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(x^2y^2 + xy + 1)ydx + (x^2y^2 - xy + 1)xdy = 0

Question

(x^2y^2 + xy + 1)ydx + (x^2y^2 - xy + 1)xdy = 0

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Solution

This is a first order differential equation. To solve it, we can try to express it in the form of an exact differential equation, i.e., Mdx + Ndy = 0, where M and N are functions of x and y.

The given equation is:

(x^2y^2 + xy + 1)ydx + (x^2y^2 - xy + 1)xdy = 0

We can rewrite this as:

Mdx + Ndy = 0

where M = (x^2y^2 + xy + 1)y and N = (x^2y^2 - xy + 1)x

For the equation to be exact, the partial derivative of M with respect to y must be equal to the partial derivative of N with respect to x.

Let's calculate these partial derivatives:

∂M/∂y = 2x^2y^3 + x^2y + x

∂N/∂x = 2x^3y^2 + x^2y - y

As we can see, ∂M/∂y ≠ ∂N/∂x, so the equation is not exact.

Therefore, we cannot solve this differential equation using the method of exact equations. We would need to use a different method, such as the method of integrating factors, or a numerical method.

This problem has been solved

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