Understanding the Given Differential Equation

The problem asks for the general solution of the differential equation \( \frac{2y dx - 3x dy}{y} = 0 \). This is a first-order differential equation. The equation involves differentials \( dx \) and \( dy \).

The presence of \( y \) in the denominator indicates that the equation is defined for \( y \neq 0 \).

Simplifying the Differential Equation

Assuming \( y \neq 0 \), we can multiply both sides of the equation by \( y \):

This simplified form is easier to work with. This new equation \( 2y dx - 3x dy = 0 \) is equivalent to the original one for \( y \neq 0 \). We will solve this simplified equation first and then discuss the implications for the original equation.

Solving the Differential Equation by Separation of Variables

The equation \( 2y dx - 3x dy = 0 \) is a variable-separable differential equation. We need to rearrange the terms so that all \( x \) terms (including \( dx \)) are on one side and all \( y \) terms (including \( dy \)) are on the other side.

Rearrange the equation:

Now, separate the variables by dividing by \( 3xy \). This step requires assuming \( x \neq 0 \) and \( y \neq 0 \).

We can write this as:

Now, integrate both sides of the separated equation:

Combine the constants of integration into a single arbitrary constant \( C = C_2 - C_1 \):

Finding the General Solution Form

We need to manipulate the integrated equation to match the options provided. Use logarithm properties \( a \ln b = \ln(b^a) \) and \( \ln a - \ln b = \ln(a/b) \).

Exponentiate both sides with base \( e \):

Let \( K = e^C \). Since \( C \) is an arbitrary constant, \( K \) is a positive constant.

This means \( \frac{x^{2/3}}{y} = \pm K \).

Let \( A = \pm K \). Since \( K > 0 \), \( A \) is a non-zero arbitrary constant.

Let \( C' = \frac{1}{A} \). Since \( A \neq 0 \), \( C' \) is a non-zero arbitrary constant.

Now, cube both sides to get rid of the fractional exponent:

Let \( c = (C')^3 \). Since \( C' \) is a non-zero arbitrary constant, \( c \) is also a non-zero arbitrary constant.

This is the general solution obtained from integrating \( 2y dx - 3x dy = 0 \) assuming \( x \neq 0 \) and \( y \neq 0 \). The option provided, however, lists \( y^3 = cx^2 \) where \( c \) is an arbitrary constant, typically meaning \( c \) can be zero as well.

Let's check if \( y^3 = cx^2 \) is a solution to \( 2y dx - 3x dy = 0 \) for an arbitrary constant \( c \).

If \( y^3 = cx^2 \), then differentiating implicitly with respect to \( x \), we get:

From \( y^3 = cx^2 \), if \( x \neq 0 \), we have \( c = \frac{y^3}{x^2} \). Substitute this value of \( c \) into the differentiated equation:

Rearrange the terms:

Divide by \( y^2 \) (assuming \( y \neq 0 \)):

This matches the simplified equation. The original equation \( \frac{2y dx - 3x dy}{y} = 0 \) is defined only when \( y \neq 0 \). The solution \( y^3 = cx^2 \) when \( c=0 \) gives \( y=0 \), which is a case where the original equation is undefined. However, it is common practice for the "general solution" to refer to the family of curves covering most solutions, often derived from the simplified equation. Given the options, \( y^3 = cx^2 \) with an arbitrary constant \( c \) is the expected general solution form derived from \( 2y dx - 3x dy = 0 \).

Considering the standard form of the arbitrary constant from the integration step \( \frac{2}{3} \ln|x| = \ln|y| + C_0 \), we can write \( \ln|y| = \frac{2}{3} \ln|x| - C_0 \). Let \( -C_0 = \ln|c| \) where \( c \) is a non-zero constant. Then \( \ln|y| = \ln|x|^{2/3} + \ln|c| = \ln(|c||x|^{2/3}) \). So \( |y| = |c||x|^{2/3} \), which implies \( y = \pm c x^{2/3} \). Let \( C' = \pm c \), where \( C' \) is a non-zero constant. \( y = C' x^{2/3} \). Cubing both sides gives \( y^3 = (C')^3 x^2 \). Let \( c_{final} = (C')^3 \). Since \( C' \neq 0 \), \( c_{final} \neq 0 \). If we allow the constant in \( \ln|y| = \frac{2}{3} \ln|x| + C \) to be just \( C \), then \( \ln|y| - \ln|x|^{2/3} = C \), \( \ln\left|\frac{y}{x^{2/3}}\right| = C \), \( \left|\frac{y}{x^{2/3}}\right| = e^C \). Let \( e^C = k > 0 \). \( \left|\frac{y}{x^{2/3}}\right| = k \). \( \frac{y}{x^{2/3}} = \pm k \). Let \( c = \pm k \), so \( c \neq 0 \). Then \( y = c x^{2/3} \). \( y^3 = c^3 x^2 \). Renaming \( c^3 \) as \( C \), we get \( y^3 = Cx^2 \) where \( C \neq 0 \). However, often the arbitrary constant symbol absorbs signs and powers, leading to the form \( y^3 = cx^2 \) where \( c \) can be any real number.

Given the options, the form \( y^3 = cx^2 \) with \( c \) as an arbitrary constant (allowing \( c=0 \)) is the standard general solution presented for the equation \( 2y dx - 3x dy = 0 \).

Conclusion on the General Solution

Based on the derivation from the simplified equation \( 2y dx - 3x dy = 0 \), the form of the general solution is \( y^3 = cx^2 \).

Revision Table: Differential Equation Concepts

Additional Information: Separation of Variables Method

The method of separation of variables is applicable to first-order differential equations that can be written in the form \( M(x) dx + N(y) dy = 0 \) or \( \frac{dy}{dx} = f(x)g(y) \).

The steps involved are generally:

1. Rearrange the equation to separate variables, typically getting \( M(x) dx = N(y) dy \) or \( \frac{dy}{g(y)} = f(x) dx \). Ensure denominators are non-zero during separation.
2. Integrate both sides: \( \int M(x) dx = \int N(y) dy \) or \( \int \frac{dy}{g(y)} = \int f(x) dx \).
3. Include an arbitrary constant of integration on one side.
4. Solve the resulting algebraic equation for the dependent variable (usually \( y \)) in terms of the independent variable (usually \( x \)) and the arbitrary constant.

In our case, the simplified equation \( 2y dx - 3x dy = 0 \) was rearranged to \( \frac{2}{3} \frac{dx}{x} = \frac{dy}{y} \), which is in the form \( f(x) dx = g(y) dy \) where \( f(x) = \frac{2}{3x} \) and \( g(y) = \frac{1}{y} \). Integrating these forms led to the general solution \( y^3 = cx^2 \).