Solving the Differential Equation x dy – y dx = 0

The given problem asks us to find the solution to the differential equation \(x \, dy - y \, dx = 0\).

This is a first-order differential equation. We can attempt to solve it using the method of separation of variables.

The equation is:

\(x \, dy - y \, dx = 0\)

To separate the variables \(x\) and \(y\), we can move the term \(y \, dx\) to the right side:

\(x \, dy = y \, dx\)

Now, we want to get all terms involving \(y\) and \(dy\) on one side and all terms involving \(x\) and \(dx\) on the other side. We can divide both sides by \(xy\), assuming \(x \neq 0\) and \(y \neq 0\):

\(\frac{x \, dy}{xy} = \frac{y \, dx}{xy}\)

This simplifies to:

\(\frac{dy}{y} = \frac{dx}{x}\)

Now that the variables are separated, we can integrate both sides of the equation:

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

Integrating \(\frac{1}{y}\) with respect to \(y\) gives \(\ln|y|\), and integrating \(\frac{1}{x}\) with respect to \(x\) gives \(\ln|x|\). We need to add a constant of integration, let's call it \(C'\), to one side (usually the right side):

\(\ln|y| = \ln|x| + C'\)

To solve for \(y\), we can exponentiate both sides of the equation:

\(e^{\ln|y|} = e^{\ln|x| + C'}\)

\(|y| = e^{\ln|x|} \cdot e^{C'}\)

\(|y| = |x| \cdot e^{C'}\)

Let \(e^{C'} = A\), where \(A\) is a positive constant. So, \(|y| = A|x|\). This means \(y = \pm Ax\). Let \(C = \pm A\). Since \(A\) is a positive constant, \(C\) can be any non-zero real number. If we also consider the case where \(y=0\) is a solution (substituting \(y=0\) into the original equation gives \(x(0) - 0(dx) = 0\), which is true), the solution \(y=Cx\) includes the case \(y=0\) when \(C=0\). Therefore, the general solution can be written as:

\(y = Cx\)

Where \(C\) is an arbitrary constant.

Comparing Solution with Options

Let's compare our derived solution \(y = Cx\) with the given options:

• Option 1: \(xy = c\). This is not in the form \(y = Cx\). Dividing by \(x\) gives \(y = c/x\).
• Option 2: \(y = cx\). This matches our derived general solution, where \(c\) is the arbitrary constant.
• Option 3: \(x + y = c\). This is a linear equation, not matching our solution.
• Option 4: \(x – y = c\). This is also a linear equation, not matching our solution.

The solution that matches our result is \(y = cx\).

The final answer is the one that represents the general solution \(y = Cx\).

Revision Table: Key Concepts

Additional Information: Types of First-Order Differential Equations

Besides separation of variables, first-order differential equations can often be solved using other methods, depending on their form. Some common types include:

• Exact Differential Equations: Equations of the form \(M(x, y) \, dx + N(x, y) \, dy = 0\) where \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\).
• Linear First-Order Equations: Equations of the form \(\frac{dy}{dx} + P(x)y = Q(x)\). These are solved using an integrating factor.
• Homogeneous Equations: Equations that can be written in the form \(\frac{dy}{dx} = f\left(\frac{y}{x}\right)\). These can be solved using the substitution \(y = vx\).
• Bernoulli Equations: Equations of the form \(\frac{dy}{dx} + P(x)y = Q(x)y^n\). These can be transformed into linear equations using a suitable substitution.

The equation \(x \, dy - y \, dx = 0\) can be rewritten as \(x \, dy = y \, dx\), or \(\frac{dy}{dx} = \frac{y}{x}\). This form shows it is also a homogeneous differential equation, solvable using the substitution \(y=vx\), which would yield the same general solution \(y=Cx\).