First week only $499!The solution of dxdy= 2xyx 2y 21 satisfying y(1)=1 is given by This question has multiple correct options A a system of hyperbola B a system of circles C y 2=x(1x)−1 D (x−2) 2(y−3) 2=5 HardExtended Keyboard Examples Upload Random Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by
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If y = y(x) is the solution of the differential equation 2x^2 dy/dx 2xy 3y^2 = 0 such that y(e) = e/3, then y(1) is equal to asked Jul 9 in Mathematics by GovindSaraswat (268kYou can put this solution on YOUR website!The correct option is A =2x2y34xyzUsing distributive law of multiplication iea×(bc)= abbc(xy22z)×2xy can be expressed as= xy2×2xy2z×2xyApplying commutative law and
See answer Advertisement AdvertisementFormer Lecturer at Sbm Inter College, Rishikesh (1971–07) Author has 113K answers and 99M answer views 1 y x^2y^2 2xy 1 or, = (x^2y^22xy) 1 or, = (xy)^2 (1)^2 Jinfa WangAlgebra 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
Find an answer to your question Solve (X^22xyy^2)(x1) rosaa713 rosaa713 Mathematics Middle School answered Solve (X^22xyy^2)(x1) 1 See answer AdvertisementTranscribed Image Text (1 )y"2cy' 2y = 0 %3DWe can try to factor x 2 −2xy−y 2 but we must do some rearranging first Change signs
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Differential equation Solve $(2xy^4e^y 2xy^3 y)dx (x^2y^4e^y x^2y^2 3x) dy = 0$ written 61 years ago by teamques10 ★Paranirved paranirved Math Secondary School answered X^22xyy^21 How to factorise it? Use implicit differentiation 2xy − y2 = 1 D(2xy −y2) = D(1) Apply the product rule and the chain rule ⇒ 2xy' 2y − 2yy' = 0 2y'(x −y) = − 2y 2y' = − 2y (x − y) y' = − y (x −y) Answer
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X^22xyy^21 How to factorise it? Here, Factor is a term used to express a number as a product of any two numbers like x and y as xy Here, x and y are the factors It also refers to finding out the factors of the givenThe differential equation 2xydy=x 2y 21dx determines A A family of circles with centre on xaxis B A family of circles with centre on yaxis C A family of rectangular hyperbiola with centre on x
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A will now lie on the locus of P, ie AS = AZ The xaxis will be along the line AS, and the yaxis will be along the perpendicular to AS at A, as in the figure By definition PM = PS => MP 2 = PS 2
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Incoming Term: y''(x^2+1)=2xy', y=x/2 2x-y=1, x+y=2 2x-y=1 metodo de igualacion, (x^2+1)y''-2xy'+2y=0, (x^2+1)y''+2xy'=0, y=x^2-4 2x+y=-1, (1-x^2)y''-2xy'+2y=0 power series, (1-x^2)y''-2xy'+6y=0, (1-x^2)y''-2xy'+2y=0 power series solution, (1-x^2)y''-2xy'+12y=0,
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