Find all functions that satisfy to (E): ∀ x∈R xf(x)+∫


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Question Number 80293 by ~blr237~ last updated on 01/Feb/20

$F i n d a l l f u n c t i o n s t h a t s a t i s f y t o$ $(E) : \forall x \in R x f (x) + \int_{0}^{x} f (x - t) c o s (2 t) d t = s i n (2 x)$
Commented by john santu last updated on 02/Feb/20

$2 y + {x y}^{'} = 2 \cos 2 x \Rightarrow d i f f e r e n t i a l$ $e q u a t i o n$
Commented by ~blr237~ last updated on 02/Feb/20

$j u s t c h e c k y o u r m i s t a k e o n t h e (3 - 4) l i n e s c r o s s i n g$
Commented by john santu last updated on 02/Feb/20

$\frac{d}{d x} [\underset{0}{\int^{x}} f (u) \cos (x - u) d u] =$ $f (x) \cos (x) + f (0) \cos (0) ?$
Commented by ~blr237~ last updated on 02/Feb/20

$N o$
Commented by jagoll last updated on 02/Feb/20

$w h a t t h e r i g h t ?$
Commented by ~blr237~ last updated on 02/Feb/20

$l i k e t h a t . . . . ? w h a t i s n o t v a l i d ? b e c l e a r ?$
Commented by jagoll last updated on 02/Feb/20

$w e k n o w t h a t$ $\int_{a}^{x} f (u) d u \Rightarrow \frac{d}{d x} \int_{a}^{x} f (u) d u = f (x)$ $i t c l e a r b y f u n d a m e n t a l$ $c a l c u l u s 1 .$ $s o \frac{d}{d x} \int_{1}^{x} f (u) \cos (x - u) d u =$ $f (x) \cos (x - x) = f (x)$
Commented by ~blr237~ last updated on 02/Feb/20

$s i r w h a t i s g o i n g o n ?$ $l e t t a k e f (t) = t$ $F (x) = \int_{0}^{x} (x - t) c o s 2 t d t$ $b y p a r t u^{'} = c o s 2 t a n d v = x - t$ $F (x) = {[\frac{1}{2} (x - t) s i n 2 t]}_{0}^{x} + \frac{1}{2} \int_{0}^{x} s i n 2 t d t$ $= \frac{1}{2} {[\frac{- c o s 2 t}{2}]}_{0}^{x} = \frac{1 - c o s 2 x}{4}$ $\frac{d F}{d x} = \frac{s i n 2 x}{2} \neq f (x) = x$
Commented by jagoll last updated on 02/Feb/20

$o k a y s i r$
Commented by ~blr237~ last updated on 02/Feb/20

$W h e n y o u w a n t t o d e r i v a t e F (x) = \int_{0}^{u (x)} f (x, t) d t m a k e$ $s u r e f i r s t l y (b y a g o o d s t a t i n g v a r i a b l e : m a y b e z = \frac{t}{u (x)}) t o r e m o v e x a t t h e w h o l e b o u n d a r y$ $w e w e r e h a v i n g F (x) = \int_{0}^{x} f (x - t) c o s 2 t d t$ $F (x) = \int_{0}^{1} x [f (x - x v) c o s (2 x v)] d v w i t h v = \frac{t}{x}$ $N o w y o u c a n u s e \frac{d F}{d x} = \int_{0}^{1} \frac{\partial}{\partial x} g (x, v) d v w h e r e g (x, v) = x f (x - x v) c o s (2 x v)$
Commented by mr W last updated on 02/Feb/20

$t o j a g o l l s i r :$ $\int_{a}^{x} f (u) d u \Rightarrow \frac{d}{d x} \int_{a}^{x} f (u) d u = f (x)$ $i s c o r r e c t . b u t i t c a n n o t b e a p p l i e d t o$ $\int_{a}^{x} f (u, x) d u \Rightarrow \frac{d}{d x} \int_{a}^{x} f (u, x) d u \neq f (x, x)$
Commented by mr W last updated on 02/Feb/20

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