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Y=-x^2-4x y=1 x=-3 x=-1-Login Create Account Class9 » Maths Number Systems show that x1 y1 / x1 x1 y1 / y1 =x 2 y 2 /xy Share with your friends Share 9 30 ; Show activity on this post Find the second solution First solution is 1 1 − x Solve by Frobenius Method x ( 1 − x) y ″ 2 ( 1 − 2 x) y ′ − 2 y = 0 The first solution I am able to get is 1 1 − x Other solution is 1 x, but I am getting − 1 x ( 1 − x)
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1 = x Then put your answer back into one of the equations the first looks easier to me, so im going to use that one y = 2x 2 y = 2(1) 2 y = 2 2 y = 0 So again yourThe differential equation has the differential form given in Problem 51 Its solution is given in (2) of Problem 52 as latexx^{2} yy=c_{2}/latexExperts are tested by Chegg as specialists in their subject area We review their content and use your feedback to keep the quality high Previous question Next question
Y^2 xy(1x) x^3 Please answer it quickly 2 See answers Advertisement AdvertisementAnswer to Solved dy/dx = xy^2/(1 x^2) 2) (y^2 x^2) dy (2xy x) This problem has been solved!This website uses cookies to ensure you get the best experience
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X 2 x y y 2 = 6 x 2 x y y 2 = 6 x2 xyy2 = 6 x 2 x y y 2 = 6 Since the equation is not identical to the original equation, it is not symmetric to the yaxis Not symmetric to the yaxis Check if the graph is symmetric about the origin by plugging in −x x for x x and −y y for y yView 03docx from RESEARCH M SPT730 at Virtual University of Pakistan 0305 Leandro Tavares Part A 1(xy)^2x^22xyy^2 2(xy)(xy) x^2 2xy y^2 3x^2xyxyy^2x^22xyy^2 4Adding equation (1) and (2) x y = 2xy x y = 6xy 2x = 8xy 2 = 8y y = 1 4 From (1) x 1 4 = 2 × ( 1 4) 1 2 x = 1 4
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If xyyzzx=0 then find the value of 1/x^2yz 1/y^2zx 1/z^2xy Dear student xy yz zx = 0 so yz = x(yz) x^2yz = x(xyz) so 1/(x^2yz) = 1/(x(xyz) Thank you for registering One of our academic counsellors will contact you within 1 working dayAnswer By definition, Legendre polynomials P_n(x) (n is the degree) form a family of orthogonal* polynomials that all verify the differential equations (1x^2)P''_n(x)2xP'_n(x)n(n1)P_n(x)=0 So, yes, y=P_2(x)=(3x^2–1)/2 is a particular solution of your differential equation But what's the geSolution for 1) xy' (1 – x)y = xe* 2) (x2 – 1)y' 2y = (x 1)² %3D
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Proof Let G be a group, and x, y ∈ G Then, we have x y ( x y) − 1 = e, where e is the identity element of G Multiplying on the left by x − 1, we get y ( x y) − 1 = x − 1 Multiplying once again on the left by y − 1 gives ( x y) − 1 = y − 1 x − 1, which was what was to be shownThe solution of the differential equation (dy/dx)=(x2y21/2xy) satisfying y( 1) = 1, is (A) a hyperbola (B) a circle y2 = x (1 x ) 10 (D) (x Now dividing both the sides we get, xy as cancelled from both the numerator and denominator (x y)(x −y) x y Cancelling x y from both numerator and denominator we get, Ans = (x −y) Answer link Mark D Looking at the numerator x y − y x if we put them over a common denominator
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The general solution of the differential equation ` (dy)/(dx) = ((1y^(2)))/(xy(1x^(2)))` isX^{2}xy(1y)y^{3} Snapsolve SnapsolveMath check please 1 Solve the system of equations y=2x^23 y=3x1 a)no solution b)(1/2,5),(2,5/2) c)(1/2,5/2),(2,5)**** d)(1/2,5/2),(2,5) 2
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The inequalities are, 2x #x2212;y #x003E;Answer (1 of 5) Got some news for you the curve x^2 xy y^2 = 15 doesn't even pass through the point (2, 5) 2^2 (2)(5) 5^2 = 4 10 25 = 31;See the answer See the answer See the answer done loading
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