Question

If $y=a{{e}^{mx}}+b{{e}^{-mx}}$ , then $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-{{m}^{2}}y=$
1. $1$
2. $0$
3. $-1$
4. None of these.

Answer 1

In this problem we need to calculate the value of the given expression which is $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-{{m}^{2}}y$. In the given expression we can observe the term $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ which is nothing but the second order derivative of the function $y$ with respect to $x$ . So we will first calculate the first order derivative of the function $y$ with respect to $x$ by using the differentiation formulas. Now we will consider the first order derivative and differentiate it with respect to $x$. Here also we will use some differentiation formulas to get the value of second order differentiation. After having the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ , substitute the value in the given expression and simplify it to get the required result.

Complete step by step answer:
Given function $y=a{{e}^{mx}}+b{{e}^{-mx}}$ and the expression is $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-{{m}^{2}}y$.
Consider the value $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ which is a second order derivative of the function $y$ with respect to $x$.
So differentiating the function $y=a{{e}^{mx}}+b{{e}^{-mx}}$ with respect to $x$, then we will have
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( a{{e}^{mx}}+b{{e}^{-mx}} \right)$
Apply the differentiation to each term individually, then we will get
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( a{{e}^{mx}} \right)+\dfrac{d}{dx}\left( b{{e}^{-mx}} \right)$
Differentiation for constants is not defined, so write them outside of the differentiation, then we will have
$\dfrac{dy}{dx}=a\dfrac{d}{dx}\left( {{e}^{mx}} \right)+b\dfrac{d}{dx}\left( {{e}^{-mx}} \right)$
Applying the differentiation formula $\dfrac{d}{dx}\left( {{e}^{ax}} \right)=a{{e}^{ax}}$ in the above equation.
$\begin{aligned} & \dfrac{dy}{dx}=a\left( m{{e}^{mx}} \right)+b\left( -m{{e}^{-mx}} \right) \\\ & \Rightarrow \dfrac{dy}{dx}=ma\left( {{e}^{mx}} \right)-mb{{e}^{-mx}} \\\ & \Rightarrow \dfrac{dy}{dx}=m\left( a{{e}^{mx}}-b{{e}^{-mx}} \right) \\\ \end{aligned}$
Now we have the first order derivative of the given function $y=a{{e}^{mx}}+b{{e}^{-mx}}$ as $\dfrac{dy}{dx}=m\left( a{{e}^{mx}}-b{{e}^{-mx}} \right)$ .
Again, differentiate the first order derivative with respect to $x$ to calculate the second order derivative, then we will have
$\begin{aligned} & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left[ m\left( a{{e}^{mx}}-b{{e}^{-mx}} \right) \right] \\\ & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=m\left[ \dfrac{d}{dx}\left( a{{e}^{mx}}-b{{e}^{-mx}} \right) \right] \\\ & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=m\left[ \dfrac{d}{dx}\left( a{{e}^{mx}} \right)-\dfrac{d}{dx}\left( b{{e}^{-mx}} \right) \right] \\\ & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=m\left[ a\dfrac{d}{dx}\left( {{e}^{mx}} \right)-b\dfrac{d}{dx}\left( {{e}^{-mx}} \right) \right] \\\ \end{aligned}$
Applying the differentiation formula $\dfrac{d}{dx}\left( {{e}^{ax}} \right)=a{{e}^{ax}}$ in the above equation, then we will get
$\begin{aligned} & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=m\left[ a\left( m{{e}^{x}} \right)-b\left( -m{{e}^{-mx}} \right) \right] \\\ & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=m\left[ ma{{e}^{mx}}+mb{{e}^{-mx}} \right] \\\ & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}={{m}^{2}}\left( a{{e}^{mx}}+b{{e}^{-mx}} \right) \\\ \end{aligned}$
We have the value $y=a{{e}^{mx}}+b{{e}^{-mx}}$. Substituting this value in the above equation, then we will have
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}={{m}^{2}}y$
Subtracting the value ${{m}^{2}}y$ from both sides of the above equation, then we will get
$\begin{aligned} & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}-{{m}^{2}}y={{m}^{2}}y-{{m}^{2}}y \\\ & \therefore \dfrac{{{d}^{2}}y}{d{{x}^{2}}}-{{m}^{2}}y=0 \\\ \end{aligned}$

So, the correct answer is “Option 2”.

Note: In this problem we have asked to calculate the value of the expression $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-{{m}^{2}}y$ only. In some cases they may ask to calculate the value of a second order derivative only. Then also we can follow the above mentioned procedure to calculate the second order derivative.