Simple and best practice solution for (2xy)dy(x^2y^21)dx=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 Equation SOLVE Solution for (2xy)dy(x^2y^21)dx=0Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, historyHere is the equation $$(2xyx^2x^4)\,dx(1x^2)\,dy=0$$ It is not exact since partial derivatives are not equal Any help would be appreciated ordinarydifferentialequations Share Cite Follow edited Feb 13 '17 at 2316 Michael Hardy 251k 28 28 gold badges 251 251 silver badges 533 533 bronze badges asked Feb 13 '17 at 2258 krszyoscezio krszyoscezio 10 10 bronze badges
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Solution for (A) (xy)* dx(2xy x² –1)dy = 0, y(1) =1 Q Perform the indicated computations Write the answer in scientific notation · What is a solution to the differential equation #dy/dx=x^2(1y)# with y=3 when x=0?Simple and best practice solution for 2xy(4y^2)dx(y1)(x^22)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
This is almost certainly a problem in a section of the text on "exact" equations, ie ones of the form dF= F,x dx F,y dy =0 The solution is F(x,y)=constant You need to look for an integrating factor A such that F,x = A 2 x y F,y = A (x^2 y^22305 · Solve dy/dx=2xy/(x^2y^2) Question Solve dy/dx=2xy/(x^2y^2) check_circle Expert Answer Want to see the stepbystep answer?Solve first grade ecuation with bernoulli dy/dx = (y^22xy)/x^2general ecuation and particular when y(1)=1 1 Educator answer eNotescom will help you with any book or any question
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?Note that mathy'=\dfrac{x^2y^21}{x(2yx)}/math The substitution mathu=(2yx)^2/math transforms this to a linear DE math\begin{align*}u' &= 2(2yx)(2y'1Solve the differential equation dy/dx = (x^2 y^2)/2xy asked Jan 1, in Differential equations by AmanYadav (556k points) differential equations;
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· 1/y^2 * dy = 2x * dx Integrate both sides 1/y = x^2 C Make y the subject, invert both sides y = 1/(x^2 C) Now sub in the initial condition to find C 1/4 = 1/(1 C) 4 = 1 C C = 3 y = 1/(x^2 3) 0 0 musa a Lv 5 7 years ago dy/dx = 2xy^2 dy/y^2 = 2x dxy^1 = x^2 C 1/y = x^2 C1/025 = (1)^2 C C = 5 1/y = x^2 5 0 0 grunfeld Lv 7 7 years ago (dy / dx ) 0yDivide 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 square2613 · dy/dx = (x*y)/(1x²) y ∫dy/y=∫ (x)/(1x²) dx rechts 2/2 ergänzen, dann hat man die Form f'/f ln(y)=1/2 ln(1x^2) 1/2c y = √(1x^2) * d Wobei das d eine neue Konstante ist Wir haben ja den Logarithmus ersetzt durch eFunktion anwenden Da muss dann auch das 1/2*c mit Das kann mittels der Potenzgesetze letztlich zur Konstanten d umgewandelt werden Bei
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2xy dx (x^2 1) dy =0 => 2xydx = (x^2–1)dy => 2x/(x^2–1)dx = dy/y after integrating both sides => ln(x^2–1) = ln(y) constant => ln(x^2–1)ln(ySolve the differential equation (x^2 1) dy/dx 2xy = 1/(x^2 1) asked Sep 21, in Differential Equations by Chandan01 (512k points) differential equations;Solved Solve 2xy dx (x^2 1) dy = 0 By signing up, you'll get thousands of stepbystep solutions to your homework questions You can also ask
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0 votes 1 answer Find the equation of a curve passing through origin and satisfying the differential equation (1 x^2) dy/dx 2xy = 4x^2 asked Sep 21, in Differential EquationsSee Answer Check out a sample Q&A here Want to see this answer and more?Dy/dx = y 2 /(xy x 2) (1) Put y = vx, dy/dx = v1 x(dv/dx) From (1), v x(dv/dx) = v 2 x 2 /(vvx x 2) where c is constant of integration ← Prev Question Next Question → Related questions 0 votes 1 answer Find the general solution of y^2dx (x^2 – xy y^2) dy = 0 asked Sep 21, in Differential Equations by Chandan01 (512k points) differential equations;
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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 itSimple and best practice solution for (2xy)dx(x^21)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 itConsider the equation (y^2 2xy)dx x^2 dy = 0 Show the equation is not exact, but becomes exact if we multiply it by y^2 Use the resulting exact equation to find the general (implicit) solution of the original equation
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Calculus Applications of Definite Integrals Solving Separable Differential Equations 1 Answer Eddie Aug 22, 16 #y = 4e^ (x^3/3) 1# Explanation #dy/dx=x^2(1y)# This is separable #1/(1y) dy/dx=x^2# #int \ 1/(1y) dy/dx \ dx = int \ x^2 \ dx# #int \ 1/(1y) \ dy = int \ x^2 \ dx# #ln (1y) = xFind dy/dx y^2=1/(1x^2) Differentiate both sides of the equation Differentiate the left side of the equation Tap for more steps Differentiate using the chain rule, which states that is where and Tap for more steps To apply the Chain Rule, set as Differentiate using the Power Rule which states that is where Replace all occurrences of with Rewrite as Differentiate the rightWelcome 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
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Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and moreI assume that by math(2xyy)dx(x^2x)dy/math, you mean math\int (2xyy)dx\int (x^2x)dy/math To solve this equation, it should be considered a single08 · (y^22xy)dx (x^2) dy = 0 Integrating factor y^2 solution xx^2/y = C Finding the integrating factorGiving these things some names names M(x,y) = y^22xy N(x,y) = x^2 M(x,y) dx N(x,y) = 0 We're looking to make this an exact equation, because if we do, it can be solved rather systematically In order to be exact, by claurait's theoem (think that's
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Thanks ordinarydifferentialequations Share Cite Follow edited Feb 17 '13 at 2240 Michael Hardy 249k 28 28 goldI'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 $$\mu_1(x)=x^{2} \ \ , \ \ \mu_2(y/x)=\left(1\frac{y^2}{x^2}\right)^{2}$$ Then, I've used $\muSo we have an equation of the form, mathM(x,y)\text{d}xN(x,y)\text{d}y=0/math Where mathM=(xy)^2,N=(2xyx^21)/math Thus this is an exact differential
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0 votes 1 answer Find the particular solution of the differential equation dy/dx = 1 x y xy given that y = 0 when x = 1 asked Mar 13 in Differential Equations by Yaad (351k points) differential equations;Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and moreFind the particular solution of the differential equation (1 y^2)(1 log x) dx 2xy dy = 0 given that y = 0 when x = 1 asked Mar 17 in Differential Equations by Takshii (
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Exactdifferentialequationcalculator 2xy9x^2(2yx^21)\frac{dy}{dx}=0 en Related Symbolab blog posts Advanced Math Solutions – Ordinary Differential Equations Calculator, Separable ODE Last post, we talked about linear firstThe equation is also not separable The equation is also not homogenous, I don't think So what do I do? · 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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