Question and Answers Forum

All Questions      Topic List

Differential Equation Questions

Previous in All Question      Next in All Question      

Previous in Differential Equation      Next in Differential Equation      

Question Number 113060 by bemath last updated on 12/Sep/20

(1) (√(dy/dx)) = ((d((√y)))/dx) (2) x^2 ≡ 73 (mod 216) (3) If 2^x_1 +2^x_2 +2^x_3 +...+2^x_n =80,000 where x_1 ,x_2 ,x_3 ,...,x_n are distinct whole number . find the value of n

$$\:\left(\mathrm{1}\right)\:\sqrt{\frac{\mathrm{dy}}{\mathrm{dx}}}\:=\:\frac{\mathrm{d}\left(\sqrt{\mathrm{y}}\right)}{\mathrm{dx}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{x}^{\mathrm{2}} \:\equiv\:\mathrm{73}\:\left(\mathrm{mod}\:\mathrm{216}\right) \\ $$$$\left(\mathrm{3}\right)\:\mathrm{If}\:\mathrm{2}^{\mathrm{x}_{\mathrm{1}} } +\mathrm{2}^{\mathrm{x}_{\mathrm{2}} } +\mathrm{2}^{\mathrm{x}_{\mathrm{3}} } +...+\mathrm{2}^{\mathrm{x}_{\mathrm{n}} } =\mathrm{80},\mathrm{000} \\ $$$$\:\mathrm{where}\:\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} ,\mathrm{x}_{\mathrm{3}} ,...,\mathrm{x}_{\mathrm{n}} \:\mathrm{are}\:\mathrm{distinct} \\ $$$$\mathrm{whole}\:\mathrm{number}\:.\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{n} \\ $$

Answered by john santu last updated on 11/Sep/20

solve (√(dy/dx)) = ((d((√y)))/dx) (sol)(√(dy/dx)) = (dy/(2(√y) dx )) squaring both sides (dy/dx) = (1/(4y)) ((dy/dx))^2 ⇒ p^2 = 4yp p(p−4y) = 0 { ((p=0⇒(dy/dx) =0, y = C)),((p=4y⇒(dy/dx) = 4y ⇒ln y = 4x+k , y=C_1 e^(4x) )) :} ((JS)/(a math farmer))

$$\:\:\:{solve}\:\sqrt{\frac{{dy}}{{dx}}}\:=\:\frac{{d}\left(\sqrt{{y}}\right)}{{dx}} \\ $$$$\left({sol}\right)\sqrt{\frac{{dy}}{{dx}}}\:=\:\frac{{dy}}{\mathrm{2}\sqrt{{y}}\:{dx}\:} \\ $$$${squaring}\:{both}\:{sides} \\ $$$$\:\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{4}{y}}\:\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \Rightarrow\:{p}^{\mathrm{2}} \:=\:\mathrm{4}{yp} \\ $$$$\:{p}\left({p}−\mathrm{4}{y}\right)\:=\:\mathrm{0}\:\begin{cases}{{p}=\mathrm{0}\Rightarrow\frac{{dy}}{{dx}}\:=\mathrm{0},\:{y}\:=\:{C}}\\{{p}=\mathrm{4}{y}\Rightarrow\frac{{dy}}{{dx}}\:=\:\mathrm{4}{y}\:\Rightarrow\mathrm{ln}\:{y}\:=\:\mathrm{4}{x}+{k}\:,\:{y}={C}_{\mathrm{1}} {e}^{\mathrm{4}{x}} }\end{cases}\: \\ $$$$\:\:\:\frac{{JS}}{{a}\:{math}\:{farmer}} \\ $$

Commented by bemath last updated on 11/Sep/20

✓thank you

$$\checkmark\mathrm{thank}\:\mathrm{you} \\ $$

Answered by bobhans last updated on 11/Sep/20

(2) x^2 ≡73 (mod 216) because 216 = 2^3 .3^3 , we first solve the congruences → { ((x^2 ≡ 73 (mod 2^3 ))),((x^2 ≡73 (mod 3^3 ))) :} (a) regarding x^2 ≡73≡1 (mod 8) we obtain x≡ ±1 , ±3 (mod 8) (b) regarding x^2 ≡73≡100 (mod 27) we obtain x≡±10 (mod 27) using the Chinese Remainder Theorem with result from (a) &(b) we conclude x ≡ ±17, ±37, ±71, ± 91 (mod 216)

$$\left(\mathrm{2}\right)\:\mathrm{x}^{\mathrm{2}} \equiv\mathrm{73}\:\left(\mathrm{mod}\:\mathrm{216}\right) \\ $$$$\mathrm{because}\:\mathrm{216}\:=\:\mathrm{2}^{\mathrm{3}} .\mathrm{3}^{\mathrm{3}} \:,\:\mathrm{we}\:\mathrm{first}\:\mathrm{solve}\:\mathrm{the} \\ $$$$\mathrm{congruences}\:\:\rightarrow\begin{cases}{\mathrm{x}^{\mathrm{2}} \equiv\:\mathrm{73}\:\left(\mathrm{mod}\:\mathrm{2}^{\mathrm{3}} \right)}\\{\mathrm{x}^{\mathrm{2}} \equiv\mathrm{73}\:\left(\mathrm{mod}\:\mathrm{3}^{\mathrm{3}} \right)}\end{cases} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{regarding}\:\mathrm{x}^{\mathrm{2}} \equiv\mathrm{73}\equiv\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{8}\right) \\ $$$$\:\mathrm{we}\:\mathrm{obtain}\:\mathrm{x}\equiv\:\pm\mathrm{1}\:,\:\pm\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{8}\right) \\ $$$$\left(\mathrm{b}\right)\:\mathrm{regarding}\:\mathrm{x}^{\mathrm{2}} \equiv\mathrm{73}\equiv\mathrm{100}\:\left(\mathrm{mod}\:\mathrm{27}\right) \\ $$$$\:\:\mathrm{we}\:\mathrm{obtain}\:\mathrm{x}\equiv\pm\mathrm{10}\:\left(\mathrm{mod}\:\mathrm{27}\right) \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{Chinese}\:\mathrm{Remainder}\:\mathrm{Theorem} \\ $$$$\mathrm{with}\:\mathrm{result}\:\mathrm{from}\:\left(\mathrm{a}\right)\:\&\left(\mathrm{b}\right)\:\mathrm{we}\:\mathrm{conclude}\: \\ $$$$\mathrm{x}\:\equiv\:\pm\mathrm{17},\:\pm\mathrm{37},\:\pm\mathrm{71},\:\pm\:\mathrm{91}\:\left(\mathrm{mod}\:\mathrm{216}\right) \\ $$

Answered by bobhans last updated on 11/Sep/20

(3) consider 80,000_(10) =10,011,100,010,000,000_2 that say 80000=2^(16) +2^(13) +2^(12) +2^(11) +2^7 so n = 5

$$\left(\mathrm{3}\right)\:\mathrm{consider}\:\mathrm{80},\mathrm{000}_{\mathrm{10}} =\mathrm{10},\mathrm{011},\mathrm{100},\mathrm{010},\mathrm{000},\mathrm{000}_{\mathrm{2}} \\ $$$$\mathrm{that}\:\mathrm{say}\:\mathrm{80000}=\mathrm{2}^{\mathrm{16}} +\mathrm{2}^{\mathrm{13}} +\mathrm{2}^{\mathrm{12}} +\mathrm{2}^{\mathrm{11}} +\mathrm{2}^{\mathrm{7}} \\ $$$$\mathrm{so}\:\mathrm{n}\:=\:\mathrm{5} \\ $$

Commented by Rasheed.Sindhi last updated on 11/Sep/20

e^X cellent!

$${e}^{\mathcal{X}} {cellent}! \\ $$

Commented by 1549442205PVT last updated on 12/Sep/20

Number in the question is 8000?

$$\mathrm{Number}\:\mathrm{in}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{8000}? \\ $$

Commented by bemath last updated on 12/Sep/20

80,000 sir

$$\mathrm{80},\mathrm{000}\:\mathrm{sir} \\ $$

Commented by Rasheed.Sindhi last updated on 12/Sep/20

But the number is 8000.

$${But}\:{the}\:{number}\:{is}\:\mathrm{8000}. \\ $$

Commented by bemath last updated on 12/Sep/20

sorry sir typo

$$\mathrm{sorry}\:\mathrm{sir}\:\mathrm{typo} \\ $$

Commented by bemath last updated on 12/Sep/20

wkwkwkek..=====

$$\mathrm{wkwkwkek}..===== \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com