4 min read Jun 17, 2024

Solving the Differential Equation: x^(2y-2xy^(2))dx - (x^(3)-3x^(2y)dy = 0

This differential equation is a non-linear first-order ordinary differential equation, which means it involves both the dependent variable (y) and its first derivative (dy/dx). We can solve it using the following steps:

1. Identifying the Form of the Equation

The equation is in the form:

M(x, y)dx + N(x, y)dy = 0


  • M(x, y) = x^(2y-2xy^(2))
  • N(x, y) = -(x^(3)-3x^(2y))

2. Checking for Exactness

A differential equation is exact if:

∂M/∂y = ∂N/∂x

Let's calculate these partial derivatives:

  • ∂M/∂y = 2x^(2y-2xy^(2)) * (ln(x) - 2xy)
  • ∂N/∂x = -3x^2 + 6x^(2y-1)

As ∂M/∂y ≠ ∂N/∂x, the equation is not exact.

3. Finding an Integrating Factor

Since the equation is not exact, we need to find an integrating factor, μ(x, y), to make it exact. There are two common approaches:

  • μ(x) depending only on x: If (∂N/∂x - ∂M/∂y) / M is a function of x only, then μ(x) = exp(∫(∂N/∂x - ∂M/∂y) / M dx).
  • μ(y) depending only on y: If (∂M/∂y - ∂N/∂x) / N is a function of y only, then μ(y) = exp(∫(∂M/∂y - ∂N/∂x) / N dy).

Let's calculate (∂N/∂x - ∂M/∂y) / M:

((∂N/∂x - ∂M/∂y) / M) = (-3x^2 + 6x^(2y-1) - 2x^(2y-2xy^(2)) * (ln(x) - 2xy)) / x^(2y-2xy^(2))

This expression is not a function of x only.

Now, let's calculate (∂M/∂y - ∂N/∂x) / N:

((∂M/∂y - ∂N/∂x) / N) = (2x^(2y-2xy^(2)) * (ln(x) - 2xy) + 3x^2 - 6x^(2y-1)) / -(x^(3)-3x^(2y))

This expression is not a function of y only.

Therefore, neither of the common approaches for finding an integrating factor directly applies in this case. Finding an integrating factor might require more advanced techniques or might not be possible.

4. Solving the Equation (if possible)

If we were able to find an integrating factor, we would multiply the original equation by it, making the equation exact. Then, we could solve it as follows:

  • Find a function F(x, y) such that ∂F/∂x = M and ∂F/∂y = N.
  • The general solution would be given by F(x, y) = C, where C is an arbitrary constant.


While we were able to identify the form of the equation and determine it's not exact, finding an integrating factor for this specific equation is not straightforward using the standard methods. Further investigation or alternative techniques might be required to solve this differential equation.

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