((⊂BeMath⊃)/∩) (1)lim_(x→0) ((1−cos x (√(cos 2x)) (√(cos 3x))...(√(cos nx)))/x^2 ) ? (2)x^2 y′′+xy′−4y=0; y(1)=2 and y′(1)=0 (3)find the probability that a person throwing three coins at once will get all the face or everything back for second time at 5 the throws.


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Question Number 108469 by bemath last updated on 17/Aug/20

$\frac{\subset B e M a t h \supset}{\cap}$ $(1) \lim_{x \to 0} \frac{1 - \cos x \sqrt{\cos 2 x} \sqrt{\cos 3 x} . . . \sqrt{\cos n x}}{x^{2}} ?$ $(2) x^{2} y^{″} + {x y}^{'} - 4 y = 0; y (1) = 2 a n d$ $y^{'} (1) = 0$ $(3) f i n d t h e p r o b a b i l i t y t h a t a p e r s o n t h r o w i n g t h r e e$ $c o i n s a t o n c e w i l l g e t a l l t h e f a c e o r$ $e v e r y t h i n g b a c k f o r s e c o n d t i m e a t$ $5 t h e t h r o w s .$
	Answered by john santu last updated on 17/Aug/20
	$((⊸JS⊸)/∼) (2) let x = e^r ⇒ (dy/dr) = (dy/dx). (dx/dr) ⇒(dy/dr) = x (dy/dx) ⇒(d^2 y/dr^2 ) = x^2 (d^2 y/dx^2 )+x (dy/dx) so we get x^2 (d^2 y/dx^2 )+x (dy/dx) −4y = 0 ⇒(d^2 y/dr^2 )−4y = 0 homogenous equation λ^2 −4=0 → λ=±2 general solution y = C_1 e^(−2r) + C_2 e^(2r) =C_1 (e^r )^(−2) +C_2 (e^r )^2 y = C_1 x^(−2) +C_2 x^2 and y′(x)=−2C_1 x^(−3) +2C_2 x (i)y(1)= C_1 +C_2 =2 (ii)y′(1)=−2C_1 +2C_2 =0 →C_1 =C_2 { ((C_1 =1)),((C_2 =1)) :} ∴ y = x^(−2) +x^2$
	$\frac{⊸ J S ⊸}{\sim}$ $(2) l e t x = e^{r}$ $\Rightarrow \frac{d y}{d r} = \frac{d y}{d x} . \frac{d x}{d r}$ $\Rightarrow \frac{d y}{d r} = x \frac{d y}{d x}$ $\Rightarrow \frac{d^{2} y}{{d r}^{2}} = x^{2} \frac{d^{2} y}{{d x}^{2}} + x \frac{d y}{d x}$ $s o w e g e t$ $x^{2} \frac{d^{2} y}{{d x}^{2}} + x \frac{d y}{d x} - 4 y = 0$ $\Rightarrow \frac{d^{2} y}{{d r}^{2}} - 4 y = 0$ $h o m o g e n o u s e q u a t i o n$ $λ^{2} - 4 = 0 \to λ = \pm 2$ $g e n e r a l s o l u t i o n$ $y = C_{1} e^{- 2 r} + C_{2} e^{2 r} = C_{1} {(e^{r})}^{- 2} + C_{2} {(e^{r})}^{2}$ $y = C_{1} x^{- 2} + C_{2} x^{2}$ $a n d y^{'} (x) = - 2 C_{1} x^{- 3} + 2 C_{2} x$ $(i) y (1) = C_{1} + C_{2} = 2$ $(i i) y^{'} (1) = - 2 C_{1} + 2 C_{2} = 0$ $\to C_{1} = C_{2} {\begin{cases} C_{1} = 1 \\ C_{2} = 1 \end{cases}$ $∴ y = x^{- 2} + x^{2}$
	Answered by mathmax by abdo last updated on 17/Aug/20

	$2) x^{2} y^{^{″}} + {xy}^{^{'}} - 4 y = 0 with y (1) = 2 and y^{^{'}} (1) = 0$ $let y = x^{m} \Rightarrow y^{^{'}} = {mx}^{m - 1} \Rightarrow y^{(2)} = m (m - 1) x^{m - 2}$ $e \Rightarrow m (m - 1) x^{m} + {mx}^{m} - {4 x}^{m} = 0 \Rightarrow (m^{2} - m + m - 4) x^{m} = 0 \Rightarrow$ $m^{2} - 4 = 0 \Rightarrow m = \bar{+} 2 \Rightarrow y (x) = {ax}^{2} + {bx}^{- 2}$ $y (1) = 2 \Rightarrow a + b = 2$ $y^{^{'}} (x) = 2 ax - 2 b x^{- 3} wehave y^{^{'}} (1) = 0 \Rightarrow 2 a - 2 b = 0 \Rightarrow a = b \Rightarrow$ $2 a = 2 \Rightarrow a = b = 1 \Rightarrow y (x) = x^{2} + \frac{1}{x^{2}}$
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