Find dy/dx x^2yxy^2=6 Differentiate both sides of the equation Differentiate the left side of the equation Tap for more steps By the Sum Rule, the derivative of with respect to is Evaluate Tap for more steps Differentiate using the Product Rule which states that is where andDifferential equation Solve $(2xy^4e^y 2xy^3 y)dx (x^2y^4e^y x^2y^2 3x) dy = 0$ written 51 years ago by aksh_31 ♦ 22k • modified 51 years ago Mumbai University > First Year Engineering > sem 2 > Applied Maths 2 Marks 6 Year 13 applied mathematics ADD COMMENT FOLLOW SHAREFactor out the Greatest Common Factor (GCF), 'dx 2 ' dx 2 (y 2 2x) = 0 Subproblem 1 Set the factor 'dx 2 ' equal to zero and attempt to solve Simplifying dx 2 = 0 Solving dx 2 = 0 Move all terms containing d to the left, all other terms to the right
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Y xy 2x 2y 2 dx x xyx- 2y 2 dy 0-Factor out the Greatest Common Factor (GCF), 'dx' dx(3 4y 2xy 2 2x 3) = 0 Subproblem 1 Set the factor 'dx' equal to zero and attempt to solve Simplifying dx = 0 Solving dx = 0 Move all terms containing d to the left, all other terms to the right Simplifying dx = 0 The solution to this equation could not be determinedOther Math questions and answers;
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Factor out the Greatest Common Factor (GCF), 'dxy 2 ' dxy 2 (y 2x) = 0 Subproblem 1 Set the factor 'dxy 2 ' equal to zero and attempt to solve Simplifying dxy 2 = 0 Solving dxy 2 = 0 Move all terms containing d to the left, all other terms to the right So, the given differential equation is exact On integrating M wrt x, treating y as a constant, On integrating N wrt y, treating x as a constant, (omitting 2xy2 2x2y which already occur in ∫M dx) Therefore, the solution of the given equation is x3/3 2x2y 2xy2 y3/3 = λ ⇒ x3 y3 6xy (x y) = 3λFor the differential equation `(x^2y^2)dx2xy dy=0`, which of the following are true (A) solution is `x^2y^2=cx` (B) `x^2y^2=cx` `x^2y^2=xc` (D) `y
Question (2xy^22y)dx(2x^2y2x)dy=0 This problem has been solved!Simple and best practice solution for (xy2xy2)dx (x^22x)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 itMathy^2 x^2 (\dfrac{dy}{dx}) = xy(\dfrac{dy}{dx})/math math\dfrac{y}{x} \dfrac{x}{y}(\dfrac{dy}{dx}) = (\dfrac{dy}{dx})/math math\dfrac{y}{x} (\dfrac
Use the distributive property to multiply x 2 y 3 d x d by y To find the opposite of x^ {2}dy^ {4}xdy, find the opposite of each term To find the opposite of x 2 d y 4 x d y, find the opposite of each term Combine ydx and xdy to get 2ydx Combine − y d x and − x d y to get − 2 y d xThe issue is that you integrated y with respect to x, and concluded that it was equal to y This is only viable if y = aex for some constant a, which we have no reason to suspect Solve y ^2x (\frac {dy} {dx})^2 = 1 using proposed change of variables Solve y2 −x(dxdy )2 = 1 using proposed change of variablesRewrite 2xy dxx2 dy−1 dy = 0 2 x y d x x 2 d y − 1 d y = 0 Change the sides $$2 xy \ dx x^2 \ dy = 1 \ See full answer below
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Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history$(xy^22x^2y^3)dx(x^2yx^3y^2)dy=0$ $xy(y2xy^2)dxxy(xx^2y)dy=0$ $f(x,y)=xy(y2xy^2)$ $g(x,y)=xy(xx^2y)$ Here, xy is a common term for both, f(x,y) and g(x,y)It is homogeneous equation
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Solve the differential equation `xy dy/dx = x^2 2y^2` by integrating the side attached to dx with respect to x x^5/5 x^2y^2xy^4k(y) this k represent function of y as you differentiated x you removed y as it wads constant with respect to x this is F(x,y) so you have to diferentiate it with respect to y in order to find k(y) 2xy^24xy^3k'(y) you equal it with the side attached to dy in theSolve the following differential equations 1 (xy 1)dx x(x 4y 2)dy = 0 6 (x y 1)dx (2x 2y 1)dy = 0 2
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I'm at the beggining of a differential equations course, and I'm stuck solving this equation $$(x^2y^2)dx2xy\ dy=0$$ I'm asked to solve it using 2 different methods I proved I can find integrating factors of type $\mu_1(x)$ and $\mu_2(y/x)$If I'm not wrong, these two integrating factors are $$\mu_1(x)=x^{2} \ \ , \ \ \mu_2(y/x)=\left(1\frac{y^2}{x^2}\right)^{Steps for Solving Linear Equation ( x y ^ { 2 } x ) d x ( y x ^ { 2 } y ) d y = 0 ( x y 2 x) d x ( y x 2 y) d y = 0 Use the distributive property to multiply xy^ {2}x by d Use the distributive property to multiply x y 2 x by d \left (xy^ {2}dxd\right)x\left (yx^ {2}y\right)dy=0Click here👆to get an answer to your question ️ Solution of the differential equation y(xy 2x^2y^2)dx x(xy x^2y^2)dy = 0 is given by plogx Qlogy 1/xy = C
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Solve $(2x^2 y^2)\,dx xy \, dy = 0$ 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? The solution of the differential equation (3xy y^2)dx (x^2 xy)dy = 0 is (A) x^2(2xy y^2) = c^2 asked in Differential equations by AmanYadav ( 556k points) differential equations1 Answer1 Active Oldest Votes 1 The solution (below) is obtained on the form of an implicit equation Then, it is not very difficult to explicit x ( y) But it seems not possible to get a closed form for the inverse function y ( x) with the available standard functions Share answered Dec 16 '16 at 846 JJacquelin
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Resolver para x (x^2y^2)dx(x^2xy)dy=0 Factorizar la ecuación Toca para ver más pasos Factoriza a partir de Toca para ver más pasos Factoriza a partir de Dividir cada término por y simplificar Toca para ver más pasos Dividir cada término de por Reduce la expresión anulando los factores comunes 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 Visit Stack ExchangeHow do I solve this differential equation y (y^22x^2) dxx (2y^2x^2) dy=0
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The equation is also not separable The equation is also not homogenous, I don't think So what do I do?See the answer (2xy^22y)dx(2x^2y2x)dy=0 Expert Answer 100% (2 ratings) Previous question Next questionDifferential equation mathM(x,y) \, dx N(x,y) \, dy = 0 \tag*{}/math is nonexact when math\dfrac{\partial M}{\partial y} \ne \dfrac{\partial N}{\partial x
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The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction y^ {2}2xyx^ {2}=0 y 2 2 x y x 2 = 0 This equation is in standard form ax^ {2}bxc=0 Substitute 1 for a, 2x for b, and x^ {2} for c in the quadratic formula,Simple and best practice solution for y^2dx(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 itAnswer to Solve the homogeneous equations x^2 dy ( y^2 xy) dx =0 By signing up, you'll get thousands of stepbystep solutions to your
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Math advanced math advanced math questions and answers Solve The DE (x^2yxyy)dx (x^2y2x^2)dy=0 The Solution Should Be Xlnxx^1y2lny=c` (x^(2)y^(2)) dx 2xy dy = 0`The ODE is homogeneous ODE of order one This is because the coefficients of dx and dy are both homogeneous two variables functions of the same order I suggest you write the ODE as y′ = 32t2t2−t−2 = f (t), (x = 0,t = y/x) Find the solution of (xy^22x^2y^3)dx (x^2yx^3y^2)dy=0
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Explanation x dy dx = 2x2y y, separating the variables 1 y dy dx = 2x2 1 x so dy y = 2x2dx dx x, integrating ∫ 1 y dy = ∫2x2dx ∫ 1 x dx, we have lny = x2 lnx C Which gives ln y x = x2 C eln y x = ex2c theory of logs ie, y x = ex2c and so,yHow do I solve (X*22xyy*2) dx (y*22xyx*2) dy=0? y^2 = x^2(2lnx c) We can rewrite this Ordinary Differential Equation in differential form (x^2 y^2) \ dx xy \ dy = 0 A as follows \ \ \ \ dy/dx = (x^2 y^2)/(xy) dy/dx = x/y y/x B Leading to a suggestion of a substitution of the form u = y/x iff y = ux And differentiating wrt x whilst applying the product rule dy/dx = u x(du)/dx Substituting into the
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Solution of the differential equation y(xy 2x2y2)dx x(xy x2y2)dy = 0 is given by (A) 2 logxlogy(1/xy)=C (B) 2 logylogx(1/xy)=C 2Equations Tiger Algebra gives you not only the answers, but also the complete step by step method for solving your equations (2x^2y)dx(x^2yx)dy=0 so that you understand betterIn order to complete the square, the equation must first be in the form x^ {2}bx=c Divide y, the coefficient of the x term, by 2 to get \frac {y} {2} Then add the square of \frac {y} {2} to both sides of the equation This step makes the left hand side of the equation a perfect square Square \frac {y
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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 −Solution for (Xy^2x)dx (yx^2y)dy=0 equation Simplifying (X 1y 2 x) * dx (y 1x 2 y) * dy = 0 Reorder the terms for easier multiplication dx (X 1xy 2) (y 1x 2 y) * dy = 0 (X * dx 1xy 2 * dx) (y 1x 2 y) * dy = 0 (dxX 1dx 2 y 2) (y 1x 2 y) * dy = 0 Reorder the terms dxX 1dx 2 y 2 (1x 2 y y) * dy = 0 Reorder the terms for easier multiplication dxX 1dx 2 y 2 dy (1x 2 y y) = 0 dxX 1dx 2 y 2 (1x 2 y * dy y * dy) = 0 dxX 1dx 2 yThe differential equation xydx (x^2 2y^2)dy =0 can be rewritten as the following Bernulli equation, inthe unknown x(y) dx/dy x/y = 2y/x The equation can be turned to normal form taking x= V(y)^1/2 Obtain V' 2V/y = 4y The integrating factor is e^Integral of (–2dy/y) = e^(lny^2) = 1/y^2 Then, the solution for V(y) is
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(xy^22x^2y^3)dx(x^2yx^3y^2)dy=0 (y2xy^2)dx=(x^2yx)dy y(12xy)dx=x(xy1)dy Substitute t=yx and t'=yxy' 3\int \frac {dx} x=\int \frac {(t1)dt} {t^2} This is an example of an exact DE — that is, it has the form F(x,y)dxG(x,y)dy=0, where \frac{\partial F}{\partial y}=\frac{\partial G}{\partial x} The solution to such a DE is givenSolution for (x^2y^2)dx (x^2xy)dy=0 equation Simplifying (x 2 y 2) * dx (x 2 1xy) * dy = 0 Reorder the terms for easier multiplication dx (x 2 y 2) (x 2 1xy) * dy = 0 (x 2 * dx y 2 * dx) (x 2 1xy) * dy = 0 Reorder the terms (dxy 2 dx 3) (x 2 1xy) * dy = 0 (dxy 2 dx 3) (x 2 1xy) * dy = 0 Reorder the terms dxy 2 dx 3 (1xy x 2) * dy = 0 Reorder the terms for easier multiplication dxy 2 dx 3 dy (1xy x 2) = 0 dxy 2 dx
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