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Exact Differential Equations: Solving (3xy^2-y^3)dx-(2x^2y-xy^2)dy=0
In this article, we will explore how to solve the exact differential equation (3xy^2-y^3)dx-(2x^2y-xy^2)dy=0. Exact differential equations are a type of differential equation that can be solved using the concept of exact differentials.
What is an Exact Differential Equation?
A differential equation of the form:
M(x,y)dx + N(x,y)dy = 0
is said to be exact if there exists a function F(x,y) such that:
∂F/∂x = M(x,y) and ∂F/∂y = N(x,y)
In other words, an exact differential equation is one that can be expressed as the total differential of a function F(x,y).
Solving the Given Equation
The given equation is:
(3xy^2-y^3)dx - (2x^2y-xy^2)dy = 0
To determine if this equation is exact, we need to check if there exists a function F(x,y) such that:
∂F/∂x = 3xy^2-y^3 ... (1)
and
∂F/∂y = -(2x^2y-xy^2) ... (2)
Finding the Function F(x,y)
To find the function F(x,y), we can integrate equation (1) with respect to x, treating y as a constant:
F(x,y) = ∫(3xy^2-y^3)dx + g(y)
where g(y) is an arbitrary function of y.
F(x,y) = x^2y^2 - xy^3 + g(y) ... (3)
Now, we can differentiate equation (3) with respect to y, treating x as a constant:
∂F/∂y = 2x^2y - 3xy^2 + g'(y)
Comparing this with equation (2), we get:
2x^2y - 3xy^2 + g'(y) = -(2x^2y-xy^2)
This implies that g'(y) = 0, which means g(y) is a constant.
The General Solution
Substituting g(y) = C, where C is an arbitrary constant, into equation (3), we get the general solution:
F(x,y) = x^2y^2 - xy^3 + C
Conclusion
In this article, we have shown that the differential equation (3xy^2-y^3)dx-(2x^2y-xy^2)dy=0 is exact and can be solved using the concept of exact differentials. The general solution is F(x,y) = x^2y^2 - xy^3 + C, where C is an arbitrary constant.
