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Differential EquationQuestion and Answers: Page 1

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Reduce to canonical form: sin^2 (x)(∂^2 u/∂x^2 )+sin^2 (2x)(∂^2 u/(∂x∂y))+cos^2 (x)(∂^2 u/∂y^2 )=0

$$\mathrm{Reduce}\:\mathrm{to}\:\mathrm{canonical}\:\mathrm{form}: \\ $$$$\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\frac{\partial^{\mathrm{2}} \mathrm{u}}{\partial\mathrm{x}^{\mathrm{2}} }+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2x}\right)\frac{\partial^{\mathrm{2}} \mathrm{u}}{\partial\mathrm{x}\partial\mathrm{y}}+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)\frac{\partial^{\mathrm{2}} \mathrm{u}}{\partial\mathrm{y}^{\mathrm{2}} }=\mathrm{0} \\ $$

Question Number 226961 Answers: 1 Comments: 0

σ = ∫_0 ^( ∞) xe^(−x) J_2 (x)dx=?

$$\:\:\: \\ $$$$\:\:\:\:\:\sigma\:=\:\int_{\mathrm{0}} ^{\:\infty} {xe}^{−{x}} \:{J}_{\mathrm{2}} \left({x}\right){dx}=? \\ $$$$\:\:\:\: \\ $$

Question Number 226778 Answers: 0 Comments: 0

Solve the following D.E (a) (dy/dx)+2y=xy^2 (b) (dy/dx)+3(y/x)=2x^4 y^4

$${Solve}\:{the}\:{following}\:{D}.{E} \\ $$$$\left({a}\right)\:\frac{{dy}}{{dx}}+\mathrm{2}{y}={xy}^{\mathrm{2}} \\ $$$$\left({b}\right)\:\frac{{dy}}{{dx}}+\mathrm{3}\frac{{y}}{{x}}=\mathrm{2}{x}^{\mathrm{4}} {y}^{\mathrm{4}} \\ $$

Question Number 226777 Answers: 4 Comments: 0

Show that ∫_0 ^1 x^2 (1+x^2 )^(−1) dx=((4−π)/4) Hence by using Simpson^′ s rule find the value of π with eleven ordinates. correct to 4 decimal places

$${Show}\:{that} \\ $$$$\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}=\frac{\mathrm{4}−\pi}{\mathrm{4}} \\ $$$${Hence}\:{by}\:{using}\:{Simpson}^{'} {s} \\ $$$${rule}\:{find}\:{the}\:{value}\:\:{of}\:\pi\:{with}\: \\ $$$${eleven}\:{ordinates}. \\ $$$${correct}\:{to}\:\mathrm{4}\:{decimal}\:{places} \\ $$

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(dy/dx)=((y^6 −2x^2 )/(2xy^5 +x^2 y^2 ))

$$\:\frac{{dy}}{{dx}}=\frac{{y}^{\mathrm{6}} −\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}{xy}^{\mathrm{5}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} } \\ $$

Question Number 224025 Answers: 0 Comments: 0

Resuelve la ecuacio^ n diferencial [4x^3 y − (e^(xy) /x) + y ln(x) + x ((x − 4))^(1/3) ]dx + [x^4 − (e^(xy) /y) + x ln(x) − x]dy Help ....

$${Resuelve}\:{la}\:{ecuaci}\acute {{o}n}\:{diferencial} \\ $$$$\left[\mathrm{4}{x}^{\mathrm{3}} {y}\:−\:\frac{{e}^{{xy}} }{{x}}\:+\:{y}\:\mathrm{ln}\left({x}\right)\:+\:{x}\:\sqrt[{\mathrm{3}}]{{x}\:−\:\mathrm{4}}\right]{dx}\:+\:\left[{x}^{\mathrm{4}} −\:\frac{{e}^{{xy}} }{{y}}\:+\:{x}\:\mathrm{ln}\left({x}\right)\:−\:{x}\right]{dy} \\ $$$${Help}\:.... \\ $$

Question Number 223988 Answers: 0 Comments: 0

Solve the DE using the method of Frobenius : (1−x^2 )y′′−2xy′+n(n+1)y=0

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{DE}\:\mathrm{using}\:\mathrm{the}\:\mathrm{method}\:\mathrm{of}\:\mathrm{Frobenius}\::\: \\ $$$$\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\mathrm{y}''−\mathrm{2xy}'+\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\mathrm{y}=\mathrm{0} \\ $$

Question Number 223054 Answers: 0 Comments: 0

find y y^dy = x^dx

$$\mathrm{find}\:{y} \\ $$$${y}^{{dy}} =\:{x}^{{dx}} \\ $$

Question Number 222296 Answers: 0 Comments: 0

(1+x^4 )y′−x^3 y = x^5 −x^3 +2x+1

$$\left(\mathrm{1}+{x}^{\mathrm{4}} \right){y}'−{x}^{\mathrm{3}} {y}\:=\:{x}^{\mathrm{5}} −{x}^{\mathrm{3}} +\mathrm{2}{x}+\mathrm{1} \\ $$

Question Number 221973 Answers: 1 Comments: 0

(d^2 y/dx^2 )+y=k−(1/x^2 )−(6/x^4 ) Find y(x) (k is constant).

$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+{y}={k}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{6}}{{x}^{\mathrm{4}} }\:\:\:\:\:\: \\ $$$${Find}\:{y}\left({x}\right)\:\:\:\:\left({k}\:{is}\:{constant}\right). \\ $$

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Solve: 5x^2 y′′ + x(1 + x) y′ − y = 0

$$\mathrm{Solve}: \\ $$$$\:\:\:\:\:\mathrm{5x}^{\mathrm{2}} \:\mathrm{y}''\:\:+\:\:\:\mathrm{x}\left(\mathrm{1}\:\:+\:\:\mathrm{x}\right)\:\mathrm{y}'\:\:−\:\:\mathrm{y}\:\:\:=\:\:\:\mathrm{0} \\ $$

Question Number 217245 Answers: 1 Comments: 0

find the following differential equation by eliminating the arbritrary constant (1)y=Ae^x +Bcosx (2) xy=Ae^x +Be^(−x) +x^2

$${find}\:{the}\:{following}\:{differential}\:{equation}\: \\ $$$${by}\:{eliminating}\:{the}\:{arbritrary}\:{constant} \\ $$$$\left(\mathrm{1}\right){y}={Ae}^{{x}} +{Bcosx} \\ $$$$\left(\mathrm{2}\right)\:{xy}={Ae}^{{x}} +{Be}^{−{x}} +{x}^{\mathrm{2}} \\ $$$$ \\ $$

Question Number 217046 Answers: 1 Comments: 0

form the differential equation by eliminating the arbritrary constant y^2 =Ax^2 +Bx+C

$${form}\:{the}\:{differential}\:{equation}\:{by}\: \\ $$$${eliminating}\:{the}\:{arbritrary}\:{constant} \\ $$$${y}^{\mathrm{2}} ={Ax}^{\mathrm{2}} +{Bx}+{C} \\ $$

Question Number 216787 Answers: 1 Comments: 0

form the differential equationfrom the following 1) y=Ae^(3x) +Be^(5x) 2) y^2 =(x−1) 3) c(y+c)^2 +x^3 =0

$${form}\:{the}\:{differential}\:{equationfrom}\:{the}\:{following} \\ $$$$\left.\mathrm{1}\right)\:{y}={Ae}^{\mathrm{3}{x}} +{Be}^{\mathrm{5}{x}} \\ $$$$\left.\mathrm{2}\right)\:{y}^{\mathrm{2}} =\left({x}−\mathrm{1}\right) \\ $$$$\left.\mathrm{3}\right)\:{c}\left({y}+{c}\right)^{\mathrm{2}} +{x}^{\mathrm{3}} =\mathrm{0} \\ $$

Question Number 215468 Answers: 2 Comments: 0

((∂xω)/(∂yω))∙(∂e/∂ω)=? x=f(y) ω=g(y) e=h(ω)

$$\frac{\partial{x}\omega}{\partial{y}\omega}\centerdot\frac{\partial{e}}{\partial\omega}=? \\ $$$${x}={f}\left({y}\right) \\ $$$$\omega={g}\left({y}\right) \\ $$$${e}={h}\left(\omega\right) \\ $$

Question Number 214813 Answers: 1 Comments: 1

Help me solve this (dy/dx)+(a/y)+b(√x)=0

$${Help}\:{me}\:{solve}\:{this} \\ $$$$\frac{{dy}}{{dx}}+\frac{{a}}{{y}}+{b}\sqrt{{x}}=\mathrm{0} \\ $$

Question Number 213097 Answers: 1 Comments: 0

Uhhhh. can you guys solve Partial differantial equation ▽^2 𝛗=0 Cylinderical Laplacian case ▽^2 =(1/ρ)∙((∂ )/∂ρ)(ρ((∂ )/∂ρ))+((1/ρ))^2 (∂^2 /∂φ^2 )+((∂^2 )/∂z^2 ) Spherical Laplacian case ▽^2 =((1/r))^2 ((∂ )/∂r)(r^2 ((∂ )/∂r))+(1/(r^2 sin(θ)))∙((∂ )/∂θ)(sin(θ)((∂ )/∂θ))+(1/(r^2 sin^2 (θ)))∙(∂^2 /∂ϕ^2 )

$$\mathrm{Uhhhh}. \\ $$$$\mathrm{can}\:\mathrm{you}\:\mathrm{guys}\:\mathrm{solve}\:\mathrm{Partial}\:\mathrm{differantial}\:\mathrm{equation} \\ $$$$\bigtriangledown^{\mathrm{2}} \boldsymbol{\phi}=\mathrm{0} \\ $$$$\mathrm{Cylinderical}\:\mathrm{Laplacian}\:\mathrm{case} \\ $$$$\bigtriangledown^{\mathrm{2}} =\frac{\mathrm{1}}{\rho}\centerdot\frac{\partial\:\:}{\partial\rho}\left(\rho\frac{\partial\:\:}{\partial\rho}\right)+\left(\frac{\mathrm{1}}{\rho}\right)^{\mathrm{2}} \frac{\partial^{\mathrm{2}} \:}{\partial\phi^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} \:\:}{\partial{z}^{\mathrm{2}} } \\ $$$$\mathrm{Spherical}\:\mathrm{Laplacian}\:\mathrm{case} \\ $$$$\bigtriangledown^{\mathrm{2}} =\left(\frac{\mathrm{1}}{{r}}\right)^{\mathrm{2}} \frac{\partial\:\:}{\partial{r}}\left({r}^{\mathrm{2}} \frac{\partial\:\:}{\partial{r}}\right)+\frac{\mathrm{1}}{{r}^{\mathrm{2}} \mathrm{sin}\left(\theta\right)}\centerdot\frac{\partial\:\:}{\partial\theta}\left(\mathrm{sin}\left(\theta\right)\frac{\partial\:\:}{\partial\theta}\right)+\frac{\mathrm{1}}{{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\theta\right)}\centerdot\frac{\partial^{\mathrm{2}} \:}{\partial\varphi^{\mathrm{2}} } \\ $$

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I=∫_0 ^∞ (((x−arctan x)^2 )/x^4 )dx.

$$ \\ $$$$\:\:{I}=\int_{\mathrm{0}} ^{\infty} \frac{\left({x}−\mathrm{arctan}\:{x}\right)^{\mathrm{2}} }{{x}^{\mathrm{4}} }{dx}. \\ $$

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