Question

If we have an expression as $y=A{{e}^{mx}}+B{{e}^{nx}}$ , show that $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-\left( m+n \right)\dfrac{dy}{dx}+mny=0$

Answer 1

First we will take the expression given in the question that is $y=A{{e}^{mx}}+B{{e}^{nx}}$ and then we will find the first order derivative and the second order derivative using $f\left( x \right)={{e}^{x}}\Rightarrow f'\left( x \right)={{e}^{x}}$ and the power rule that is $f\left( x \right)={{x}^{n}}\Rightarrow f'\left( x \right)=n{{x}^{n-1}}$ , then we will put the values of the first derivative and the second order derivative in the Left hand side expression of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-\left( m+n \right)\dfrac{dy}{dx}+mny=0$, and then ultimately prove it equal to $0$ .

Complete step-by-step solution:
Let’s take $y=A{{e}^{mx}}+B{{e}^{nx}}$ , now we will differentiate this expression with respect to the variable $x$
$\dfrac{dy}{dx}=~\dfrac{d\left( A{{e}^{mx}}+B{{e}^{nx}} \right)}{dx}$ ,
Now we know that, $\dfrac{d\left( f\left( x \right)+g\left( x \right) \right)}{dx}=\dfrac{d\left( f\left( x \right) \right)}{dx}+\dfrac{d\left( g\left( x \right) \right)}{dx}$
Therefore we have
$\dfrac{dy}{dx}=~\dfrac{d\left( A{{e}^{mx}}+B{{e}^{nx}} \right)}{dx}=\dfrac{d\left( A{{e}^{mx}} \right)}{dx}+\dfrac{d\left( B{{e}^{nx}} \right)}{dx}$ ,
Now we will take out the constant:
$\dfrac{d\left( A{{e}^{mx}} \right)}{dx}+\dfrac{d\left( B{{e}^{nx}} \right)}{dx}=A\dfrac{d\left( {{e}^{mx}} \right)}{dx}+B\dfrac{d\left( {{e}^{nx}} \right)}{dx}$ ,
We already know that the differentiation of: $f\left( x \right)={{e}^{x}}\Rightarrow f'\left( x \right)={{e}^{x}}$
Therefore, we can write as
$A\dfrac{d\left( {{e}^{mx}} \right)}{dx}+B\dfrac{d\left( {{e}^{nx}} \right)}{dx}=A{{e}^{mx}}\dfrac{d\left( mx \right)}{dx}+B{{e}^{nx}}\dfrac{d\left( nx \right)}{dx}$
It is standard according to the power rule that is $f\left( x \right)={{x}^{n}}\Rightarrow f'\left( x \right)=n{{x}^{n-1}}$
Therefore, applying it, we get
$\begin{aligned} & A{{e}^{mx}}\dfrac{d\left( mx \right)}{dx}+B{{e}^{nx}}\dfrac{d\left( nx \right)}{dx}=A{{e}^{mx}}m+B{{e}^{nx}}n \\\ & \Rightarrow Am{{e}^{mx}}+Bn{{e}^{nx}} \\\ \end{aligned}$
Therefore, $\dfrac{dy}{dx}=Am{{e}^{mx}}+Bn{{e}^{nx}}\text{ }......\text{ Equation 1}\text{.}$
Now, we will again differentiate in order to find the second derivative:
$\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d\left( Am{{e}^{mx}}+Bn{{e}^{nx}} \right)}{dx}$
Now we know that, $\dfrac{d\left( f\left( x \right)+g\left( x \right) \right)}{dx}=\dfrac{d\left( f\left( x \right) \right)}{dx}+\dfrac{d\left( g\left( x \right) \right)}{dx}$
Therefore,
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=~\dfrac{d\left( Am{{e}^{mx}}+Bn{{e}^{nx}} \right)}{dx}=\dfrac{d\left( Am{{e}^{mx}} \right)}{dx}+\dfrac{d\left( Bn{{e}^{nx}} \right)}{dx}$ ,
Now we will take out the constant:
$\dfrac{d\left( Am{{e}^{mx}} \right)}{dx}+\dfrac{d\left( Bn{{e}^{nx}} \right)}{dx}=Am\dfrac{d\left( {{e}^{mx}} \right)}{dx}+Bn\dfrac{d\left( {{e}^{nx}} \right)}{dx}$ ,
We already know that the differentiation of: $f\left( x \right)={{e}^{x}}\Rightarrow f'\left( x \right)={{e}^{x}}$
Therefore,
$Am\dfrac{d\left( {{e}^{mx}} \right)}{dx}+Bn\dfrac{d\left( {{e}^{nx}} \right)}{dx}=Am{{e}^{mx}}\dfrac{d\left( mx \right)}{dx}+Bn{{e}^{nx}}\dfrac{d\left( nx \right)}{dx}$
It is standard according to the power rule that $f\left( x \right)={{x}^{n}}\Rightarrow f'\left( x \right)=n{{x}^{n-1}}$
Therefore:
$\begin{aligned} & Am{{e}^{mx}}\dfrac{d\left( mx \right)}{dx}+Bn{{e}^{nx}}\dfrac{d\left( nx \right)}{dx}=A{{e}^{mx}}{{m}^{2}}+B{{e}^{nx}}{{n}^{2}} \\\ & \Rightarrow A{{m}^{2}}{{e}^{mx}}+B{{n}^{2}}{{e}^{nx}} \\\ \end{aligned}$
Therefore, $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=A{{m}^{2}}{{e}^{mx}}+B{{n}^{2}}{{e}^{nx}}\text{ }..........\text{Equation 2}\text{.}$
Now we are given in the question that: $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-\left( m+n \right)\dfrac{dy}{dx}+mny=0$
We will now solve the left hand side of the expression that is:
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-\left( m+n \right)\dfrac{dy}{dx}+mny$ ,
We will now put the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ from equation 2 and the value of $\dfrac{dy}{dx}$ from equation 1,
$\begin{aligned} & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}-\left( m+n \right)\dfrac{dy}{dx}+mny \\\ & =\left( A{{m}^{2}}{{e}^{mx}}+B{{n}^{2}}{{e}^{nx}} \right)-\left( m+n \right)\left( Am{{e}^{mx}}+Bn{{e}^{nx}} \right)+mn\left( A{{e}^{mx}}+B{{e}^{nx}} \right) \\\ & =\left( A{{m}^{2}}{{e}^{mx}}+B{{n}^{2}}{{e}^{nx}} \right)-m\left( Am{{e}^{mx}}+Bn{{e}^{nx}} \right)-n\left( Am{{e}^{mx}}+Bn{{e}^{nx}} \right)+mnA{{e}^{mx}}+mnB{{e}^{nx}} \\\ & =A{{m}^{2}}{{e}^{mx}}+B{{n}^{2}}{{e}^{nx}}-A{{m}^{2}}{{e}^{mx}}-Bmn{{e}^{nx}}-Amn{{e}^{mx}}-B{{n}^{2}}{{e}^{nx}}+Amn{{e}^{mx}}+Bmn{{e}^{nx}} \\\ & =0=R.H.S. \\\ \end{aligned}$
Hence Proved.

Note: Students can make the mistake while doing the calculation, you must be careful about the signs in the calculation as it can be a bit messy. Also note that in functions like $f\left( x \right)={{e}^{g\left( x \right)}}\Rightarrow f'\left( x \right)={{e}^{g\left( x \right)}}g'\left( x \right)$ . Remember to mention the properties that you use while proving any expression.