$\text{General solution of the differential equation}\frac{2ydx-3xdy}{2}=0\text{is}\left(c\text{is an arbitrary constant}\right)$

The correct option is y³ = cx².

To find the general solution of the given differential equation, we start with the equation:
$$\frac{2ydx-3xdy}{2}=0$$

Multiplying both sides by 2, we simplify the equation to:
$$2yd-3xdy=0$$

Rearranging the terms to separate the variables, we get:
$$2yd=3xdy$$

Now, we divide both sides by xy (assuming x, y ≠ 0) to separate the variables completely:
$$2\frac{dx}{x}=3\frac{dy}{y}$$

Integrating both sides of the equation:
$$\int 2\frac{dx}{x}=\int 3\frac{dy}{y}$$

This yields:
$$2\ln \left|x\right|=3\ln \left|y\right|+c_1$$
where c₁ is an arbitrary constant of integration.

Using the logarithmic property a ln(b) = ln(b^a), we can rewrite the terms:
$$\ln \left(x^2\right)=\ln \left(y^3\right)+c_1$$

Letting the constant of integration c₁ be represented as ln(c'), where c' > 0 is another constant, we get:
$$\ln \left(x^2\right)=\ln \left(y^3\right)+\ln \left(c^{\prime }\right)$$

Using the property of logarithms ln(a) + ln(b) = ln(ab):
$$\ln \left(x^2\right)=\ln \left(c^{\prime }{y}^3\right)$$

By taking the exponential of both sides, we obtain:
$$x^2=c^{\prime }{y}^3$$

Dividing both sides by the constant c', we get:
$$y^3=\frac{1}{c^{\prime }}{x}^2$$

Since c' is an arbitrary constant, we can define a new arbitrary constant c = 1 / c'. Thus, the equation simplifies to:
$$y^3=c{x}^2$$

This matches the given correct option.