x^2 (d^2 y/dx^2 ) + x(dy/dx) + y=0 please solve this Euler equation


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Question Number 60921 by necx1 last updated on 27/May/19

$x^{2} \frac{d^{2} y}{{d x}^{2}} + x \frac{d y}{d x} + y = 0$ $p l e a s e s o l v e t h i s E u l e r e q u a t i o n$
	Answered by tanmay last updated on 27/May/19

	$x = e^{t} \to \frac{d x}{d t} = e^{t} [t = l n x]$ $\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x} = \frac{1}{e^{t}} \times \frac{d y}{d t}$ $\frac{d y}{d t} = e^{t} \times \frac{d y}{d x} \to x \frac{d y}{d x}$ $\frac{d}{d x} (\frac{d y}{d x})$ $= \frac{d}{d t} (\frac{d y}{d x}) \times \frac{d t}{d x}$ $= \frac{d}{d t} (\frac{1}{e^{t}} \times \frac{d y}{d t}) \times \frac{1}{e^{t}}$ $\frac{1}{e^{t}} \times [\frac{1}{e^{t}} \times \frac{d^{2} y}{{d t}^{2}} + e^{- t} \times - 1 \times \frac{d y}{d t}]$ $= {(\frac{1}{e^{t}})}^{2} \times \frac{d^{2} y}{{d t}^{2}} - {(\frac{1}{e^{t}})}^{2} \times \frac{d y}{d t}$ $s o \frac{d^{2} y}{{d x}^{2}} = {(\frac{1}{e^{t}})}^{2} [\frac{d^{2} y}{{d t}^{2}} - \frac{d y}{d t}]$ $x^{2} \times \frac{d^{2} y}{{d x}^{2}} = \frac{d^{2} y}{{d t}^{2}} - \frac{d y}{d t}$ $s o {e q}^{n}$ $x^{2} \frac{d^{2} y}{{d x}^{2}} + x \frac{d y}{d x} + y = 0$ $\frac{d^{2} y}{{d t}^{2}} - \frac{d y}{d t} + \frac{d y}{d t} + y = 0$ $\frac{d^{2} y}{{d t}^{2}} + y = 0$ $y = {A e}^{m t} \to \frac{d^{2} y}{{d t}^{2}} = {A m}^{2} e^{m t}$ ${A m}^{2} e^{m t} + {A e}^{m t} = 0$ ${A e}^{m t} (m^{2} + 1) = 0$ ${A e}^{m t} \neq 0$ $m^{2} + 1 = 0$ $m = \pm i$ $y = C_{1} e^{i t} + C_{2} e^{- i t}$ $y = C_{1} (c o s t + i s i n t) + C_{2} (c o s t - i s i n t)$ $= (C_{1} + C_{2}) c o s t + ({i C}_{1} - {i C}_{2}) s i n t$ $= A_{1} c o s t + B_{1} s i n t$ $= A_{1} c o s (l n x) + B_{1} s i n (l n x)$
	Commented by necx1 last updated on 27/May/19

	$M h e n . . . . I n e e d t o f u r t h e r m y k n o w l e d g e$ $o n t h i s . T h a n k s f o r t h e h e l p e v e n t h o u g h$ $I d o n t f u l l y u n d e r s t a n d .$
	Commented by necx1 last updated on 27/May/19

	$M h e n . . . . I n e e d t o f u r t h e r m y k n o w l e d g e$ $o n t h i s . T h a n k s f o r t h e h e l p e v e n t h o u g h$ $I d o n t f u l l y u n d e r s t a n d .$
	Commented by tanmay last updated on 27/May/19

	$r e a d d i f f e q n b o o k b y D a n i e l a n d m u r r y$
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