Question

If $y=a{{e}^{x}}+b{{e}^{-x}}$ , by eliminating a and b, find a differential equation.

Answer 1

To find a differential equation of $y=a{{e}^{x}}+b{{e}^{-x}}$ by eliminating a and b, we have differentiate this equation and try to make the RHS or LHS similar to the given function. If we are not able to eliminate a and b, then differentiate the result again. Perform this process till we can substitute the given function in the differentiated result in such a way that a and b are eliminated.

Complete step by step solution:
We have to find a differential equation of $y=a{{e}^{x}}+b{{e}^{-x}}$ by eliminating a and b. Let us consider $y=a{{e}^{x}}+b{{e}^{-x}}...\left( i \right)$ .
We have to differentiate equation (i) with respect to x.
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( a{{e}^{x}}+b{{e}^{-x}} \right)$
We know that $\dfrac{d}{dx}\left( ax+bx \right)=\dfrac{d}{dx}\left( ax \right)+\dfrac{d}{dx}\left( bx \right)$ . Hence, the above equation can be written as
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( a{{e}^{x}} \right)+\dfrac{d}{dx}\left( b{{e}^{-x}} \right)$
We know that $\dfrac{d}{dx}\left( ax \right)=a\times \dfrac{d}{dx}\left( x \right)$ . Hence, we can write the above equation as
$\Rightarrow \dfrac{dy}{dx}=a\dfrac{d}{dx}\left( {{e}^{x}} \right)+b\dfrac{d}{dx}\left( {{e}^{-x}} \right)$
We know that $\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}}$ .
$\Rightarrow \dfrac{dy}{dx}=a{{e}^{x}}+b\dfrac{d}{dx}\left( {{e}^{-x}} \right)$
We have to use the chain rule in the second term of the RHS.
$\Rightarrow \dfrac{dy}{dx}=a{{e}^{x}}+b{{e}^{-x}}\dfrac{d}{dx}\left( -x \right)$
We know that $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ .
$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=a{{e}^{x}}+b{{e}^{-x}}\cdot -1 \\\ & \Rightarrow \dfrac{dy}{dx}=a{{e}^{x}}-b{{e}^{-x}} \\\ \end{aligned}$
We can see that we cannot eliminate a and b. Thus, we have to differentiate the above equation again.
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( a{{e}^{x}}-b{{e}^{-x}} \right)$
The above equation can be written as
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( a{{e}^{x}} \right)-\dfrac{d}{dx}\left( b{{e}^{-x}} \right)$
On differentiating the RHS, we will get
$\begin{aligned} & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=a{{e}^{x}}-\left( b{{e}^{-x}}\cdot -1 \right) \\\ & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=a{{e}^{x}}+b{{e}^{-x}} \\\ \end{aligned}$
Using (i), we can write the above equation as
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=y$

Hence, the differential equation of $y=a{{e}^{x}}+b{{e}^{-x}}$ is $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=y$ .

Note: Students must know how to differentiate an equation and the results of basic differentiation. In these types of questions, we have to differentiate the given equation until we can substitute the given function in the result of the differentiation.