Solved Separable Differential Equations 1 2 2xy 6x Chegg Com
∫1 〖(2−𝑦)𝑑𝑦=〗 ∫1 (𝑥1)𝑑𝑥 2y − 𝑦^2/2 = 𝑥^2/2 x c 〖4𝑦 − 𝑦〗^2/2 = (𝑥^2 2𝑥 2𝑐)/2 4y − y2 = x2 2x 2c Accounts Tax GST GST All topics Solve ( 1 x 2) dy/dx 2xy = 4x 2, given that y(0) = 0 The answer 3 y ( 1 x 2) = 4x 3 Share with your friends Share 4
Y(2xy+1)dx+x(1+2xy+x^(2)y^(2))dy=0
Y(2xy+1)dx+x(1+2xy+x^(2)y^(2))dy=0-Math Input Use Math Input Mode to directly enter textbook math notation Try itThe solution of the differential equation (dy / dx) = x y / x satisfying the condition y (1) = 1 is Leave a Comment Cancel reply Your Mobile number and Email id will not be published
Differential Equations
Hence, du/dx = dy/dx * (2y) This is the derivative of the second component4) For the third component, let w = 2xy We use the product rule dw/dx = 2x*(dy/dx * 1) 2*(y) = dy/dx (2x) 2y This is the derivative of the third component5) Overall the equation differentiates to 2 x ln(2) dy/dx(2y) = dy/dx*(2x) 2y This can be rearranged to dy/dx = 2xy/(x^2y) y＝Ce^(x^2) 进行分离变量可得：dy/y＝2xdx 同时两边积分为：lny＝x^2＋lnC 所以通解是y＝Ce^(x^2) 定义 对于一个微分方程而言，其解往往不止一个，而是有一组，可以表示这一组中所有解或者部分解的统一形式，称为通解（general solution）。Simple 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 it
Answer (1 of 5) We need to get the integration of the derivative, y' = dy/dx = 2xy We divide both sides by y, 1/y (dy/dx) = 2x and then integrate both sides, ∫ (1/y) dy = ∫ 2x dx ∫ (1/y) dy = 2 * ∫ x dx the solution on both sides is given by, ln y = 2 * ((1/2) x^2 C) ln y = (x^2 C)Solve the following differential equation (x2 y2)dx 2xy dy = 0Dx 2xy 24 X2 (zy dax ( x x 2 dy y ( 2 x ) ) = 0 , 24 dy dy _* 2 2 9
Y(2xy+1)dx+x(1+2xy+x^(2)y^(2))dy=0のギャラリー
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 Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
 Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 | Solve Y 1 2xy Dx X 1 Xy Dy 0 |
 Solve Y 1 2xy Dx X 1 Xy Dy 0 | Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
「Y(2xy+1)dx+x(1+2xy+x^(2)y^(2))dy=0」の画像ギャラリー、詳細は各画像をクリックしてください。
 Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 | Solve Y 1 2xy Dx X 1 Xy Dy 0 |
Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
「Y(2xy+1)dx+x(1+2xy+x^(2)y^(2))dy=0」の画像ギャラリー、詳細は各画像をクリックしてください。
 Solve Y 1 2xy Dx X 1 Xy Dy 0 | Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
 Solve Y 1 2xy Dx X 1 Xy Dy 0 | Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
 Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
「Y(2xy+1)dx+x(1+2xy+x^(2)y^(2))dy=0」の画像ギャラリー、詳細は各画像をクリックしてください。
 Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
 Solve Y 1 2xy Dx X 1 Xy Dy 0 | Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
「Y(2xy+1)dx+x(1+2xy+x^(2)y^(2))dy=0」の画像ギャラリー、詳細は各画像をクリックしてください。
 Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
 Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
 Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 | Solve Y 1 2xy Dx X 1 Xy Dy 0 |
「Y(2xy+1)dx+x(1+2xy+x^(2)y^(2))dy=0」の画像ギャラリー、詳細は各画像をクリックしてください。
 Solve Y 1 2xy Dx X 1 Xy Dy 0 | Solve Y 1 2xy Dx X 1 Xy Dy 0 | Solve Y 1 2xy Dx X 1 Xy Dy 0 |
 Solve Y 1 2xy Dx X 1 Xy Dy 0 | Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
 Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
「Y(2xy+1)dx+x(1+2xy+x^(2)y^(2))dy=0」の画像ギャラリー、詳細は各画像をクリックしてください。
Solve Y 1 2xy Dx X 1 Xy Dy 0 | Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
Solve Y 1 2xy Dx X 1 Xy Dy 0 | Solve Y 1 2xy Dx X 1 Xy Dy 0 | Solve Y 1 2xy Dx X 1 Xy Dy 0 |
Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
「Y(2xy+1)dx+x(1+2xy+x^(2)y^(2))dy=0」の画像ギャラリー、詳細は各画像をクリックしてください。
 Solve Y 1 2xy Dx X 1 Xy Dy 0 | Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
 Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
「Y(2xy+1)dx+x(1+2xy+x^(2)y^(2))dy=0」の画像ギャラリー、詳細は各画像をクリックしてください。
 Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
 Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 | Solve Y 1 2xy Dx X 1 Xy Dy 0 |
Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
「Y(2xy+1)dx+x(1+2xy+x^(2)y^(2))dy=0」の画像ギャラリー、詳細は各画像をクリックしてください。
 Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 | Solve Y 1 2xy Dx X 1 Xy Dy 0 |
 Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 | Solve Y 1 2xy Dx X 1 Xy Dy 0 |
「Y(2xy+1)dx+x(1+2xy+x^(2)y^(2))dy=0」の画像ギャラリー、詳細は各画像をクリックしてください。
Solve Y 1 2xy Dx X 1 Xy Dy 0 | Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
 Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
 Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
「Y(2xy+1)dx+x(1+2xy+x^(2)y^(2))dy=0」の画像ギャラリー、詳細は各画像をクリックしてください。
Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
Solve Y 1 2xy Dx X 1 Xy Dy 0 |  Solve Y 1 2xy Dx X 1 Xy Dy 0 |
The equation is not exact, because you'd actually need d (2xy)/dy=d (x 2 y 2 )/dx, which means that 2x=2x Letting (x 2 y 2 )=M, 2xy=N, then Mx=2x, Ny=2x So MxNy=4x Now, (MxNy)/N=4x/ (2xy)=2/y, so this is just a function of y v (y) That means that we can let u (y)=e integral of vdy u (y)=e 2ln y =1/y 2Find the particular solution of the differential equation x 2 dy = (2xy y 2) dx, given that y = 1 when x = 1 Advertisement Remove all ads Solution Show Solution
Incoming Term: (x^(2)+y^(2)+2xy+1)(dy)/(dx)=x+y, y(2xy+1)dx+x(1+2xy+x^(2)y^(2))dy=0, (dy)/(dx)=(x^(2)+y^(2)-2x-2y+2)/(xy-x-y+1), y(2xy+1)dx+x(1+2xy-x^(3)y^(3))dy=0,
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