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# If ${{y}^{\dfrac{1}{m}}}+{{y}^{\dfrac{-1}{m}}}=2x$ then prove that $\left( {{x}^{2}}-1 \right){{y}_{3}}+3x{{y}_{2}}+\left( 1-{{m}^{2}} \right){{y}_{1}}=0$.

Last updated date: 11th Aug 2024
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Hint: Here ${{y}_{1}},{{y}_{2}},{{y}_{3}}$ are the first, second and third derivatives. First convert the given expression into simpler $y$ and $x$ terms and then start differentiating.

The given expression is,
${{y}^{\dfrac{1}{m}}}+{{y}^{\dfrac{-1}{m}}}=2x$
This can be re-written as,
$\Rightarrow {{y}^{\dfrac{1}{m}}}+\dfrac{1}{{{y}^{\dfrac{1}{m}}}}=2x$
Now taking the LCM and solving, we get
$\Rightarrow \dfrac{{{\left( {{y}^{\dfrac{1}{m}}} \right)}^{2}}+1}{{{y}^{\dfrac{1}{m}}}}=2x$
On Cross multiplying, we get
$\Rightarrow {{\left( {{y}^{\dfrac{1}{m}}} \right)}^{2}}+1=2x{{y}^{\dfrac{1}{m}}}$
$\Rightarrow {{\left( {{y}^{\dfrac{1}{m}}} \right)}^{2}}-2x{{y}^{\dfrac{1}{m}}}+1=0$
Let ${{y}^{\dfrac{1}{m}}}=z$, then above equation becomes
$\Rightarrow {{z}^{2}}-2xz+1=0$
This is a quadratic equation. The general quadratic equation is $a{{x}^{2}}+bx+c=0$, comparing the above equation with this we get
$a=1,b=-2x,c=1$
The root of this quadratic equation is given by
$z=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting the corresponding values, we get
$z=\dfrac{-(-2x)\pm \sqrt{{{\left( -2x \right)}^{2}}-4\left( 1 \right)\left( 1 \right)}}{2\left( 1 \right)}$
$\Rightarrow z=\dfrac{2x\pm \sqrt{4{{x}^{2}}-4}}{2}$
Taking $4$ common under the root and taking out, we get
$\Rightarrow z=\dfrac{2x\pm 2\sqrt{{{x}^{2}}-1}}{2}$
Taking out $2$ common, we get
$\therefore z=x\pm \sqrt{{{x}^{2}}-1}$
Substituting back the value of $z$, we get
${{y}^{\dfrac{1}{m}}}=z=x\pm \sqrt{{{x}^{2}}-1}$
$\Rightarrow {{y}^{\dfrac{1}{m}}}=x\pm \sqrt{{{x}^{2}}-1}$
Powering both sides by $m$, we get
${{\left( {{y}^{\dfrac{1}{m}}} \right)}^{m}}={{\left( x\pm \sqrt{{{x}^{2}}-1} \right)}^{m}}$
$\therefore y={{\left( x\pm \sqrt{{{x}^{2}}-1} \right)}^{m}}.........(i)$
Now differentiating the above expression with respect to $x$, we get
${{y}_{1}}=\dfrac{d}{dx}{{\left( x\pm \sqrt{{{x}^{2}}-1} \right)}^{m}}$
We know, $\dfrac{d}{dx}({{y}^{n}})=n{{y}^{n-1}}\dfrac{d}{dx}(y)$ , so the above equation becomes,
${{y}_{1}}=\dfrac{dy}{dx}=m{{\left( x\pm \sqrt{{{x}^{2}}-1} \right)}^{m-1}}\dfrac{d}{dx}\left( x\pm \sqrt{{{x}^{2}}-1} \right)$
Applying the sum rule of differentiation, we get
${{y}_{1}}=\dfrac{dy}{dx}=m{{\left( x\pm \sqrt{{{x}^{2}}-1} \right)}^{m-1}}\left[ \dfrac{d}{dx}\left( x \right)\pm \dfrac{d}{dx}\left( \sqrt{{{x}^{2}}-1} \right) \right]$
$\Rightarrow {{y}_{1}}=m{{\left( x\pm \sqrt{{{x}^{2}}-1} \right)}^{m-1}}\times \left( 1\pm \dfrac{1}{2}\times \dfrac{1}{\sqrt{{{x}^{2}}-1}}\times 2x \right)$
Taking the cancelling the lie terms and taking the LCM, we get
$\Rightarrow {{y}_{1}}=m{{\left( x\pm \sqrt{{{x}^{2}}-1} \right)}^{m-1}}\times \left( \dfrac{\sqrt{{{x}^{2}}-1}\pm x}{\sqrt{{{x}^{2}}-1}} \right)$
Cross multiplying, we get
$\Rightarrow {{y}_{1}}\sqrt{{{x}^{2}}-1}=m{{\left( x\pm \sqrt{{{x}^{2}}-1} \right)}^{m-1}}\times \left( \sqrt{{{x}^{2}}-1}\pm x \right)$
$\Rightarrow {{y}_{1}}\sqrt{{{x}^{2}}-1}=m{{\left( x\pm \sqrt{{{x}^{2}}-1} \right)}^{m}}$
Now substituting $y={{\left( x\pm \sqrt{{{x}^{2}}-1} \right)}^{m}}$ from equation (i) in the above equation, we get
$\Rightarrow {{y}_{1}}\sqrt{{{x}^{2}}-1}=my$
Squaring on both sides, we get
$\Rightarrow {{\left( {{y}_{1}}\sqrt{{{x}^{2}}-1} \right)}^{2}}={{\left( my \right)}^{2}}$
$\Rightarrow {{y}_{1}}^{2}\left( {{x}^{2}}-1 \right)={{m}^{2}}{{y}^{2}}$
Now differentiating the above equation with respect to $'x'$ , we get
$\Rightarrow \dfrac{d}{dx}\left( {{y}_{1}}^{2}\left( {{x}^{2}}-1 \right) \right)=\dfrac{d}{dx}\left( {{m}^{2}}{{y}^{2}} \right)$
Now we know, $\dfrac{d}{dx}\left( u.v \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u$, applying this formula, we get
$\Rightarrow {{y}_{1}}^{2}\dfrac{d}{dx}\left( {{x}^{2}}-1 \right)+\left( {{x}^{2}}-1 \right)\dfrac{d}{dx}\left( {{y}_{1}}^{2} \right)={{m}^{2}}\dfrac{d}{dx}\left( {{y}^{2}} \right)$
$\Rightarrow {{y}_{1}}^{2}2x+\left( {{x}^{2}}-1 \right)2{{y}_{1}}\dfrac{d}{dx}\left( {{y}_{1}} \right)={{m}^{2}}2y\dfrac{d}{dx}(y)$
Now we know, ${{y}_{1}}=\dfrac{dy}{dx},{{y}_{2}}=\dfrac{d{{y}_{1}}}{dx}$ , so the above equation becomes,
$\Rightarrow {{y}_{1}}^{2}2x+\left( {{x}^{2}}-1 \right)2{{y}_{1}}{{y}_{2}}={{m}^{2}}2y{{y}_{1}}$
$\Rightarrow 2{{y}_{1}}\left( {{y}_{1}}x+\left( {{x}^{2}}-1 \right){{y}_{2}} \right)={{m}^{2}}2y{{y}_{1}}$
Dividing throughout by $'2{{y}_{1}}'$ , we get
$\Rightarrow {{y}_{1}}x+\left( {{x}^{2}}-1 \right){{y}_{2}}={{m}^{2}}y$
Now again we will differentiate the above equation with respect to $'x'$, we get
\begin{align} & \Rightarrow \dfrac{d}{dx}\left( {{y}_{1}}x+\left( {{x}^{2}}-1 \right){{y}_{2}} \right)=\dfrac{d}{dx}\left( {{m}^{2}}y \right) \\ & \Rightarrow \dfrac{d}{dx}\left( {{y}_{1}}x \right)+\dfrac{d}{dx}\left( \left( {{x}^{2}}-1 \right){{y}_{2}} \right)={{m}^{2}}\dfrac{d}{dx}\left( y \right) \\ \end{align}
Now we know, $\dfrac{d}{dx}\left( u.v \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u$, applying this formula, we get
$\Rightarrow {{y}_{1}}\dfrac{d}{dx}\left( x \right)+x\dfrac{d}{dx}\left( {{y}_{1}} \right)+\left( {{x}^{2}}-1 \right)\dfrac{d}{dx}\left( {{y}_{2}} \right)+{{y}_{2}}\dfrac{d}{dx}\left( {{x}^{2}}-1 \right)={{m}^{2}}\dfrac{d}{dx}\left( y \right)$
Now we know, ${{y}_{1}}=\dfrac{dy}{dx},{{y}_{2}}=\dfrac{d{{y}_{1}}}{dx}$ , so the above equation becomes,
$\Rightarrow {{y}_{1}}(1)+x{{y}_{2}}+({{x}^{2}}-1){{y}_{3}}+{{y}_{2}}2x={{m}^{2}}{{y}_{1}}$
$\Rightarrow {{y}_{1}}+x{{y}_{2}}+({{x}^{2}}-1){{y}_{3}}+{{y}_{2}}2x-{{m}^{2}}{{y}_{1}}=0$
On regrouping, we get
$\Rightarrow ({{x}^{2}}-1){{y}_{3}}+3x{{y}_{2}}+\left( 1-{{m}^{2}} \right){{y}_{1}}=0$
Hence proved.

Note: In the given question we are asked to prove $\left( {{x}^{2}}-1 \right){{y}_{3}}+3x{{y}_{2}}+\left( 1-{{m}^{2}} \right){{y}_{1}}=0$
We should not confuse with the ${{y}_{1}},{{y}_{2}}\,and\,{{y}_{3}}$. ${{y}_{1}},{{y}_{2}}\, and \,{{y}_{3}}$ are 1st, 2nd, and 3rd derivative of the given function.
If we directly apply differentiation to the given expression, it becomes lengthy and complicated.