The solution of the equation (2y – 1) dx-(2x + 3)dy = 0 is

The correct option is "2x+3/2y−1 = k" (which represents $\frac{2x+3}{2y-1}=k$).

To find the solution to the given differential equation, we can use the method of separation of variables. Let's start with the given equation:
$(2y-1)dx-(2x+3)dy=0$

First, we rearrange the equation by moving the term containing $dy$ to the right side of the equation:
$(2y-1)dx=(2x+3)dy$

Next, we separate the variables by dividing both sides by $(2x+3)(2y-1)$ (assuming $2x+3\ne 0$ and $2y-1\ne 0$):
$\frac{dx}{2x+3}=\frac{dy}{2y-1}$

Now, we integrate both sides of the equation:
$\int \frac{dx}{2x+3}=\int \frac{dy}{2y-1}$

Using the integration formula $\int \frac{du}{au+b}=\frac{1}{a}ln|au+b|+C$, we evaluate the integrals:
$\frac{1}{2}ln|2x+3|=\frac{1}{2}ln|2y-1|+\mathrm{C\text{'}}$

To simplify, multiply the entire equation by 2:
$ln|2x+3|=ln|2y-1|+2\mathrm{C\text{'}}$

Let $2\mathrm{C\text{'}}=ln\left|c\right|$ where $c$ is an arbitrary constant. Substitute this back into the equation:
$ln|2x+3|=ln|2y-1|+ln\left|c\right|$

Apply the logarithmic property $lnA+lnB=ln\left(AB\right)$ on the right side:

By taking the exponential of both sides, we eliminate the logarithms:
$2x+3=k(2y-1)$
where $k$ is a constant (representing $\pm c$).

Finally, dividing both sides by $2y-1$ gives:
$\frac{2x+3}{2y-1}=k$

Thus, the solution matches the option: 2x+3/2y−1 = k.