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Differential EquationQuestion and Answers: Page 1

Question Number 206616 Answers: 1 Comments: 0

Question Number 206449 Answers: 1 Comments: 0

solve the first order differential equation: xdy − ydx = (xy)^(1/2) dx

$${solve}\:{the}\:{first}\:{order}\:{differential} \\ $$$${equation}: \\ $$$$ \\ $$$${xdy}\:−\:{ydx}\:=\:\left({xy}\right)^{\mathrm{1}/\mathrm{2}} {dx} \\ $$

Question Number 206142 Answers: 1 Comments: 0

let (d^2 y/dx^2 )+p(x)(dy/dx)+q(x)y=0 , x∈R where p(x) and q(x) are continuous function if y_1 = sinx−2cosx and y_2 = 2sinx +cosx are L.I (linearly independent) solution then ∣4p(0)+2q(1)∣ = ?

$$\:\:\:\:\mathrm{let}\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+{p}\left({x}\right)\frac{{dy}}{{dx}}+{q}\left({x}\right){y}=\mathrm{0}\:,\:{x}\in\mathbb{R}\:\mathrm{where}\: \\ $$$$\:\:\:\:{p}\left({x}\right)\:\mathrm{and}\:{q}\left({x}\right)\:\mathrm{are}\:\mathrm{continuous}\:\mathrm{function}\:\mathrm{if} \\ $$$$\:\:\:\:{y}_{\mathrm{1}} =\:\mathrm{sin}{x}−\mathrm{2cos}{x}\:{and}\:{y}_{\mathrm{2}} \:=\:\mathrm{2sin}{x}\:+\mathrm{cos}{x} \\ $$$$\:\:\:\:\mathrm{are}\:{L}.{I}\:\left(\mathrm{linearly}\:\mathrm{independent}\right)\:\mathrm{solution} \\ $$$$\:\:\:\:\:\mathrm{then}\:\:\mid\mathrm{4}{p}\left(\mathrm{0}\right)+\mathrm{2}{q}\left(\mathrm{1}\right)\mid\:=\:?\:\:\: \\ $$

Question Number 206047 Answers: 1 Comments: 0

Question Number 205184 Answers: 0 Comments: 0

Resuelve la siguiente ecuacio^ n diferencial (dx/dy) + x^2 = (1/y^4 )

$${Resuelve}\:{la}\:{siguiente}\:{ecuaci}\acute {{o}n}\:{diferencial} \\ $$$$\frac{{dx}}{{dy}}\:+\:{x}^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{{y}^{\mathrm{4}} } \\ $$

Question Number 203898 Answers: 0 Comments: 2

Let f(W) be a function of vector W ∈ R^N , i.e. f(W) = (1/(1 + e^(−W^T x) )) Determine the first derivative and matrix of second derivatives of f with respect to W

$${Let}\:{f}\left({W}\right)\:{be}\:{a}\:{function}\:{of}\:{vector}\:{W}\:\in\: {R}^{{N}} , \\ $$$${i}.{e}.\:{f}\left({W}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{e}^{−{W}^{{T}} {x}} } \\ $$$${Determine}\:{the}\:{first}\:{derivative}\:{and} \\ $$$${matrix}\:{of}\:{second}\:{derivatives}\:{of}\:{f}\:{with} \\ $$$${respect}\:{to}\:{W} \\ $$$$ \\ $$

Question Number 203694 Answers: 1 Comments: 0

Question Number 203498 Answers: 0 Comments: 0

(d^(3 ) y/dx^3 )=4(x+(1/4))^2 −4y

$$\frac{{d}^{\mathrm{3}\:} {y}}{{dx}^{\mathrm{3}} }=\mathrm{4}\left({x}+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} −\mathrm{4}{y} \\ $$

Question Number 201555 Answers: 1 Comments: 0

Question Number 201421 Answers: 1 Comments: 5

Question Number 200902 Answers: 2 Comments: 0

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Question Number 200022 Answers: 0 Comments: 0

solve the associated legendre equation λ=l (l+1)η^2 ;l=0,1,2... and m^2 ≤ l(l+1) which requires −l≤m≤l using power series

$$\mathrm{solve}\:\mathrm{the}\:\mathrm{associated}\:\mathrm{legendre}\:\mathrm{equation} \\ $$$$\lambda={l}\:\left({l}+\mathrm{1}\right)\eta^{\mathrm{2}} \:;{l}=\mathrm{0},\mathrm{1},\mathrm{2}...\:\:\:{and}\:{m}^{\mathrm{2}} \leqslant\:{l}\left({l}+\mathrm{1}\right)\: \\ $$$${which}\:{requires}\:−{l}\leqslant{m}\leqslant{l}\:\mathrm{using}\:\mathrm{power}\:\mathrm{series} \\ $$

Question Number 199996 Answers: 0 Comments: 0

Calculate the first order energy correction for 1−dimensional non−degenerate anharmonic oscillator whose harmiltonian is HL

$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{first}\:\mathrm{order}\:\mathrm{energy}\:\mathrm{correction}\:\mathrm{for} \\ $$$$\mathrm{1}−\mathrm{dimensional}\:\mathrm{non}−\mathrm{degenerate}\:\mathrm{anharmonic} \\ $$$$\mathrm{oscillator}\:\mathrm{whose}\:\mathrm{harmiltonian}\:\mathrm{is}\:\mathscr{H}\underline{\mathscr{L}} \\ $$

Question Number 199876 Answers: 0 Comments: 0

Question Number 199891 Answers: 0 Comments: 1

solve by laplce transform y^(′′) −y^′ +y =(x+1)e^x

$${solve}\:{by}\:{laplce}\:{transform} \\ $$$${y}^{''} −{y}^{'} +{y}\:=\left({x}+\mathrm{1}\right){e}^{{x}} \\ $$

Question Number 199844 Answers: 2 Comments: 0

solve (d^2 /dx^2 ) x

$$\mathrm{solve}\:\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{dx}^{\mathrm{2}} }\:\mathrm{x} \\ $$

Question Number 198646 Answers: 1 Comments: 0

Question Number 198104 Answers: 1 Comments: 0

Question Number 197389 Answers: 0 Comments: 0

please check my answer (x−2y+5)dx+(2x−y+4)dy=0 X=x+a & Y=y+b (X−a−2Y+2b+5)dX+(2X−2a−Y+b+4)dY=0 -a+2b=-5 -2a+b=-4 a=1,b=-2 (X−2Y)dX+(2X−Y)dY=0 Y=XV⇒(dY/dX)=V+X(dV/dx) (X−2XV)+(2X−XV)(V+X(dV/dX))=0 (2X−XV)(V+X(dV/dX))=-X+2XV (V+X(dV/dX))=((-1+2V)/(2−V)) X(dV/dX)=((-1+2V)/(2−V))−V X(dV/dX)=((-1+2V−2V+V^( 2) )/(2−V)) X(dV/dX)=((V^( 2) −1)/(2−V)) ∫((2−V)/(V^( 2) −1))dV=∫(dX/X) ∫(1/(2(V−1)))+(3/(2(V+1)))dV=ln x+C (1/2)ln(V−1)+(3/2)ln(V+1) (1/2)ln(((y−2)/(x+1))−1)+(3/2)ln(((y−2)/(x+1))+1)=ln x+C

$$ \\ $$$${please}\:{check}\:{my}\:{answer} \\ $$$$\:\left({x}−\mathrm{2}{y}+\mathrm{5}\right){dx}+\left(\mathrm{2}{x}−{y}+\mathrm{4}\right){dy}=\mathrm{0} \\ $$$$\:{X}={x}+{a}\:\&\:{Y}={y}+{b} \\ $$$$ \\ $$$$\:\left({X}−{a}−\mathrm{2}{Y}+\mathrm{2}{b}+\mathrm{5}\right){dX}+\left(\mathrm{2}{X}−\mathrm{2}{a}−{Y}+{b}+\mathrm{4}\right){dY}=\mathrm{0} \\ $$$$\:-{a}+\mathrm{2}{b}=-\mathrm{5} \\ $$$$\:-\mathrm{2}{a}+{b}=-\mathrm{4} \\ $$$$\:{a}=\mathrm{1},{b}=-\mathrm{2} \\ $$$$\:\left({X}−\mathrm{2}{Y}\right){dX}+\left(\mathrm{2}{X}−{Y}\right){dY}=\mathrm{0} \\ $$$$\:{Y}={XV}\Rightarrow\frac{{dY}}{{dX}}={V}+{X}\frac{{dV}}{{dx}} \\ $$$$\:\left({X}−\mathrm{2}{XV}\right)+\left(\mathrm{2}{X}−{XV}\right)\left({V}+{X}\frac{{dV}}{{dX}}\right)=\mathrm{0} \\ $$$$\:\left(\mathrm{2}{X}−{XV}\right)\left({V}+{X}\frac{{dV}}{{dX}}\right)=-{X}+\mathrm{2}{XV} \\ $$$$\:\left({V}+{X}\frac{{dV}}{{dX}}\right)=\frac{-\mathrm{1}+\mathrm{2}{V}}{\mathrm{2}−{V}} \\ $$$$\:{X}\frac{{dV}}{{dX}}=\frac{-\mathrm{1}+\mathrm{2}{V}}{\mathrm{2}−{V}}−{V} \\ $$$$\:{X}\frac{{dV}}{{dX}}=\frac{-\mathrm{1}+\mathrm{2}{V}−\mathrm{2}{V}+{V}^{\:\mathrm{2}} }{\mathrm{2}−{V}} \\ $$$$\:{X}\frac{{dV}}{{dX}}=\frac{{V}^{\:\mathrm{2}} −\mathrm{1}}{\mathrm{2}−{V}} \\ $$$$\:\int\frac{\mathrm{2}−{V}}{{V}^{\:\mathrm{2}} −\mathrm{1}}{dV}=\int\frac{{dX}}{{X}} \\ $$$$\:\int\frac{\mathrm{1}}{\mathrm{2}\left({V}−\mathrm{1}\right)}+\frac{\mathrm{3}}{\mathrm{2}\left({V}+\mathrm{1}\right)}{dV}={ln}\:{x}+{C} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({V}−\mathrm{1}\right)+\frac{\mathrm{3}}{\mathrm{2}}{ln}\left({V}+\mathrm{1}\right) \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{{y}−\mathrm{2}}{{x}+\mathrm{1}}−\mathrm{1}\right)+\frac{\mathrm{3}}{\mathrm{2}}{ln}\left(\frac{{y}−\mathrm{2}}{{x}+\mathrm{1}}+\mathrm{1}\right)={ln}\:{x}+{C} \\ $$$$ \\ $$$$ \\ $$

Question Number 197113 Answers: 1 Comments: 0

Prove that ∫^( +∞) _( 0) (((ln(t+(√(1+t^2 ))))/t))dt=(π^2 /2)

$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\underset{\:\mathrm{0}} {\int}^{\:+\infty} \left(\frac{{ln}\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)}{{t}}\right){dt}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$

Question Number 196934 Answers: 1 Comments: 0

Question Number 196458 Answers: 1 Comments: 0

Question Number 196199 Answers: 1 Comments: 0

Solve: y′′′=((3y′(y′′)^2 )/(1+(y′)^2 ))

$$\mathrm{Solve}: \\ $$$${y}'''=\frac{\mathrm{3}{y}'\left({y}''\right)^{\mathrm{2}} }{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} } \\ $$

Question Number 195518 Answers: 1 Comments: 0

y=((x^3 −27)/(x^2 +3x+9)) y=((sin^2 x)/(1+cosx)) y=((sinx+cosx)/( (√(1+cos2x))))

$${y}=\frac{{x}^{\mathrm{3}} −\mathrm{27}}{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{9}} \\ $$$${y}=\frac{{sin}^{\mathrm{2}} {x}}{\mathrm{1}+{cosx}} \\ $$$${y}=\frac{{sinx}+{cosx}}{\:\sqrt{\mathrm{1}+{cos}\mathrm{2}{x}}} \\ $$

Question Number 195301 Answers: 1 Comments: 0

{ ((x^2 +y=11)),((x+y^2 =7)) :}⇒ x,y=?

$$\begin{cases}{{x}^{\mathrm{2}} +{y}=\mathrm{11}}\\{{x}+{y}^{\mathrm{2}} =\mathrm{7}}\end{cases}\Rightarrow\:{x},{y}=? \\ $$

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