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Question Number 49357 by peter frank last updated on 06/Dec/18

Commented by maxmathsup by imad last updated on 01/Jan/19

$2) l e t p r o v e t h a t y^{(n)} (x) = \frac{{(- 1)}^{n - 1} (n - 1)!}{{(1 + x)}^{n}} w i t h y (x) = l n (1 + x)$ $b y r e c u r r e n c e o n n f o r n = 1 y^{(1)} (x) = \frac{1}{1 + x} = \frac{{(- 1)}^{1 - 1} 0!}{{(1 + x)}^{1}} (t r u e) l e t s u p p o s e$ $y^{(n)} (x) = \frac{{(- 1)}^{n - 1} (n - 1)!}{{(1 + x)}^{n}} w e h a v e$ $y^{(n + 1)} (x) = \frac{d}{d x} (y^{(n)} (x)) = {(- 1)}^{n - 1} (n - 1)! \times \frac{- n {(1 + x)}^{n - 1}}{{(1 + x)}^{2 n}}$ $= {(- 1)}^{n} n! \frac{1}{{(1 + x)}^{n + 1}} = \frac{{(- 1)}^{n + 1 - 1} (n + 1 - 1)!}{{(1 + x)}^{n + 1}} s o t h e r e s u l t i s t r u e a t t e r m$ $(n + 1) .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18

	$1) \frac{d^{2} x}{{d t}^{2}} + k \frac{d x}{d t} + n^{2} x = p c o s w t$ $s t e p 1$ $\frac{d^{2} x}{{d t}^{2}} + k \frac{d x}{d t} + n^{2} x = 0$ $l e t x = e^{α t} s o \frac{d x}{d t} = α e^{α t} a n d \frac{d^{2} x}{{d t}^{2}} = α^{2} e^{α t}$ $e^{α t} (α^{2} + k α + n^{2}) = 0 e^{α t} \neq 0$ $α^{2} + k α + n^{2} = 0$ $α = \frac{- k \pm \sqrt{k^{2} - 4 n^{2}}}{2}$ $C o m p l e m e n t a r y f u n c t i o n$ $x = {A e}^{(\frac{- k + \sqrt{k^{2} - 4 n^{2}}}{2}) t} + {B e}^{{(\frac{- k - \sqrt{k^{2} - 4 n^{2}}}{2})}^{} t}$ $s t e p 2$ $d e t e r m i n a t i o n o f p a r t i c u l a r i n t r e g a l$ $\frac{d^{2} x}{{d t}^{2}} + k \frac{d x}{d t} + n^{2} x = p c o s w t$ $l e t \frac{d}{d t} = D a n d \frac{d^{2}}{{d x}^{2}} = D^{2}$ $s o (D^{2} + k D + n^{2}) x = p c o s w t$ $x = \frac{1}{D^{2} + n^{2} + k D} \times p c o s w t$ $= \frac{(D^{2} + n^{2} - k D)}{{(D^{2} + n^{2})}^{2} - k^{2} D^{2}} p c o s w t$ $D (p c o s w t) = - w p s i n w t$ $D^{2} (p c o s w t) = - w^{2} p c o s w t$ $= \frac{(- w^{2} p c o s w t + n^{2} p c o s w t + k p w s i n w t)}{{(- w^{2} + n^{2})}^{2} - k^{2} (- w^{2})}$ $= p \times [\frac{(n^{2} - w^{2}) c o s w t + k w s i n w t}{{(n^{2} - w^{2})}^{2} + k^{2} w^{2}}]$
	Answered by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18
	$2)y=ln(1+x) (dy/dx)=(1/((1+x))) (d^2 y/dx^2 )=((−1)/((1+x)^2 )) so when n=2 LHS=((−1)/((1+x)^2 )) and RHS=(((−1)^(2−1) (2−1)!)/((1+x)^2 ))=((−1)/((1+x)^2 )) so for n=2 the statement is true now let assume for n=k the statement is true we have to prove that it is true for n=k+1 (d^k y/dx^k )=(((−1)^(k−1) (k−1)!)/((1+x)^k )) so (d/dx)((d^k y/dx^k ))=(d^(k+1) y/dx^(k+1) )←LHS (d/dx)[(((−1)^(k−1) (k−1)!)/((1+x)^k ))] =(((−1)^(k−1) (k−1)!)/1)×(d/dx){(1/((1+x)^k ))} =(−1)^(k−1) (k−1)!×(−k)×(1+x)^(−k−1) =(−1)^(k−1) (k−1)!(−1)k×(1/((1+x)^(k+1) )) =(−1)^(k−1+1) ×k(k−1)!×(1/((1+x)^(k+1) )) =(((−1)^k ×k!)/((1+x)^(k+1) ))←this expression is same if you put n=k+1 in expression (((−1)^(n−1) ×(n−1)!)/((1+x)^n )) hence proved...$
	$2) y = l n (1 + x)$ $\frac{d y}{d x} = \frac{1}{(1 + x)} \frac{d^{2} y}{{d x}^{2}} = \frac{- 1}{{(1 + x)}^{2}}$ $s o w h e n n = 2$ $L H S = \frac{- 1}{{(1 + x)}^{2}} a n d R H S = \frac{{(- 1)}^{2 - 1} (2 - 1)!}{{(1 + x)}^{2}} = \frac{- 1}{{(1 + x)}^{2}}$ $s o f o r n = 2 t h e s t a t e m e n t i s t r u e$ $n o w l e t a s s u m e f o r n = k t h e s t a t e m e n t i s t r u e$ $w e h a v e t o p r o v e t h a t i t i s t r u e f o r n = k + 1$ $\frac{d^{k} y}{{d x}^{k}} = \frac{{(- 1)}^{k - 1} (k - 1)!}{{(1 + x)}^{k}}$ $s o \frac{d}{d x} (\frac{d^{k} y}{{d x}^{k}}) = \frac{d^{k + 1} y}{{d x}^{k + 1}} \leftarrow L H S$ $\frac{d}{d x} [\frac{{(- 1)}^{k - 1} (k - 1)!}{{(1 + x)}^{k}}]$ $= \frac{{(- 1)}^{k - 1} (k - 1)!}{1} \times \frac{d}{d x} {\frac{1}{{(1 + x)}^{k}}}$ $= {(- 1)}^{k - 1} (k - 1)! \times (- k) \times {(1 + x)}^{- k - 1}$ $= {(- 1)}^{k - 1} (k - 1)! (- 1) k \times \frac{1}{{(1 + x)}^{k + 1}}$ $= {(- 1)}^{k - 1 + 1} \times k (k - 1)! \times \frac{1}{{(1 + x)}^{k + 1}}$ $= \frac{{(- 1)}^{k} \times k!}{{(1 + x)}^{k + 1}} \leftarrow t h i s e x p r e s s i o n i s s a m e i f y o u$ $p u t n = k + 1 i n e x p r e s s i o n$ $\frac{{(- 1)}^{n - 1} \times (n - 1)!}{{(1 + x)}^{n}}$ $h e n c e p r o v e d . . .$
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