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

Jasonbradley

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Jun 17, 2024

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

This differential equation is an example of a non-exact differential equation. This means that it cannot be solved directly by integrating both sides. To solve it, we need to find an integrating factor.

Here's how we can solve this differential equation:

1. Check if the equation is exact:

A differential equation of the form M(x, y)dx + N(x, y)dy = 0 is exact if ∂M/∂y = ∂N/∂x.

In our case:

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

Therefore:

• ∂M/∂y = -2y
• ∂N/∂x = 2x

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

2. Find an integrating factor:

We can find an integrating factor μ(x, y) that makes the equation exact. There are two common approaches:

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

• Finding μ(xy): If (∂M/∂y - ∂N/∂x)/(xN - yM) is a function of xy, then μ(xy) = exp(∫(∂M/∂y - ∂N/∂x)/(xN - yM) d(xy)).

In our case, (∂M/∂y - ∂N/∂x)/N = (-2y - 2x)/(x^2 - 2xy) = -2(y+x)/(x(x-2y)) is a function of x and y. We can simplify this to -2/x + 2/(x-2y), which is a function of x only.

Therefore, we can find an integrating factor μ(x) as:

μ(x) = exp(∫(-2/x + 2/(x-2y)) dx) = exp(-2ln(x) + 2ln(x-2y)) = (x-2y)^2/x^2

3. Multiply the equation by the integrating factor:

Multiplying the original equation by μ(x), we get:

((x-2y)^2/x^2) * (x^2 - y^2) dx + ((x-2y)^2/x^2) * (x^2 - 2xy) dy = 0

Simplifying:

(x^2 - 2xy + 4y^2 - y^2) dx + (x^2 - 2xy) dy = 0

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

Now, the equation is exact, since ∂/∂y(x^2 - 2xy + 3y^2) = -2x + 6y = ∂/∂x(x^2 - 2xy).

4. Solve the exact equation:

Let's find the potential function Φ(x, y) such that:

• ∂Φ/∂x = x^2 - 2xy + 3y^2
• ∂Φ/∂y = x^2 - 2xy

Integrating the first equation with respect to x, we get:

Φ(x, y) = (1/3)x^3 - x^2y + 3xy^2 + g(y)

where g(y) is an arbitrary function of y.

Differentiating Φ(x, y) with respect to y, we get:

∂Φ/∂y = -x^2 + 6xy + g'(y)

Comparing this with the second equation, we get:

-x^2 + 6xy + g'(y) = x^2 - 2xy

Therefore, g'(y) = 0, which implies g(y) = C (a constant).

5. The general solution:

Thus, the general solution to the differential equation is:

(1/3)x^3 - x^2y + 3xy^2 + C = 0

where C is an arbitrary constant.