If y=a x+ x^2+1 ^n+b x- x^2+1 ^ -n, prove that (x^2-1) d^2y dx^2 +x dy dx -n^2y=0.

We have y=a x+ x^2+1 ^n+b x- x^2+1 ^ -n, Differentiating y with respect to x, we get dy dx =an x+ x^2+1 ^ n-1 (1+1 2 x^2+1 2x )-bn x- x^2+1 ^ -n-1 (1-1 2 x^2+1 2x ) =an x+ x^2+1 ^ n-1 (1+ x x^2+1 )-bn x- x^2+1 ^ -n-1 (1- x x^2+1 ) =an x+ x^2+1 ^ n-1 ( x^2+1 +x x^2+1 )-bn x- x^2+1 ^ n-1 ( x^2+1 -x x^2+1 ) =an x+ x^2+1 ^ n-1 ( x+ x^2+1 x^2+1 )+bn x- x^2+1 ^ -n-1 ( x x^2+1 x^2+1 ) = a x+ x^2+1 ^n ( n x^2+1 )+b x- x^2+1 ^ -n ( n x^2+1 ) = ( n x^2+1 )y x^2+1 dy dx =ny Squaring both sides, we get (x^2+1) ( dy dx )^2=n^2y^2...(2) Differentiating (2) with respect to x , we get (x^2+1)2 dy dx d^2y dx^2 +2x ( dy dx )^2=n^2 (2y dy dx ) (x^2+1) d^2y dx^2 +x ( dy dx )=n^2(y) (x^2+1) d^2y dx^2 +x ( dy dx )=n^2y=0 Hence, (x^2+1) d^2y dx^2 +x ( dy dx )=n^2y=0

If y=a x+ x^2+1 ^n+b x- x^2+1 ^ -n, prove that (x^2-1) d^2y dx^2 …

Download App

Question

If $\text{y}=\text{a}\{\text{x}+\sqrt{\text{x}^2+1}\}^\text{n}+\text{b}\{\text{x}-\sqrt{\text{x}^2+1}\}^{-\text{n},}$ prove that $(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}-\text{n}^2\text{y}=0.$

✓

Answer

We have
$\text{y}=\text{a}\{\text{x}+\sqrt{\text{x}^2+1}\}^\text{n}+\text{b}\{\text{x}-\sqrt{\text{x}^2+1}\}^{-\text{n},}$
Differentiating y with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{an}\{\text{x}+\sqrt{\text{x}^2+1}\}^{\text{n}-1}\Big(1+\frac{1}{2\sqrt{\text{x}^2+1}\times2\text{x}}\Big)-\text{bn}\{\text{x}-\sqrt{\text{x}^2+1}\}^{-\text{n}-1}\Big(1-\frac{1}{2\sqrt{\text{x}^2+1}\times2\text{x}}\Big)$
$=\text{an}\{\text{x}+\sqrt{\text{x}^2+1}\}^{\text{n}-1}\Big(1+\frac{\text{x}}{\sqrt{\text{x}^2+1}}\Big)-\text{bn}\{\text{x}-\sqrt{\text{x}^2+1}\}^{-\text{n}-1}\Big(1-\frac{\text{x}}{\sqrt{\text{x}^2+1}}\Big)$
$=\text{an}\{\text{x}+\sqrt{\text{x}^2+1}\}^{\text{n}-1}\Big(\frac{\sqrt{\text{x}^2+1}+\text{x}}{\sqrt{\text{x}^2+1}}\Big)-\text{bn}\{\text{x}-\sqrt{\text{x}^2+1}\}^{\text{n}-1}\Big(\frac{\sqrt{\text{x}^2+1}-\text{x}}{\sqrt{\text{x}^2+1}}\Big)$
$=\text{an}\{\text{x}+\sqrt{\text{x}^2+1}\}^{\text{n}-1}\Big(\frac{\text{x}+\sqrt{\text{x}^2+1}}{\sqrt{\text{x}^2+1}}\Big)+\text{bn}\{\text{x}-\sqrt{\text{x}^2+1}\}^{-\text{n}-1}\Big(\frac{\text{x}\sqrt{\text{x}^2+1}}{\sqrt{\text{x}^2+1}}\Big)$
$=\Big\{\text{a}\{\text{x}+\sqrt{\text{x}^2+1}\}^\text{n}\Big(\frac{\text{n}}{\sqrt{\text{x}^2+1}}\Big)+\text{b}\{\text{x}-\sqrt{\text{x}^2+1}\Big\}^{-\text{n}}\Big(\frac{\text{n}}{\sqrt{\text{x}^2+1}}\Big)$
$=\Big(\frac{\text{n}}{\sqrt{\text{x}^2+1}}\Big)\text{y}$
$\Rightarrow\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}}=\text{ny}$
Squaring both sides, we get
$(\text{x}^2+1)\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\text{n}^2\text{y}^2...(2)$
Differentiating (2) with respect to x , we get
$(\text{x}^2+1)2\frac{\text{dy}}{\text{dx}}\times\frac{\text{d}^2\text{y}}{\text{dx}^2}+2\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\text{n}^2\Big(2\text{y}\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow(\text{x}^2+1)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{n}^2(\text{y})$
$\Rightarrow(\text{x}^2+1)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{n}^2\text{y}=0$
Hence, $(\text{x}^2+1)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{n}^2\text{y}=0$