Understanding the Differential Equation ydx - xdy = 0

The given equation is a first-order differential equation: \(ydx - xdy = 0\). Our goal is to find its general solution, which means finding a function \(y(x)\) (or a relation between \(y\) and \(x\)) that satisfies this equation for any arbitrary constant.

Solving the Differential Equation using Separation of Variables

This differential equation is a good candidate for the method of separation of variables. This method involves rearranging the equation so that all terms involving \(y\) and \(dy\) are on one side, and all terms involving \(x\) and \(dx\) are on the other side.

Let's start with the given equation:

\(ydx - xdy = 0\)

Move the \(xdy\) term to the right side:

\(ydx = xdy\)

Now, we want to separate variables. To do this, we can divide both sides by \(xy\), assuming \(x \neq 0\) and \(y \neq 0\):

\(\frac{ydx}{xy} = \frac{xdy}{xy}\)

This simplifies to:

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

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

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

Integration Steps

The integral of \(\frac{1}{u}\) with respect to \(u\) is \(\ln|u|\) plus a constant of integration. Applying this to both sides:

\(\ln|x| + C_1 = \ln|y| + C_2\)

We can combine the constants of integration into a single constant. Let \(C = C_2 - C_1\). Then:

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

To isolate \(\ln|y|\) or \(\ln|x|\), let's move \(\ln|y|\) to the left side (or \(\ln|x|\) to the right). Let's move \(\ln|y|\) to the left:

\(\ln|x| - \ln|y| = C\)

Using the logarithm property \(\ln a - \ln b = \ln(a/b)\):

\(\ln\left|\frac{x}{y}\right| = C\)

Now, exponentiate both sides with base \(e\) to remove the logarithm:

\(e^{\ln\left|\frac{x}{y}\right|} = e^C\)

\(\left|\frac{x}{y}\right| = e^C\)

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

\(\left|\frac{x}{y}\right| = K\)

This means \(\frac{x}{y} = \pm K\). Let's denote a new arbitrary constant \(C'\) where \(C' = \pm K\). Note that \(C'\) cannot be zero at this step because \(K > 0\).

\(\frac{x}{y} = C'\)

Rearranging this to solve for \(y\):

\(x = C'y\)

\(y = \frac{1}{C'} x\)

Let \(C = \frac{1}{C'}\). Since \(C' \neq 0\), \(C\) can be any non-zero constant.

\(y = Cx\)

Considering Cases where \(x=0\) or \(y=0\)

When we divided by \(xy\), we assumed \(x \neq 0\) and \(y \neq 0\). Let's check if the solution \(y = Cx\) covers the cases where \(x=0\) or \(y=0\).

• If \(y=0\), the equation \(y = Cx\) becomes \(0 = Cx\). This is true for any \(x\) if \(C=0\). So, \(y=0\) (the x-axis) is a solution, and it is included in the form \(y=Cx\) when \(C=0\).
• If \(x=0\), the equation \(y = Cx\) becomes \(y = C \cdot 0\), which means \(y=0\). This corresponds to the point \((0,0)\). However, the y-axis (\(x=0\) for all \(y\)) is also a potential solution to the original differential equation \(ydx - xdy = 0\). If \(x=0\), then \(dx=0\), and the equation becomes \(y \cdot 0 - 0 \cdot dy = 0\), which is \(0=0\). So the y-axis is also a solution. The form \(y=Cx\) represents a family of lines passing through the origin. While \(y=0\) is covered by \(C=0\), \(x=0\) is not covered unless we write the solution as \(\frac{y}{x} = C\). However, the form \(y=Cx\) is standard and often considered the general solution, implying all lines through the origin. The case \(x=0\) is sometimes seen as a singular solution or is implicitly included in the family of lines depending on the context and how the integration constant is handled from \(\ln|y| = \ln|x| + \ln|C'|\) leading to \(|y| = |C'x|\) which means \(y = \pm C'x\), covering all lines \(y=Cx\), including \(x=0\) and \(y=0\) as specific instances within the family of lines through the origin.

Given the options, the form \(y = Cx\) is the most appropriate general solution derived from separating variables \(\frac{dy}{y} = \frac{dx}{x}\).

Checking the Options

Let's examine the provided options based on our derived general solution \(y = Cx\):

1. \(y = Cx\): This matches our derived solution.
2. \(y = Ce^x\): This is a solution to \(\frac{dy}{dx} = y\). It does not match.
3. \(ax + by = C\): This represents general lines, not necessarily passing through the origin. Our solution represents lines passing through the origin. It does not match.
4. \(x^2 + y^2 = C\): This represents circles centered at the origin. It does not match.

Therefore, the general solution among the options is \(y = Cx\).

Conclusion

The differential equation \(ydx - xdy = 0\) can be solved by separating variables, leading to \(\frac{dy}{y} = \frac{dx}{x}\). Integrating both sides gives \(\ln|y| = \ln|x| + \text{constant}\), which simplifies to the general solution \(y = Cx\), representing a family of lines passing through the origin.

Revision Table: Key Concepts

Let's review the essential concepts used in solving this differential equation problem:

• Differential Equation: An equation that relates a function with its derivatives. The given equation is a first-order ODE.
• General Solution: A solution to a differential equation that contains an arbitrary constant (or constants, equal to the order of the equation). This represents a family of functions that satisfy the equation.
• Separation of Variables: A technique to solve first-order ODEs where the equation can be written in the form \(f(y)dy = g(x)dx\).
• Integration: The process of finding the antiderivative. Integrals of \(\frac{1}{u}\) are essential here.
• Logarithm Properties: Properties like \(\ln a - \ln b = \ln(a/b)\) and \(e^{\ln u} = u\) are crucial for simplifying after integration.

Additional Information: Related Differential Equations

The differential equation \(ydx - xdy = 0\) is a special case of several types of differential equations:

• Homogeneous Differential Equation: A first-order ODE of the form \(\frac{dy}{dx} = f(\frac{y}{x})\). Our equation \(ydx = xdy\) can be written as \(\frac{dy}{dx} = \frac{y}{x}\). Here \(f(\frac{y}{x}) = \frac{y}{x}\). Homogeneous equations can be solved using the substitution \(y = vx\), where \(v\) is a function of \(x\). If \(y=vx\), then \(\frac{dy}{dx} = v + x\frac{dv}{dx}\). Substituting into \(\frac{dy}{dx} = \frac{y}{x}\) gives \(v + x\frac{dv}{dx} = \frac{vx}{x} = v\). This simplifies to \(x\frac{dv}{dx} = 0\). For \(x \neq 0\), this means \(\frac{dv}{dx} = 0\), so \(v = C\) (a constant). Substituting back \(v = y/x\), we get \(\frac{y}{x} = C\), or \(y = Cx\), which is the same general solution obtained by separation of variables.
• Exact Differential Equation: A first-order ODE of the form \(M(x, y)dx + N(x, y)dy = 0\) is exact if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). In our case, \(M(x, y) = y\) and \(N(x, y) = -x\). We have \(\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(y) = 1\) and \(\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(-x) = -1\). Since \(1 \neq -1\), the equation \(ydx - xdy = 0\) is NOT exact in its current form. However, multiplying by an integrating factor can make it exact. For example, multiplying by \(1/x^2\) gives \(\frac{y}{x^2}dx - \frac{1}{x}dy = 0\). Here \(M = y/x^2, N = -1/x\). \(\frac{\partial M}{\partial y} = \frac{1}{x^2}\) and \(\frac{\partial N}{\partial x} = \frac{1}{x^2}\). Now it is exact. Integrating \(\frac{y}{x^2}dx\) with respect to \(x\) gives \(-\frac{y}{x} + f(y)\). Differentiating with respect to \(y\) gives \(-\frac{1}{x} + f'(y)\). Comparing with \(N = -1/x\), we get \(f'(y)=0\), so \(f(y)=C\). The solution is \(-\frac{y}{x} = C\), or \(\frac{y}{x} = -C\). Redefining the constant, we get \(\frac{y}{x} = C_{new}\), or \(y = C_{new}x\). Other integrating factors like \(1/y^2\), \(1/xy\), or \(1/(x^2+y^2)\) could also be used.

This problem demonstrates that a simple differential equation can often be solved using multiple methods, all leading to the same general solution.