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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-

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Question Number 108469 by bemath last updated on 17/Aug/20
((⊂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.
$$\:\:\:\frac{\subset\mathcal{B}{e}\mathcal{M}{ath}\supset}{\cap} \\ $$$$\left(\mathrm{1}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}\:\sqrt{\mathrm{cos}\:\mathrm{2}{x}}\:\sqrt{\mathrm{cos}\:\mathrm{3}{x}}…\sqrt{\mathrm{cos}\:{nx}}}{{x}^{\mathrm{2}} }\:? \\ $$$$\left(\mathrm{2}\right){x}^{\mathrm{2}} {y}''+{xy}'−\mathrm{4}{y}=\mathrm{0};\:{y}\left(\mathrm{1}\right)=\mathrm{2}\:{and} \\ $$$$\:\:\:{y}'\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(\mathrm{3}\right){find}\:{the}\:{probability}\:{that}\:{a}\:{person}\:\:{throwing}\:{three} \\ $$$${coins}\:{at}\:{once}\:{will}\:{get}\:{all}\:{the}\:{face}\:{or}\: \\ $$$${everything}\:{back}\:{for}\:{second}\:{time}\:{at} \\ $$$$\mathrm{5}\:{the}\:{throws}. \\ $$
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
$$\:\:\:\frac{\multimap{JS}\multimap}{\sim} \\ $$$$\left(\mathrm{2}\right)\:{let}\:{x}\:=\:{e}^{{r}} \: \\ $$$$\Rightarrow\:\frac{{dy}}{{dr}}\:=\:\frac{{dy}}{{dx}}.\:\frac{{dx}}{{dr}}\: \\ $$$$\Rightarrow\frac{{dy}}{{dr}}\:=\:{x}\:\frac{{dy}}{{dx}} \\ $$$$\Rightarrow\frac{{d}^{\mathrm{2}} {y}}{{dr}^{\mathrm{2}} }\:=\:{x}^{\mathrm{2}} \:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+{x}\:\frac{{dy}}{{dx}} \\ $$$${so}\:{we}\:{get}\: \\ $$$${x}^{\mathrm{2}} \:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+{x}\:\frac{{dy}}{{dx}}\:−\mathrm{4}{y}\:=\:\mathrm{0} \\ $$$$\Rightarrow\frac{{d}^{\mathrm{2}} {y}}{{dr}^{\mathrm{2}} }−\mathrm{4}{y}\:=\:\mathrm{0} \\ $$$${homogenous}\:{equation}\: \\ $$$$\lambda^{\mathrm{2}} −\mathrm{4}=\mathrm{0}\:\rightarrow\:\lambda=\pm\mathrm{2} \\ $$$${general}\:{solution}\: \\ $$$${y}\:=\:{C}_{\mathrm{1}} {e}^{−\mathrm{2}{r}} \:+\:{C}_{\mathrm{2}} {e}^{\mathrm{2}{r}} \:={C}_{\mathrm{1}} \left({e}^{{r}} \right)^{−\mathrm{2}} +{C}_{\mathrm{2}} \left({e}^{{r}} \right)^{\mathrm{2}} \\ $$$${y}\:=\:{C}_{\mathrm{1}} {x}^{−\mathrm{2}} +{C}_{\mathrm{2}} {x}^{\mathrm{2}} \\ $$$${and}\:{y}'\left({x}\right)=−\mathrm{2}{C}_{\mathrm{1}} {x}^{−\mathrm{3}} +\mathrm{2}{C}_{\mathrm{2}} {x} \\ $$$$\left({i}\right){y}\left(\mathrm{1}\right)=\:{C}_{\mathrm{1}} +{C}_{\mathrm{2}} =\mathrm{2} \\ $$$$\left({ii}\right){y}'\left(\mathrm{1}\right)=−\mathrm{2}{C}_{\mathrm{1}} +\mathrm{2}{C}_{\mathrm{2}} =\mathrm{0} \\ $$$$\rightarrow{C}_{\mathrm{1}} ={C}_{\mathrm{2}} \:\begin{cases}{{C}_{\mathrm{1}} =\mathrm{1}}\\{{C}_{\mathrm{2}} =\mathrm{1}}\end{cases} \\ $$$$\therefore\:{y}\:=\:{x}^{−\mathrm{2}} +{x}^{\mathrm{2}} \\ $$
Answered by mathmax by abdo last updated on 17/Aug/20
2) x^2 y^(′′) +xy^′ −4y =0 with y(1)=2 and y^′ (1)=0 let y =x^m ⇒y^′ =mx^(m−1) ⇒y^((2)) =m(m−1)x^(m−2) e⇒m(m−1)x^m +mx^m −4x^m =0 ⇒(m^2 −m+m−4)x^m =0 ⇒ m^2 −4 =0 ⇒m =+^− 2 ⇒y (x)=ax^2 +bx^(−2) y(1)=2 ⇒a+b =2 y^′ (x) =2ax−2b x^(−3) wehave y^′ (1) =0 ⇒2a−2b =0 ⇒a=b ⇒ 2a=2 ⇒a=b=1 ⇒y(x) =x^2 +(1/x^2 )
$$\left.\mathrm{2}\right)\:\mathrm{x}^{\mathrm{2}} \mathrm{y}^{''} \:+\mathrm{xy}^{'} −\mathrm{4y}\:=\mathrm{0}\:\:\mathrm{with}\:\mathrm{y}\left(\mathrm{1}\right)=\mathrm{2}\:\mathrm{and}\:\mathrm{y}^{'} \left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{let}\:\mathrm{y}\:=\mathrm{x}^{\mathrm{m}} \:\Rightarrow\mathrm{y}^{'} \:=\mathrm{mx}^{\mathrm{m}−\mathrm{1}} \:\Rightarrow\mathrm{y}^{\left(\mathrm{2}\right)} \:=\mathrm{m}\left(\mathrm{m}−\mathrm{1}\right)\mathrm{x}^{\mathrm{m}−\mathrm{2}} \\ $$$$\mathrm{e}\Rightarrow\mathrm{m}\left(\mathrm{m}−\mathrm{1}\right)\mathrm{x}^{\mathrm{m}} \:+\mathrm{mx}^{\mathrm{m}} −\mathrm{4x}^{\mathrm{m}} \:=\mathrm{0}\:\Rightarrow\left(\mathrm{m}^{\mathrm{2}} −\mathrm{m}+\mathrm{m}−\mathrm{4}\right)\mathrm{x}^{\mathrm{m}} \:=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{m}^{\mathrm{2}} −\mathrm{4}\:=\mathrm{0}\:\Rightarrow\mathrm{m}\:=\overset{−} {+}\mathrm{2}\:\Rightarrow\mathrm{y}\:\left(\mathrm{x}\right)=\mathrm{ax}^{\mathrm{2}} \:+\mathrm{bx}^{−\mathrm{2}} \\ $$$$\mathrm{y}\left(\mathrm{1}\right)=\mathrm{2}\:\Rightarrow\mathrm{a}+\mathrm{b}\:=\mathrm{2} \\ $$$$\mathrm{y}^{'} \left(\mathrm{x}\right)\:=\mathrm{2ax}−\mathrm{2b}\:\mathrm{x}^{−\mathrm{3}} \:\:\mathrm{wehave}\:\mathrm{y}^{'} \left(\mathrm{1}\right)\:=\mathrm{0}\:\Rightarrow\mathrm{2a}−\mathrm{2b}\:=\mathrm{0}\:\Rightarrow\mathrm{a}=\mathrm{b}\:\Rightarrow \\ $$$$\mathrm{2a}=\mathrm{2}\:\Rightarrow\mathrm{a}=\mathrm{b}=\mathrm{1}\:\Rightarrow\mathrm{y}\left(\mathrm{x}\right)\:=\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$ \\ $$

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