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

We have, y=x^n a ( )+b ( ) ,...(1) Differentiating y with respect to x, we get dy dx =nx^ n-1 a ( )+b ( ) +x^n -a ( ) 1 x +b ( ) 1 x = n x x^n a ( )+b ( ) +x^ n-1 -a ( )+b ( ) n x y+x^ n-1 -a ( )+b ( ) [from(1)] ^ n-1 -a ( )+b ( ) = dy dx - n x y...(2) Differentiating dy dx with respect to x, we get d^2y dx^2 = n x dy dx = ny x^2 +(n-1)x^ n-2 -a ( )+b ( ) +x^ n-1 -a ( ) 1 x -b ( ) 1 x = n x dx dy - ny x^2 +(n-1) x^ n-1 x -a ( )+b ( ) - x^n x^2 a ( )+b ( ) n x dy dx - ny x^2 + ( n-1 x ) ( dy dx - n x y )- y x^2 = dy dx ( n+n-1 x )- (n+n^2+n+1)y x^2 = ( 2n-1 x ) dy dx = (n^2+1)y x^2 ^2 d^2y dx^2 -x(2n-1) dy dx +(n^2+1)y=0 Hence, x^2 d^2y dx^2 +(1-2n)x dy dx +(1+n^2)y=0

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

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Question

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

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Answer

We have,
$\text{y}=\text{x}^\text{n}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text{x})\},...(1)$
Differentiating y with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{nx}^{\text{n}-1}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text{x})\}+\text{x}^\text{n}\\\Big\{-\text{a}\sin(\log\text{x})\times\frac{1}{\text{x}}+\text{b}\cos(\log\text{x})\times\frac{1}{\text{x}}\Big\}$
$=\frac{\text{n}}{\text{x}}\text{x}^\text{n}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text {x})\}+\text{x}^{\text{n}-1}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}$
$\Rightarrow\frac{\text{n}}{\text{x}}\text{y}+\text{x}^{\text{n}-1}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}\ [\text{from}(1)]$
$\Rightarrow\text{x}^{\text{n}-1}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}=\frac{\text{dy}}{\text{dx}}-\frac{\text{n}}{\text{x}}\text{y}...(2)$
Differentiating $\frac{\text{dy}}{\text{dx}}$ with respect to x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{n}}{\text{x}}\frac{\text{dy}}{\text{dx}}=\frac{\text{ny}}{\text{x}^2}+(\text{n}-1)\text{x}^{\text{n}-2}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}\\+\text{x}^{\text{n}-1}\Big\{-\text{a}\cos(\log\text{x})\times\frac{1}{\text{x}}-\text{b}\sin(\log\text{x})\times\frac{1}{\text{x}}\Big\}$
$=\frac{\text{n}}{\text{x}}\frac{\text{dx}}{\text{dy}}-\frac{\text{ny}}{\text{x}^2}+(\text{n}-1)\frac{\text{x}^{\text{n}-1}}{\text{x}}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}\\-\frac{\text{x}^\text{n}}{\text{x}^2}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text{x})\}$
$\frac{\text{n}}{\text{x}}\frac{\text{dy}}{\text{dx}}-\frac{\text{ny}}{\text{x}^2}+\Big(\frac{\text{n}-1}{\text{x}}\Big)\Big(\frac{\text{dy}}{\text{dx}}-\frac{\text{n}}{\text{x}}\text{y}\Big)-\frac{\text{y}}{\text{x}^2}$
$=\frac{\text{dy}}{\text{dx}}\Big(\frac{\text{n}+\text{n}-1}{\text{x}}\Big)-\frac{(\text{n}+\text{n}^2+\text{n}+1)\text{y}}{\text{x}^2}$
$=\Big(\frac{2\text{n}-1}{\text{x}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{(\text{n}^2+1)\text{y}}{\text{x}^2}$
$\Rightarrow\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}(2\text{n}-1)\frac{\text{dy}}{\text{dx}}+(\text{n}^2+1)\text{y}=0$
Hence, $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}+(1-2\text{n})\text{x}\frac{\text{dy}}{\text{dx}}+(1+\text{n}^2)\text{y}=0$

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