[最も欲しかった] y(4x y)dx-2(x^2-y)dy=0 317194-Y 4x y dx-2 x 2-y dy 0

Solve Y 2 2x 2y Dx 2x 3 Xy Dy 0 Mathematics Stack Exchange

4x 2 y 3 4xy 2 y = 0To find dy/dx we need to differentiate all the terms in the equation As you may notice the x's and y's are mixed together so we will have to use implicit differentiation Since we are differentiating in terms of x, all of the x terms can be differentiated as usual, whilst the y terms will be followed by a dy/dx The first two terms are more straight forward to dealTo ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW `(3x yy^2)dx(x^2x y)dy=0`

Y 4x y dx-2 x 2-y dy 0

Y 4x y dx-2 x 2-y dy 0-Stack Exchange network consists of 178 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 ExchangeAn equation composed of the product of two x terms can be differentiated with respect to x using the product rule to get dy/dx The equation is expressed in the form y=uv, where u=x^2 and v= (14x)^05By the product rule dy/dx=udv/dxvdu/dxYou get dv/dx and du/dx by differentiating the u and v terms with respect to x, hencedu/dx=2x and

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6x^2*y^2*dx=4x^3*y*dy 1/x dx=2/3*1/y*dy ln abs(x)C=2/3*ln abs(y) 1/y^(2/3)=Cx y^(2/3)=1/(Cx) xy^(2/3)=1/C C=1/(xy^(2/3)) But the answer in the book says 2x^3*y^2=C Is the book's answer correct and my answer is wrong or vice versa?Here we will look at solving a special class of Differential Equations called First Order Linear Differential Equations First Order They are "First Order" when there is only dy dx, not d 2 y dx 2 or d 3 y dx 3 etc Linear A first order differential equation is linear when it can be made to look like this dy dx P(x)y = Q(x) Where P(x) and Q(x) are functions of x To solve it there is aSimple and best practice solution for 2(2x^2y^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 Equation SOLVE Solution for 2(2x^2y^2)dx(xy)dy=0 equation

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 queriesAnswer (1 of 3) (2x y) dx (4x y 6) dy = 0 Find substitutions u, t to remove the (6) y = u a, x = t b \frac{du}{dt} = \frac{d(u a)}{d(t b)} = \frac{dy}{dx} Match the constants a, b by making a system of linear equations Each equation is an expression in parentheses set equa(b) 2 x y dx ( y 2 x 2) dy = 0 Here, M = 2 x y, M y = 2x, N = y 2 x 2, and N x = 2 xNow, ( N x M y) / M = ( 2 x 2 x ) / ( 2 x y) = 2 / yThus, μ = exp ( ∫ 2 dy / y ) = y2 is an integrating factor The transformed equation is ( 2 x / y ) dx ( 1 x 2 y2) dy = 0 Let m = 2 x / y, and n = 1 x 2 y2Then, m y = 2 x y2 = n x, and the new differential equation is exact

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21(xy2x)dx(yx2y)dy=0 One solution was found d = 0 Step by step solution Step 1 Step 2 Pulling out like terms 21 Pull out like factors y Is the solution of the math problem right? This is a first order separable differential equation dy dx = y2 −4 dy y2 −4 = dx ∫ dy y2 − 4 = ∫dx Perform partial fraction 1 y2 −4 = A y 2 B y − 2 = A(y − 2) B(y 2) (y2 − 4) Compare the numerators 1 = A(y −2) B(y 2)

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