if we have DE ((d^2 x(t))/dt^2 ) = −sinx then find x(t) 1. x(t)= acosht+bsinht 2. x(t)= a+bt 3. x(t)= ae^t + be^(2t) 4. x(t)= acost + bsint


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 128430 by sachin1221 last updated on 07/Jan/21

$i f w e h a v e D E$ $\frac{d^{2} x (t)}{{d t}^{2}} = - s i n x$ $t h e n f i n d x (t)$ $1 . x (t) = a c o s h t + b s i n h t$ $2 . x (t) = a + b t$ $3 . x (t) = {a e}^{t} + {b e}^{2 t}$ $4 . x (t) = a c o s t + b s i n t$
Commented by mr W last updated on 07/Jan/21

$i t h i n k a l l o p t i o n s a r e w r o n g .$
Commented by sachin1221 last updated on 07/Jan/21

$t h e n w h a t i s c o r r e c t a n s w e r$
	Answered by mr W last updated on 07/Jan/21

	$l e t x^{'} = u$ $x^{″} = u \frac{d u}{d x}$ $u \frac{d u}{d x} = - \sin x$ $\frac{u^{2}}{2} = \cos x + a$ $u = \frac{d x}{d t} = \pm \sqrt{2 (\cos x + a)}$ $\frac{d x}{\sqrt{\cos x + a}} = \pm \sqrt{2} d t$ $\int \frac{d x}{\sqrt{\cos x + a}} = \pm \sqrt{2} \int d t$ $\frac{2 F (\frac{x}{2} ∣ \frac{2}{a + 1})}{\sqrt{a + 1}} = \sqrt{2} (b \pm t)$ $F (\frac{x}{2} ∣ \frac{2}{a + 1}) = (b \pm t) \sqrt{\frac{a + 1}{2}}$ $o r$ $F (\frac{x}{2} ∣ \frac{1}{a^{2}}) = a (b \pm t)$ $F = i n c o m p l e t e e l l i p t i c i n t e g r a l o f$ $t h e f i r s t k i n d$
	Commented by sachin1221 last updated on 10/Jan/21

	$i {d i d n}^{'} t g e t l a s t 3 s t e p s$
	Commented by mr W last updated on 10/Jan/21

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum