To solve the differential equation \((2y-1)dx + (2x-3)dy = 0\), we can observe that it is a first-order differential equation and we can try to solve it using the method for exact equations.
Let's rewrite the differential equation:
\((2y-1)dx + (2x-3)dy = 0\)
Here, the differential equation is of the form:
\(M(x,y) \, dx + N(x,y) \, dy = 0\)
where \(M(x,y) = 2y - 1\) and \(N(x,y) = 2x - 3\).
To check if the equation is exact, compute the partial derivatives \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\):
Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the differential equation is exact.
The solution of an exact differential equation can be found by integrating \(M(x,y)\) with respect to \(x\) and \(N(x,y)\) with respect to \(y\):
Integrate \(M(x,y) = 2y - 1\) with respect to \(x\):
\(\int (2y - 1) \, dx = (2y - 1)x + g(y)\)
Integrate \(N(x,y) = 2x - 3\) with respect to \(y\):
\(\int (2x - 3) \, dy = (2x - 3)y + h(x)\)
Combining these results, we get the potential function \(\Psi(x,y)\):
\(\Psi(x,y) = (2y - 1)x + (2x - 3)y = C\)
Expanding and rearranging terms to simplify, we have:
\(\Psi(x,y) = 2xy - x - 3y = C\)
This simplifies to:
\(\frac{2x - 1}{2y + 3} = C\)
Therefore, the correct solution of the differential equation is:
Correct Answer: \(\frac{2x-1}{2y+3}=c\)