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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

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