Understanding the Differential Equation

The given problem asks for a solution to the differential equation \( \left(\frac{d y}{d x}\right)^2-x \frac{d y}{d x}=0 \). This is a first-order ordinary differential equation, but it is non-linear because the term \( \left(\frac{d y}{d x}\right)^2 \) involves the derivative raised to the power of two.

To find the solutions, we can treat this as an algebraic equation in terms of \( \frac{d y}{d x} \).

Solving the Differential Equation

Let \( p = \frac{d y}{d x} \). Substituting this into the differential equation gives:

\( p^2 - xp = 0 \)

We can factor this equation:

\( p(p-x) = 0 \)

This equation is satisfied if either \( p=0 \) or \( p-x=0 \).

Case 1: \( p = 0 \)

If \( p = 0 \), then \( \frac{d y}{d x} = 0 \).

Integrating both sides with respect to \( x \):

\( \int \frac{d y}{d x} dx = \int 0 dx \)

\( y = C_1 \)

where \( C_1 \) is an arbitrary constant. This represents a family of constant functions.

Case 2: \( p - x = 0 \)

If \( p - x = 0 \), then \( p = x \), which means \( \frac{d y}{d x} = x \).

Integrating both sides with respect to \( x \):

\( \int \frac{d y}{d x} dx = \int x dx \)

\( y = \frac{x^2}{2} + C_2 \)

where \( C_2 \) is an arbitrary constant. This represents a family of parabolic functions.

The general solution to the differential equation is the union of the solutions from Case 1 and Case 2.

Checking the Given Options

Now, let's check each of the given options to see if they satisfy the original differential equation \( \left(\frac{d y}{d x}\right)^2-x \frac{d y}{d x}=0 \).

Option 1: \( y = 2x \)

First, find the derivative \( \frac{d y}{d x} \):

\( \frac{d y}{d x} = \frac{d}{dx}(2x) = 2 \)

Substitute \( \frac{d y}{d x} = 2 \) into the differential equation:

\( (2)^2 - x(2) = 4 - 2x \)

For \( y = 2x \) to be a solution, \( 4 - 2x \) must be equal to 0 for all values of \( x \). This is only true for \( x=2 \), not for all \( x \). So, \( y = 2x \) is not a solution.

Option 2: \( y = 2x + 4 \)

First, find the derivative \( \frac{d y}{d x} \):

\( \frac{d y}{d x} = \frac{d}{dx}(2x + 4) = 2 \)

Substitute \( \frac{d y}{d x} = 2 \) into the differential equation:

\( (2)^2 - x(2) = 4 - 2x \)

For \( y = 2x + 4 \) to be a solution, \( 4 - 2x \) must be equal to 0 for all values of \( x \). This is only true for \( x=2 \), not for all \( x \). So, \( y = 2x + 4 \) is not a solution.

Option 3: \( y = x^2 - 1 \)

First, find the derivative \( \frac{d y}{d x} \):

\( \frac{d y}{d x} = \frac{d}{dx}(x^2 - 1) = 2x \)

Substitute \( \frac{d y}{d x} = 2x \) into the differential equation:

\( (2x)^2 - x(2x) = 4x^2 - 2x^2 = 2x^2 \)

For \( y = x^2 - 1 \) to be a solution, \( 2x^2 \) must be equal to 0 for all values of \( x \). This is only true for \( x=0 \), not for all \( x \). So, \( y = x^2 - 1 \) is not a solution.

Option 4: \( y=\frac{\left(x^2-2\right)}{2} \)

Rewrite the option as \( y = \frac{x^2}{2} - \frac{2}{2} = \frac{x^2}{2} - 1 \).

First, find the derivative \( \frac{d y}{d x} \):

\( \frac{d y}{d x} = \frac{d}{dx}\left(\frac{x^2}{2} - 1\right) = \frac{1}{2} \frac{d}{dx}(x^2) - \frac{d}{dx}(1) = \frac{1}{2}(2x) - 0 = x \)

Substitute \( \frac{d y}{d x} = x \) into the differential equation \( \left(\frac{d y}{d x}\right)^2-x \frac{d y}{d x}=0 \):

\( (x)^2 - x(x) = x^2 - x^2 = 0 \)

The equation \( 0 = 0 \) is true for all values of \( x \). Therefore, \( y=\frac{\left(x^2-2\right)}{2} \) is a solution to the given differential equation.

This solution \( y = \frac{x^2}{2} - 1 \) is a particular case of the general solution from Case 2, \( y = \frac{x^2}{2} + C_2 \), with \( C_2 = -1 \).

Summary of Solutions

The given differential equation \( \left(\frac{d y}{d x}\right)^2-x \frac{d y}{d x}=0 \) has general solutions of the form \( y = C_1 \) and \( y = \frac{x^2}{2} + C_2 \).

Among the given options, \( y=\frac{\left(x^2-2\right)}{2} \) matches the form \( y = \frac{x^2}{2} + C_2 \) with \( C_2 = -1 \), and we verified that it satisfies the original differential equation.

Revision Table: Key Concepts for Differential Equations

Additional Information: Solving \( \left(\frac{d y}{d x}\right)^2-x \frac{d y}{d x}=0 \)

The differential equation \( \left(\frac{d y}{d x}\right)^2-x \frac{d y}{d x}=0 \) is a Clairaut's equation if written in a specific form, but in its current structure, treating it as a quadratic in \( \frac{d y}{d x} \) is the most straightforward approach. This leads to separating the equation into simpler first-order equations.

• The approach of factoring the differential equation with respect to the derivative \( \frac{d y}{d x} \) is effective when the equation can be expressed as a polynomial in \( \frac{d y}{d x} \).
• Each factor, when set to zero, yields a simpler first-order differential equation that can often be solved by integration or other standard methods.
• The solutions obtained from each factor represent different families of curves that satisfy the original differential equation.
• When checking options, it is crucial to substitute both the function \( y \) and its derivative \( \frac{d y}{d x} \) back into the original differential equation to verify if it holds true for all relevant values of \( x \).