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Y(2xy+1)dx +x(1+2xy-x^3y^3)dy=0?
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Y(2xy+1)dx +x(1+2xy-x^3y^3)dy=0?

Given Equation:


y(2xy + 1)dx + x(1 - 2xy + x^3y^3)dy = 0






Explanation:



Step 1: Identify the type of equation


  • The given equation is a first-order linear ordinary differential equation.




Step 2: Separate the variables


  • Write the equation in the form of M(x, y)dx + N(x, y)dy = 0.

  • In this case, M(x, y) = y(2xy + 1) and N(x, y) = x(1 - 2xy + x^3y^3).




Step 3: Check if the equation is exact


  • An equation is exact if ∂M/∂y = ∂N/∂x, where ∂ denotes partial derivative.

  • Calculate the partial derivatives: ∂M/∂y = 2xy + 1 and ∂N/∂x = 1 - 2xy + x^3y^3.

  • Since ∂M/∂y is not equal to ∂N/∂x, the equation is not exact.




Step 4: Find the integrating factor


  • To make the equation exact, we need to find an integrating factor, denoted by μ(x, y).

  • Multiply the entire equation by μ(x, y): μ(x, y)[y(2xy + 1)dx + x(1 - 2xy + x^3y^3)dy] = 0.

  • We want the coefficients of dx and dy to be equal, so we choose μ(x, y) such that μ(x, y)∂M/∂y = μ(x, y)∂N/∂x.

  • In this case, μ(x, y) = 1/(xy).




Step 5: Multiply the equation by the integrating factor


  • Multiply the original equation by μ(x, y) = 1/(xy): (1/(xy))[y(2xy + 1)dx + x(1 - 2xy + x^3y^3)dy] = 0.

  • Simplifying, we get: (2 + 1/x)dx + (1 - 2y + x^2y^3)dy = 0.




Step 6: Solve the exact equation


  • Now that the equation is exact, we can solve it using integration.

  • Let F(x, y) be the potential function, such that ∂F/∂x = 2 + 1/x and ∂F/∂y = 1 - 2y + x^2y^3
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Y(2xy+1)dx +x(1+2xy-x^3y^3)dy=0?
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