• C. $\frac{d^{2}y}{d{x}^2}+1=0$
• D. )$\frac{d^{2}y}{d{x}^2}-1=0$

Answer 1

The correct option is B

$\frac{d^{2}y}{d{x}^2}-y=0$

Explanation for the correct option:

Option B.$\frac{d^{2}y}{d{x}^2}-y=0$

Consider $y=c_1{e}^x+c_2{e}^{-x}$.

Differentiate the given function on both sides with respect to $x$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}\left(c_1{e}^x+c_2{e}^{-x}\right)\\ & \Rightarrow & \frac{dy}{dx}=\frac{d}{dx}\left(c_1{e}^x\right)+\frac{d}{dx}\left(c_2{e}^{-x}\right)\\ & \Rightarrow & \frac{dy}{dx}=c_1{e}^x-c_2{e}^{-x}\left[\therefore \frac{d}{dx}\left(c_1{e}^x\right)=c_2\frac{d}{dx}\left(e^x\right)\right]\end{array}$

Again differentiate both sides with respect to $x$ to determine the second derivative:

$\begin{array}{rcl}\frac{d^{2}y}{d{x}^2}& =& \frac{d}{dx}\left(c_1{e}^x-c_2{e}^{-x}\right)\\ & \Rightarrow & \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(c_1{e}^x\right)-\frac{d}{dx}\left(c_2{e}^{-x}\right)\\ & \Rightarrow & \frac{d^{2}y}{d{x}^2}=c_1{e}^x+c_2{e}^{-x}\left[\therefore \frac{d}{\mathrm{dx}}\left(e^{-x}\right)=-e^{-x}\right]\\ & \Rightarrow & \frac{d^{2}y}{d{x}^2}=y\\ & \Rightarrow & \frac{d^{2}y}{d{x}^2}-y=0\end{array}$

Hence, option (B) is correct.

Explanation for incorrect option:

Since the required second derivative of the solution of the given function is not equivalent value mention Option (A), Option (C), and Option (D)

Hence Option (A), Option (C) and Option (D) are incorrect options

Hence Option(B) is the correct answer.