Find-a-function-f-x-satisfying-the-following-equation-a-b-1-2-f-x-2-f-x-2-d-dx-f-x-2-dx-0-b-gt-0-a-gt-0-b-a-

Question Number 1581 by 112358 last updated on 21/Aug/15

F i n d a f u n c t i o n f (x) s a t i s f y i n g

t h e f o l l o w i n g e q u a t i o n .

\int_{a}^{b} [\frac{1}{2} {f (x)}^{2} - \sqrt{{f (x)}^{2} + {\frac{d}{d x} (f (x))}^{2}}] d x = 0

b > 0, a > 0, b \neq a .

Commented by 123456 last updated on 21/Aug/15

\frac{1}{2} f^{2} - \sqrt{f^{2} + {(f^{'})}^{2}} = 0

f^{2} = 2 \sqrt{f^{2} + {(f^{'})}^{2}}

f^{4} = 4 f^{2} + 4 {(f^{'})}^{2}

Commented by 112358 last updated on 25/Aug/15

f^{4} = 4 f^{2} + 4 {(f^{^{'}})}^{2}

f^{^{'}} = \pm \frac{1}{2} \sqrt{f^{4} - 4 f^{2}}

2 f^{^{'}} = \pm f \sqrt{f^{2} - 4}

2 \frac{d f}{d x} = \pm f \sqrt{f^{2} - 4}

I n t e g r a t i o n b y s e p a r a t i o n o f v a r i a b l e s

\Rightarrow 2 \int \frac{d f}{f \sqrt{f^{2} - 4}} = \pm \int d x = K \pm x

l e t f (x) = 2 s e c u \Rightarrow d f = 2 s e c u \times t a n u d u

∴ f \sqrt{f^{2} - 4} = 2 s e c u \sqrt{4 {s e c}^{2} u - 4}

= (2 s e c u) (2 \sqrt{{s e c}^{2} u - 1})

∵ {t a n}^{2} u = {s e c}^{2} u - 1

\Rightarrow f \sqrt{f^{2} - 4} = 4 s e c u \sqrt{{t a n}^{2} u} = 4 s e c u t a n u (i f t a n u > 0)

∴ 2 \int \frac{2 s e c u t a n u}{4 s e c u t a n u} d u = K \pm x

\int d u = K \pm x \Rightarrow u = K \pm x

∵ f (x) = 2 s e c u \Rightarrow u = {c o s}^{- 1} (\frac{2}{f (x)})

∴ {c o s}^{- 1} (\frac{2}{f (x)}) = K \pm x

\frac{2}{f (x)} = c o s (K \pm x)

f (x) = 2 s e c (K \pm x)

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