X 2 y 2 = (x – y) 2 2xy Algebraic identities The algebraic equations which are valid for all values of variables in them are called algebraic identities For x^2y^2=2xy, we get (by differentiating implicitly), dy/dx =1 That's the same as the derivative of a linear function with slope, 1 Hmmmmm Let's see If we have x^2y^2=2xy The we must also have x^22xy y^2=0 Factoring gets us (xy)^2 = 0 And the only way for that to happen is to have xy=0 So y=x and dy/dx =1 1 Answer Sorted by 1 Alpha identifies it as Legendre's equation and gives the solution y ( x) = c 1 x c 2 ( − x ( log ( 1 − x) / 2 − log ( x 1)) − 1) It offers step by step if you have the right account Share
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X^(2)-y^(2)+2xy(dy)/(dx)=0-(2xy^3yCosx)dx (3x^2y^2Sinx)dy=0 find the complete solution and Check the Solution and xdxydy=a^2(xdyydy)/x^2y^2 is exact Differential EqIn this viExtended Keyboard Examples Upload Random Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music
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Selesaikan Persamaan Diferensial y''''=(x^2y^2)/(2xy) Tidak ada lagi yang bisa dilakukan dengan topik ini Silakan periksa pernyataan yang dimasukkan atau coba topik lain Cookie & Privasi Situs web ini menggunakan cookie untuk memastikan Anda mendapatkan pengalaman terbaik diCalculus Consider the equation x^2 2xy 4y^2 = 64 Write an expression of the slope of the curve at any point (y^p)= y prime My work 2x 2 (xy^p y) 8yy^p = 0 2x 2xy^p 2y 8yy^p = 0 2xy^p 8yy^p = 2y 2x factored out y^p and then divided to get (2ySolve dy/dx=2xy/(x^2y^2) Question Solve dy/dx=2xy/(x^2y^2) Expert Solution Want to see the full answer?
Click here👆to get an answer to your question ️ Solve the differential equation (x^2 y^2) dx 2xydy = 0 I'll include more steps than you might want to write when you get more experience, but here we go x^22xyy^2x=2 Differentiate both sides with respect to x d/(dx)(x^22xyy^2x) = d/(dx)(2) (Write this every time) d/(dx)(x^2)d/(dx)(2xy)d/(dx)(y^2)d/(dx)(x) = d/(dx)(2) Remember that y is the name of some function(s) of x that I haven't found expressions Q2 Find the value of k so that (3x – 2) is a factor of 15x2 – kx – 14 Q3 If α and β are the zeros of the polynomial f (x) = kx2 4x 4 such that α2 β2 = 24, then find the positive value of k Q4 If α and β are zeroes of the polynomial x2 5x
X^ {2}2yxy^ {2}=0 x 2 2 y x y 2 = 0 This equation is in standard form ax^ {2}bxc=0 Substitute 1 for a, 2y for b, and y^ {2} for c in the quadratic formula, \frac {b±\sqrt {b^ {2}4ac}} {2a} This equation is in standard form a x 2 b x c = 0Answer (1 of 46) No (xy)^2 = x^2 2xy y^2 ,because (x y)^2 means as(x y)(x y) So,(xy)^2 = (xy) * (xy) =x*(xy)y*(xy) =x^2xy xyy^2 =x^22xyy^2Prove x^2 y^2 ≥ 2xy and x^2y^2z^2 ≥ 1/3 mathinschoolcom numbers and algebraic expressions logic, sets, intervals absolute value function and its properties linear function quadratic function polynomials rational function exponential function logarithm number sequences limits of sequences and functions derivative and integral of
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Answer (1 of 8) How do I solve (x²y²) dx 2xy dy= 0?Free PreAlgebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators stepbystepHInt 2x^2x(2y)y^22y2=0 As x is real, the discriminant must be \ge0 (2y)^28(y^22y2)\ge0\iff0\ge4(y2)^2\iff(y2)^2\le0 But as y is real, (y2)^2\ge0 The equation of circle passing through (0,0) and points of intersection of the circle x^2 y^2 2x 4y 4 = 0 and y
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Extended Keyboard Examples Upload Random Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music(A) solution is `x^2y^2=cx` (B) `x^2y^2=cx` `x^2y^2=xc` (D) `y For the differential equation `(x^2y^2)dx2xy dy=0`, which of the following are trueWhen is it easier to use the polar form of an equation or a rectangular form of an equation?
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Click here👆to get an answer to your question ️ Factorise (x^2 2xy y^2) z^2Answer (1 of 10) First you recognize that x^2y^2–2xy = (xy)^2 and expression becomes a difference of squares = (xy)^2 1^2 = (xy1)(xy1) That's the easiest way to0 votes 1 answer
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Then for each case you have a simple (linear) doubleinequality Though as you can see from the graph the two cases are really just the same Your two inequalities are both lines through the origin, so when they get multiplied together you just get the alternating pattern you see inCancel out x in both numerator and denominator Cancel out x in both numerator and denominator \frac {2\left (x^ {2}2xyy^ {2}\right)} {x\left (yx\right)y^ {2}} x ( y − x) y 2 − 2 ( x 2 − 2 x y y 2) Factor the expressions that are not already factored Factor theਕਦਮਦਰਕਦਮ ਸੁਲਝਾ ਦੇ ਨਾਲ ਸਾਡੇ ਮੁਫ਼ਤ ਮੈਥ ਸੋਲਵਰ ਦੀ ਵਰਤੋਂ ਕਰਕੇ ਆਪਣੀਆਂ ਗਣਿਤਕ ਪ੍ਰਸ਼ਨਾਂ ਨੂੰ ਹੱਲ ਕਰੋ। ਸਾਡਾ ਮੈਥ ਸੋਲਵਰ ਬੁਨਿਆਦੀ ਗਣਿਤ, ਪੁਰਾਣੇਬੀਜ ਗਣਿਤ, ਬੀਜ ਗਣਿਤ
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Click here👆to get an answer to your question ️ (x^2 y^2)dx 2xy dy = 0 , the solution to this differential equation represents which curveCalculus Find dy/dx x^2y^2=2xy x2 y2 = 2xy x 2 y 2 = 2 x y Differentiate both sides of the equation d dx (x2 y2) = d dx (2xy) d d x ( x 2 y 2) = d d x ( 2 x y) Differentiate the left side of the equation Tap for more steps Differentiate Tap for more stepsClick here👆to get an answer to your question ️ Factorize x^2 y^2 x y 2xy
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Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with stepbystep explanations, just like a math tutorAlgebra Factor x^22xyy^2 x2 2xy y2 x 2 2 x y y 2 Check that the middle term is two times the product of the numbers being squared in the first term and third term 2xy = 2 ⋅x⋅y 2 x y = 2 ⋅ x ⋅ y Rewrite the polynomial x2 2⋅x⋅yy2 x 2 2 ⋅ x ⋅ y y 2 Factor using the perfect square trinomial rule a2 2abb2 = (ab)2 a 2 2 a b b 2 = ( a b) 2, where a = x a = x and bAnswer (1 of 6) \begin{align}y'&=\dfrac{2xy}{y^2x^2}\\\text{Let }y&=vx\implies \dfrac{\mathrm dy}{\mathrm dx}=vx\dfrac{\mathrm dv}{\mathrm dx}\\\hline vx\dfrac
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As you can see above, #x^2y^2=2x# and #r=2cos theta# give us the same graphs I hope that this was helpful Answer link Related questions How do you convert rectangular coordinates to polar coordinates?There is no GCF to be factor Welcome to Sarthaks eConnect A unique platform where students can interact with teachers/experts/students to get solutions to their queries Students (upto class 102) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (MainsAdvance) and NEET can ask questions from any subject and get quick answers by
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Check out a sample Q&A here See Solution star_border Students who've seen this question also like FIND Algebra & Trigonometry with Analytic Geometry 13th Edition ISBN Author SwokowskiX^22xyy^22x2y1=0 x2 2xyy2 2x− 2y −1 = 0 https//wwwtigeralgebracom/drill/x~2_2xy_y~2_2x2y1=0/ x22xyy22x2y1=0 No solutions found Step by step solution Step 1 Trying to factor by pulling out 11 Factoring x22xy2xy22y1 Thoughtfully split the expression at hand intoTrigonometry Graph x^2y^22x2y1=0 x2 − y2 − 2x − 2y − 1 = 0 x 2 y 2 2 x 2 y 1 = 0 Find the standard form of the hyperbola Tap for more steps Add 1 1 to both sides of the equation x 2 − y 2 − 2 x − 2 y = 1 x 2 y 2 2 x 2 y = 1 Complete the square for x 2 − 2 x x 2
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Click here👆to get an answer to your question ️ Factorise x^2 y^2 z^2 2xy You have x2 −y2 = (x y)(x −y) So in your case x2 − y2 x −y = (x y)(x − y) x − y = x y Answer linkSteps Using the Quadratic Formula x ^ { 2 } y ^ { 2 } 2 x y = 16 x 2 y 2 − 2 x y = 1 6 All equations of the form ax^ {2}bxc=0 can be solved using the quadratic formula \frac {b±\sqrt {b^ {2}4ac}} {2a} The quadratic formula gives two solutions,
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Extended Keyboard Examples Upload Random Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music The differential equation of the system of circles touching the xaxis at origin is (A) (x^2 y^2)dy/dx 2xy = 0 asked in Differential equations by AmanYadav (559k points) differential equations; =(xy)(x2y) x^2xy2y^2 =x^22xyxy2y^2 =x(x2y)y(x2y) =(xy)(x2y) How do you factor #8z^6#?
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You could treat $2xyy'=x^2y^2$ as homogeneous (after a bit of algebra) But this is an exact equation $$(x^2y^2)2xyy'=0 \qquad \Longrightarrow \qquad (x^2y^2)dx2xy\,dy=0$$ You have $P=x^2y^2$ and $Q=2xy$ Notice that $P_y=2y=Q_x$ Thus this is exact $\int P\,dx = \dfrac{1}{3}x^3xy^2C_1(y)$ and $\int Q\,dy = xy^2C_2(x)$
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