[10000ダウンロード済み√] 2(x 2y)dx (y-x)dy=0 746460-(1-x^2y)dx+x^2(y-x)dy=0

To ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW The solution of the differential equation `y(xy 2x^2y^2) dx x(xyx^2y^2)dy$$(x^2y^2)dx (3xy)dy = 0$$ The differential equation does not seem to be separable, and I'm having a tough time to put it in the general form of $$\frac{dy}{dx} p(x)y = f(x)$$ Would someone mind offering me some assistance?The 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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(1-x^2y)dx+x^2(y-x)dy=0

(1-x^2y)dx+x^2(y-x)dy=0- 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 −Simple and best practice solution for y(x^2xy2y^2)dxx(3y^2xyx^2)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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 Welcome 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 Exam, ICSE Board Exam, State Board Exam, JEE (MainsAdvance) and NEET can ask questions from any subject and get quick answers byAn 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 canSee the answer Solve the given homogeneous first order ODE Show transcribed image text Expert Answer Previous question Next question Transcribed Image Text from this Question 2(x 2y)dx (yx)dy = 0 Get more help from Chegg

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\2xy dx \left( x^2 2 y^2 \right) dy = 0\ \ \Rightarrow \frac{dy}{dx} = \frac{2xy}{x^2 2 y^2}\ This is a homogeneous differential equationAnswer and Explanation 1 Become a Studycom member to unlock this answer!

Create your account View this answer (x23y2) dx−2xy dy = 0 ( x 2 3 y 2) d x − 2 x y d y = 0 ⇒ dy dx 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 A solution of the differential equation (dy/dx)2 – x(dy/dx) y = 0 is (A) y = 2 (B) y = 2x2 – 4 y = 2x (D) y = 2x – 4 Welcome to Sarthaks eConnect A unique platform where students can interact with teachers/experts/students to get solutions to their queries

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Solving The Homogeneous Differential Equation Dy Dx Y X Y X Youtube

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'1Wolfram Alpha gives which looks a lot more likely to me to be correct than the other answer provided so far (which includes a pretty elementary error in theSimple 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 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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Because it doesn't follow the necessary condition of exact differential equation But if we divide the equation by $(x^2y^2)$ it follows the necessary condition Can you please explain the reason behind this?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=0See Answer Check out a sample Q&A here Want to see this answer and more?

Exact Differential Equation 2xy Sec 2 X Dx X 2 2y Dy 0 Youtube

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Question 2(x 2y)dx (yx)dy = 0 This problem has been solved!4dy^ {2}x^ {2}\left (3d\right)x4dy=0 4 d y 2 x 2 ( − 3 d) x 4 d y = 0 This equation is in standard form ax^ {2}bxc=0 Substitute 4y^ {2}d for a, 3d for b, and 4dy for c in the quadratic formula, \frac {b±\sqrt {b^ {2}4ac}} {2a} This equation is in standard form a x 2 b x c = 0Click here👆to get an answer to your question ️ Solve (x y 1)dx (2x 2y 3)dy = 0

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How To Solve This Differential Equation Y Y 2 2x 2 Dx X 2y 2 X 2 Dy 0 Quora

Click here👆to get an answer to your question ️ Solve x^2y dx (x^3 y^3) dy = 0To ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW Solve `(xy^3y)dx2(x^2y^2xy^4)dy=0`Explicit substitution of y=x into the given DEqn shows it is a soln Notice (x^23xyy^2)dxx^2dy = x^2dxx^2(dx)=0 Thus it is a solution

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Bernoulli S Equation 6y 2 Dx X 2x 3 Y Dy 0 Chegg Com

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Incoming Term: 2(x+2y)dx+(y-x)dy=0, (x^2-2y)dx+(y^2-2x)dy=0, (x-2y)dx+(2x+y)dy=0, (x-2y)dx+(2x+y)dy=0 homogeneous, (x+2y)dx+(2x+y)dy=0 exact equation, (xy^2+x)dx+(y-x^2y)dy=0, (1-x^2y)dx+x^2(y-x)dy=0, (x^2-y)dx+(y^2-x)dy=0, (2x-2y)dx+(y-x+1)dy=0, (ii) (x+2y)dx-(2x-y)dy=0,