Solve the Differential Equation dy/dx=xy^2 In this tutorial we shall evaluate the simple differential equation of the form d y d x = x y 2 by using the method of separating the variables The differential equation of the form is given as d y d x = x y 2 Separating the variables, the given differential equation can be written asSimple and best practice solution for (x^2y^25)dx(yxy)dy=0 equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework If it's not what You are looking for type in the equation solver your own equation and let us solve it Solve$$(x^22xyy^2)dy= (x^22xyy^2) dx $$ I've tried many methods but still unable to solve that problem Stack Exchange Network Stack Exchange network consists of 177 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers
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X 2 y 2 dx x 2-xy dy 0 c1 x y 2 xe y/x-Int (x^2 y^2 x y^3) dx dy, x=2 to 2, y=2 to 2 Extended Keyboard;How do I solve (X*22xyy*2) dx (y*22xyx*2) dy=0?
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1125am #1 Sydney Sales DE 2xy dx (y^2 x^2) dy = 0 2xydx ( y^2 x^2 ) dy = 0\\left( x 2y \right)dx \left( 2x y \right) dy = 0\ \ \Rightarrow \frac{dy}{dx} = \frac{x 2y}{2x y}\ This is a homogeneous differential equation y = (xlnx)/(1lnx) We have dy/dx = (x^2y^2xy)/x^2 with y(1)=0 Which is a First Order Nonlinear Ordinary Differential Equation Let us attempt a substitution of the form y = vx Differentiating wrt x and applying the product rule, we get dy/dx = v x(dv)/dx Substituting into the initial ODE we get v x(dv)/dx = (x^2(vx)^2x(vx))/x^2 Then assuming that x ne 0 this
The general solution of y^2dx (x^2 – xy y^2)dy = 0 is (A) tan^1(x/y) logy c = 0 asked in Differential equations by Sarita01 ( 535k points) differential equationsView this answer We are given (x2−y) dx(y2 −x) dy = 0 ( x 2 − y) d x ( y 2 − x) d y = 0 Rewrite x2 dx−y dxy2 dy−x dy = 0 x 2 d x − y d x y 2 d y − x d y = 0 RearrangeQuestion Verify That The Indicated Function Is A Solution Of The Givendifferential Equation Where Appropiate, C1 AndC2 Denote Constants27 (x2 Y2)dx (x2 Xy)dy= 0;
Solve the differential equation Best Answer This is the best answer based on feedback and ratings 100% (2 ratings) (xyy^2)dxx^2dy=0 y (xy) = x^2 dy / dx let y = View the full answer Previous question Next questionHomogeneous Differential Equation (y^2 yx)dx x^2dy = 0If you enjoyed this video please consider liking, sharing, and subscribingYou can also help suppor 113k views asked in Class XII Maths by nikita74 (1,017 points) Find the general solution of y 2 dx (x 2 xyy 2 )dy = 0 differential equations
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C1(x Y)2 =xe(y/x)Simple and best practice solution for (Xy^2x)dx(yx^2y)dy=0 equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework91 A product of several terms equals zero When a product of two or more terms equals zero, then at least one of the terms must be zero We shall now solve each term = 0 separately In other words, we are going to solve as many equations as there are terms in the product Any solution of term = 0 solves product = 0 as well
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It is homogeneous equationSolve $(2x^2 y^2)\,dx xy \, dy = 0$ Attempted The equation is not exact because $ M_y \ne N_x $ for $ M = 2x^2 y^2 $ and $ N = xy$ Or is it exact? From x dy dx y = x2y2, one can divide both sides by x so that it fits the Bernoulli form dy dx y x = xy2 Divide both sides by y2 y−2 dy dx 1 xy = x Then, define a function v = y1−2 = y−1 Differentiate both sides with respect to x to get dv dx = − dy dx y−2, or dy dx = − y2 dv dx
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Find dy/dx x^2xyy^2=1 Differentiate both sides of the equation Differentiate the left side of the equation Tap for more steps Differentiate Tap for more steps By the Sum Rule, the derivative of with respect to is Differentiate using the Power Rule which states that is whereThe differential equation is homogeneous Let mathy = ux/math Then mathdy = x \, du u \, dx/math and the equation becomes math(x^2 u^2 x^2) \, dx 2uxAn ordinary differential equation of first order and first degree can be written as dy dx = f(x,y) d y d x = f ( x, y) , where f(x,y) f ( x, y) is a function of two variables x,y x, y Which can
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Simple and best practice solution for (x^2xyy^2)dx(xy)dy=0 equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework If it's not what You are looking for type in the equation solver your own equation and let us solve it(c) given differential equation is (x x y^2) d x e^x^2 y d y = 0 x(1 y^2) d x = = e^x^2 y d y on spearing the variable we get y/ view the full answer Previous question Next question Transcribed Image Text from this QuestionRewrite the given equation as (y*dx x*dy) * (x^2 y^2) (y*dx x*dy) = 0 Eq(1) Divide throughout by (x^2 y^2) (y*dx x*dy) (y*dx x*dy) / (x^2 y^2
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1 Answer1 Active Oldest Votes 1 ( x y) d x − ( x 2 y 2) d y = 0 Let F ( x, y) = x y and let G ( x, y) = x 2 y 2 Now consider a function u ( x, y) = 0 Then ∂ u ∂ x d x ∂ u ∂ y d yFind dy/dx x^2xyy^2=4 Differentiate both sides of the equation Differentiate the left side of the equation Tap for more steps Differentiate Tap for more steps By the Sum Rule, the derivative of with respect to is Differentiate using the Power Rule which states that is whereCompute 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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The GS is y^2 = (A x^4)/(2x^2) Or, alternatively y = sqrt(A x^4)/(sqrt(2)x) We have (x^2 y^2) \\ dx xy \\ dy = 0 Which we can write in standard form as dy/dx = (x^2 y^2)/(xy) 1 Which is a nonseparable First Order Ordinary Differential Equation A suggestive substitution would be to perform a substitution of the form y = xv => dy/dx = v xv' \\ \\ \\ where v=v(xThe equation is also not separable The equation is also not homogenous, I don't think So what do I do? ∫∫f(x,y)dx dy =0=∫∫f(x,y)dy dx ist eben falsch Das habe ich auch bereits oben in meiner Antwort geschrieben Man kann auch ohne Polarkoordinaten rechnen und sieht, das alleine das erste Integral ∫(oo bis oo) x/(x^2 y^2)dx nicht konvergiert Lediglich der CauchyHauptwert ist 0 Weißt du, wie man uneigentliche Integrale berechnet?
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Show that the differential equation (xy)dy/dx=x2y,is homogeneous and solve it asked in Mathematics by Nisa ( 597k points) differential equationsSimple and best practice solution for (x^2y^2)dx(x^2xy)dy=0 equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework If it's not what You are looking for type in the equation solver your own equation and let us solve it Solve (y√(x^2y^2))dxxdy=0 Latest Problem Solving in Differential Equations More Questions in Differential Equations Online Questions and Answers in Differential Equations
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D x 2 ( x y) = 0 d x 2 ( x y) = 0 dx2(x y) = 0 d x 2 ( x y) = 0 Si cualquier factor individual en el lado izquierdo de la ecuación es igual a 0 0, la expresión completa será igual a 0 0 dx2 = 0 d x 2 = 0 xy = 0 x y = 0 Iguala el primer factor a 0 0 y resuelve Toca para ver más pasosAnswer to Solve the differential equation y(xy1) \, dx x(1xyx^2y^2) \, dy = 0 By signing up, you'll get thousands of stepbystep solutions for Teachers for Schools for Working ScholarsSee the answer Show transcribed image text Expert Answer Previous question Next question Transcribed Image Text from this Question Solve dy/dx = x^2 xy y^2/x^2 Get more help from Chegg
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Solve the differential equation √(1 x^2 y^2 x^2y^2) xy(dy/dx) = 0 asked Mar 13 in Differential Equations by Yaad ( 352k points) differential equationsThe equation dx/y^2 = dy/x^2 = dz/(x^2)(z^2)y is the auxiliary system to obtain the general integral of the PDE (y^2)Z_x (x^2)Z_y = (x^2)(Z^2)y From the first two equations obtain dy/dx = x^2/y^2 which leads to the first integral Z1 = y^3 x^Simple and best practice solution for (4xxy^2)dx(yx^2y)dy=0 equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework If it's not what You are looking for type in the equation solver your own equation and let us solve it
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Solve Dy/dx = X^2 Xy Y^2/x^2 Question Solve Dy/dx = X^2 Xy Y^2/x^2 This problem has been solved! But if I expand the bracket $(xy)^2$ before integrating I will get $$\varnothing_1=\int Mdx=\int (xy)^2dx=\int (x^22xyy^2)dx=\frac{x^3}{3}xy^2x^2y$$ Wich will lead to the solution $$\varnothing=\varnothing_1\varnothing_2=\frac{x^3}{3}xy^2x^2yy=Constant$$ What is the wrong step ? We can rearrange this Differential Equation as follows dy dx = − x2 y2 x2 − xy = − ( 1 x2)(x2 y2) ( 1 x2)(x2 −xy) = − 1 ( y x)2 1 − y x So Let us try a substitution, Let v = y x ⇒ y = vx Then dy dx = v x dv dx And substituting into the above DE, to eliminate y
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Find dy/dx x^2y^2=2xy Differentiate both sides of the equation Differentiate the left side of the equation Tap for more steps Differentiate Tap for more steps By the Sum Rule, the derivative of with respect to is Differentiate using the Power Rule which states that is whereResolver para x (x^2y^2)dx (x^22xy)dy=0 (x2 − y2) dx (x2 − 2xy)dy = 0 ( x 2 y 2) d x ( x 2 2 x y) d y = 0 Factorizar la ecuación Toca para ver más pasos Factoriza d d a partir de d x ( x 2 − y 2) d y ( x 2 − 2 x y) d x ( x 2 y 2) d y ( x 2 2 x y) Toca para ver más pasos 1 Multiple by xayb so Mdx Ndy = 0 with M = xayb 1 xa 1yb 2, N = xa 1yb xa 2yb 1 xa 3yb 2 We choose a, b to achieve 0 = ∂yM − ∂xN = (b 1)xayb (b 2)xa 1yb 1 − (a 1)xayb − (a 2)xa 1yb 1 − (a 3)xa 2yb 2 = xayb((b − a)(1 xy) − (a 3)(xy)2) a = b = − 3 So 0 = Mdx Ndy = (x −
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Dy/dx= (x^2y^2)/(3xy) The answer is supposed to be (x^2 4y^2)^3(x^2 )= CHere is what I have done x/(3y) y/(3x) = (1/3)(x/y y/x) = (1/3)(1/(y/x) y/x)Help is appreciated EditEcuacion diferencial de primer orden
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` (x^(2)y^(2)) dx 2xy dy = 0`View this answer Given 2xy dx(x2−1) dy = 0 2 x y d x ( x 2 − 1) d y = 0 Rewrite 2xy dxx2 dy−1 dy = 0 2 x y d x x 2 d y − 1 d y = 0 Change the sides $$2 xy \ dx x^2 \ dy = 1
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