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Determine whether the given equation is separable, exact, or homogenous eˣ(y−x)dx+(1+eˣ)dy=0

User Bob Goblin
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The equation e^x(y-x)dx + (1+e^x)dy = 0 is examined and determined not to fall within the separable, exact, or homogeneous equation categorizations due to its structure and term degrees.

The equation in question, ex(y-x)dx + (1+ex)dy = 0, must be analyzed to determine if it can be classified as separable, exact, or homogenous. A separable equation can be written in the form f(y)dy = g(x)dx, where f(y) and g(x) are separate functions of y and x, respectively. An exact equation has the form M(x, y)dx + N(x, y)dy = 0, where ∇M/∇y = ∇N/∇x. A homogeneous equation has the same degree of x and y in all terms.

After examining the given equation, we notice that it does not fit the criteria for a separable equation because we cannot isolate the terms of y from those of x. It is not exact because the partial derivatives of ex(y-x) with respect to y and 1+ex with respect to x are not equal. Lastly, it is not homogeneous because each term is of a different degree. Therefore, the equation does not fall within the given categorizations.

User Andy Gauge
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