Question

If$y={{e}^{m{{\sin }^{-1}}x}}$ , then show that $\left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-x\dfrac{dy}{dx}-{{m}^{2}}y=0$

Answer 1

Hint: To solve the above question, we have to find the derivative of the equation. At first we have to find the ${{1}^{st}}$ order derivative to the $y$ with respect to $x$ and after that we have to find the ${{2}^{nd}}$ order derivative to the $y$ with respect to$x$.


Complete step-by-step answer:
To proceed we will go through some standard Calculus results
\[\begin{array}{*{35}{l}}
   \dfrac{d}{dx}{{e}^{ax}}=a{{e}^{ax}} \\
   \dfrac{d}{dx}{{\sin }^{-1}}x=\dfrac{1}{\sqrt{1-{{x}^{2}}}} \\
\end{array}\]

It is given that
$y={{e}^{m{{\sin }^{-1}}x}}$
Now we will apply the chain rule then we get:
\[\begin{matrix}
   {y}'=m{{e}^{m{{\sin }^{-1}}x}}\cdot \dfrac{1}{\sqrt{1-{{x}^{2}}}} \\
   =m\dfrac{{{e}^{m{{\sin }^{-1}}x}}}{\sqrt{1-{{x}^{2}}}} \\
\end{matrix}\]
Where first we differentiate the \[{{e}^{m{{\sin }^{-1}}x}}\] and after that applying the chain rule we have to differentiate the term \[{{\sin }^{-1}}x\] whose differentiation is \[\dfrac{1}{\sqrt{1-{{x}^{2}}}}\] .
And now we are differentiating again and applying the quotient rule, along with the chain rule, we get:
Now here we have to keep in mind the $\dfrac{u}{v}$ form of differentiation and the differentiation is
$d\left( \dfrac{u}{v} \right)=\dfrac{vdu-udv}{{{v}^{2}}}$ .
So, now we will get the above formula and we also keep in mind that $\dfrac{d}{dx}\left( \sqrt{1-{{x}^{2}}} \right)=\dfrac{1}{2}\dfrac{1}{\sqrt{1-{{x}^{2}}}}\times (-2x)$
\[\begin{array}{*{35}{l}}
   {y}''=\dfrac{\left( \sqrt{1-{{x}^{2}}} \right)\left( \dfrac{d}{dx}m{{e}^{m{{\sin }^{-1}}x}} \right)-\left( m{{e}^{m{{\sin }^{-1}}x}} \right)\left( \dfrac{d}{dx}\sqrt{1-{{x}^{2}}} \right)}{{{\left( \sqrt{1-{{x}^{2}}} \right)}^{2}}} \\
   {} \\
\end{array}\]
If we simply it then we will get the equation as,
\[=\dfrac{\left( \sqrt{1-{{x}^{2}}} \right)\left( {{m}^{2}}\dfrac{{{e}^{\overset{2}{\mathop{\min }}\,{{x}^{-1}}x}}}{\sqrt{1-{{x}^{2}}}} \right)-\left( m{{e}^{m{{\sin }^{-1}}x}} \right)\left( \dfrac{1}{2}{{\left( 1-{{x}^{2}} \right)}^{-\dfrac{1}{2}}}\cdot (-2x) \right)}{{{\left( \sqrt{1-{{x}^{2}}} \right)}^{2}}}\]

Here after we simplify the equation and we will get,
\[\begin{array}{*{35}{l}}
   =\dfrac{{{m}^{2}}{{e}^{m{{\sin }^{-1}}x}}+\dfrac{mx{{e}^{m{{\sin }^{-1}}x}}}{\sqrt{1-{{x}^{2}}}}}{1-{{x}^{2}}} \\
   =\dfrac{{{m}^{2}}y+x{y}'}{1-{{x}^{2}}} \\
   \therefore \left( 1-{{x}^{2}} \right){y}''={{m}^{2}}y+x{y}' \\
   \therefore \left( 1-{{x}^{2}} \right){y}''-x{y}'={{m}^{2}}y \\
\end{array}\]
In this equation $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}={y}''$ and $\dfrac{dy}{dx}={y}'$ .
Now if we rewrite the equation then we will get
$\left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-x\dfrac{dy}{dx}={{m}^{2}}y$

If we take ${{m}^{2}}y$ in left side then we will get,
$\left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-x\dfrac{dy}{dx}-{{m}^{2}}y=0$

Hence, the equation is proved.

Note: Here student must take care of the concept of derivative of the equation. Students have to be well aware of differentiation formulas. Another we have to taking logarithm both side of equation,
$\begin{align}
  & y={{e}^{m{{\sin }^{-1}}x}}\Rightarrow \log y=m{{\sin }^{-1}}x \\
 & \\
\end{align}$
In this way we have to do first order differentiation and second order differentiation to get the result.