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Question Number 113243 by ShakaLaka last updated on 11/Sep/20

Commented by ShakaLaka last updated on 11/Sep/20

$Can anybody solve this$ $question ?$
Commented by mr W last updated on 11/Sep/20

$\sin ω t \cdot \sin ω a t$ $h a s n o x a s v a r i a b l e!$ $p l e a s e c h e c k t h e q u e s t i o n!$
Commented by ShakaLaka last updated on 11/Sep/20

$I will have to ask my teacher$ $t h a t w h a t i s t h e c o r r e c t q u e s t i o n .$ $because this question I wrote is$ $totally same as it is .$ $Actually, this is the question of$ $the assignmemt, w h i c h w a s$ $w r i t t e n d i r e c t l y f r o m u n i v e r s i t y,$ $n o t b y m y t e a c h e r .$ $B y t h e w a y, t h e r e i s a l s o$ $a n o t h e r m i s t a k e i n t h e q u e s t i o n$ $t h a t t h e r e i s w r i t t e n \frac{\partial^{2} v}{\partial c^{2}} i n s t e a d$ $o f \frac{\partial^{2} v}{\partial t^{2}}$
	Answered by mathdave last updated on 12/Sep/20

	$i f o n l y u (x, t) = \sin w x . \sin a t$ $t h e n i s a s o l u t i o n t h a t s a t i s f y t h e w a v e$ $e q u a t i o n \frac{\partial^{2} u}{\partial t^{2}} = a^{2} . \frac{\partial^{2} u}{\partial x^{2}}$ $b y l e t t i n g$ $\frac{\partial^{2} (\sin w x . \sin w a t)}{\partial t^{2}} = - w^{2} a^{2} \sin w x . \sin w a t$ $a n d$ $a^{2} . \frac{\partial^{2} (\sin w x . \sin w a t)}{\partial x^{2}} = - w^{2} a^{2} \sin w x . \sin w a t$ $∵ u (x, t) = \sin w x . \sin w a t i s a s o l u t i o n t o$ $w a v e e q u a t i o n$ $\frac{\partial^{2} u}{\partial t^{2}} = a^{2} . \frac{\partial^{2} u}{\partial x^{2}} Q . E . D$
	Commented by ShakaLaka last updated on 13/Sep/20

	$y e a h m a t h d a v e, i t c a n b e$ $p o s s i b l e$
	Commented by ShakaLaka last updated on 15/Sep/20

	$m a t h d a v e y o u w e r e r i g h t . m y$ $t e a c h e r i s a l s o s a y i n g t h a t t h e$ $f u n c t i o n ‘ ‘ \sin ω t \cdot \sin ω a t^{″} i s$ $i n c o r r e c t, t h e c o r r e c t f u n c t i o n$ $i s ‘ ‘ \sin ω x \cdot \sin ω a t^{″}; a s y o u s a i d .$
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