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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} \dots \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

\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}}