Question

Solve the following differential equations. \(x y d x+\left(y^{2}-x^{2}\right) d y=0\)

Short answer

\( x^2 + 2y^2 \ln|y| = C' \)

Step-by-step solution

Separate Variables

First, write the equation in a form that makes it easier to separate the variables. The given equation is: \( x y \frac{d x}{d y} + y^2 - x^2 = 0 \) Rearrange it to separate the variables: \( x y \frac{d x}{d y} = x^2 - y^2 \)

Divide by \(xy\)

Divide both sides by \(xy\) to isolate \( \frac{dx}{dy}\): \( \frac{dx}{dy} = \frac{x^2 - y^2}{xy} \) Simplify the fraction: \( \frac{dx}{dy} = \frac{x^2}{xy} - \frac{y^2}{xy} = \frac{x}{y} - \frac{y}{x} \)

Introduce a Substitution

Introduce a substitution to simplify the expression. Let \( v = \frac{x}{y} \), then \( x = vy \) and \( \frac{dx}{dy} = v + y \frac{dv}{dy} \). Substitute these into the differential equation: \( v + y \frac{dv}{dy} = v - \frac{1}{v} \)

Simplify the Resulting Equation

Rearrange to isolate \( \frac{dv}{dy} \): \( y \frac{dv}{dy} = - \frac{1}{v} \) Separate the variables: \( v dv = - \frac{dy}{y} \)

Integrate Both Sides

Integrate both sides to find the solutions: \( \int v \, dv = - \int \frac{dy}{y} \) The integrals are: \( \frac{v^2}{2} = - \ln|y| + C \), where \ C \ is the constant of integration.

Substitute Back the Original Variables

Recall the substitution \ v = \frac{x}{y} \, and substitute it back into the solution. \( \frac{\left( \frac{x}{y} \right)^2}{2} = - \ln|y| + C \) Simplify: \( \frac{x^2}{2y^2} = - \ln|y| + C \)

Multiply Through by \(2y^2\)

Clear the fraction by multiplying through by \(2y^2\): \( x^2 = -2y^2 \ln|y| + 2Cy^2 \) Rename the constant \ 2C \ as a new constant \ C' \ for simplicity.

Write the Final Solution

Thus, the general solution to the differential equation is: \( x^2 + 2y^2 \ln|y| = C' \)

Key concepts

Variable Separation

Variable separation is a technique to solve differential equations by separating the variables present in the equation. This means we manipulate the equation such that all terms involving one variable (say x) and its differential are on one side of the equation, and all terms involving the other variable (say y) and its differential are on the other side. For the equation \( x y \frac{dx}{dy} + y^2 - x^2 = 0 \), we want to isolate the terms involving \( x \) and \( dx \) on one side, and the terms involving \( y \) and \( dy \) on the other. After rearranging and simplifying the original equation, we get: \[ \frac{dx}{dy} = \frac{x}{y} - \frac{y}{x} \] This motivates the next step to make solving easier by using substitutions.

Integration

Integration is the process of finding the integral of a function, which is essentially the reverse operation of differentiation. In this problem, after separating variables and substituting, we reach a step where we need to integrate both sides of the equation: \[ \int v \, dv = - \int \frac{dy}{y} \] On the left side, we need to integrate \(v\) with respect to \(v\), and on the right side, we need to integrate \(\frac{1}{y}\) with respect to \(y\). Performing these integrations gives us: \[ \frac{v^2}{2} = - \ln|y| + C \] Here, \( C \) is the constant of integration that arises from indefinite integration. Understanding integration deeply helps in not just solving differential equations but also in various applications in calculus.

Substitution Method

The substitution method is a strategic tool to simplify and solve differential equations. By substituting a new variable, the equation can be transformed into a simpler form. In this equation, we use the substitution \( v = \frac{x}{y} \). This leads to rewriting \( x \) as \( vy \) and transforming the differential equation. The substitution reduces the complexity and makes the equation solvable by standard methods: \[ v + y \frac{dv}{dy} = v - \frac{1}{v} \] This simplifies further to: \[ y \frac{dv}{dy} = - \frac{1}{v} \] The main advantage of the substitution method is that it can transform a non-linear equation into a linear or more manageable form, making the equation easier to solve.

Logarithmic Functions (ln function)

Logarithmic functions, particularly the natural logarithm (\( \ln \)), play a crucial role in solving certain differential equations. In our case, during integration, we encounter: \( - \int \frac{dy}{y} = - \ln|y| \). The \( \ln \) function is the inverse of the exponential function and provides a straightforward way to handle integrals involving \( \frac{1}{y} \). The result from the integration step is: \[ \frac{v^2}{2} = - \ln|y| + C \] Expressing our solution in terms of \( \ln|y| \) helps us to later substitute back and simplify the final solution. Logarithmic functions are essential tools in calculus for solving integrals and understanding growth and decay processes.