-dx%2B2xy-dy-%3D-0.jpeg "(x2 – Y2) Dx+2xy Dy = 0")
Exact Differential Equations: (x2 – y2) dx + 2xy dy = 0
In this article, we will discuss the solution to the differential equation (x2 – y2) dx + 2xy dy = 0, which is an example of an exact differential equation.
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)
The Given Equation
The given equation is (x2 – y2) dx + 2xy dy = 0. We can write it in the standard form as:
M(x, y) dx + N(x, y) dy = 0
where M(x, y) = x2 – y2 and N(x, y) = 2xy.
Testing for Exactness
To check if the given equation is exact, we need to check if:
∂M/∂y = ∂N/∂x
Computing the partial derivatives, we get:
∂M/∂y = -2y and ∂N/∂x = 2y
Since ∂M/∂y ≠ ∂N/∂x, the equation is not exact. However, we can make it exact by multiplying both sides of the equation by an integrating factor.
Finding the Integrating Factor
Let μ(x, y) be the integrating factor. Then, the new equation becomes:
μ(x, y) (x2 – y2) dx + μ(x, y) 2xy dy = 0
We need to find μ(x, y) such that the new equation is exact.
After some calculations, we find that μ(x, y) = 1/(x2 + y2). Multiplying the original equation by μ(x, y), we get:
(x2 – y2)/(x2 + y2) dx + 2xy/(x2 + y2) dy = 0
Solving the Exact Equation
The new equation is exact, and we can find the solution by integrating:
∫(x2 – y2)/(x2 + y2) dx = ∫(-2xy)/(x2 + y2) dy + C
Solving the integrals, we get:
(x2 – y2)/2 ln(x2 + y2) = -xy + C
where C is the constant of integration.
Conclusion
In this article, we have seen how to solve the differential equation (x2 – y2) dx + 2xy dy = 0 by finding an integrating factor and converting it into an exact equation. The solution to the equation is (x2 – y2)/2 ln(x2 + y2) = -xy + C.
