Question

In Exercises \(28-34\) solve the given homogeneous equation implicitly. $$ y^{\prime}=\frac{x+y}{x-y} $$

Short answer

Question: Determine the implicit solution to the first-order homogeneous differential equation: $$ y'=\frac{x+y}{x-y} $$ Answer: The implicit solution to the given homogeneous differential equation is: $$ \tan^{-1}\left(\frac{y}{x}\right)=\ln|x|+C $$

Step-by-step solution

Verify the Differential Equation is Homogeneous

We can verify that the given differential equation is homogeneous, where F is: $$ F(v)=\frac{1+v}{1-v} $$ Since F is defined only in terms of the ratio \(\frac{y}{x}\), the given equation is a first-order homogeneous differential equation.

Make a Substitution

Now, we'll make a substitution to turn the given differential equation into a simpler and separable equation. Let's represent the ratio \(\frac{y}{x}\) as \(v\), which means: $$ v = \frac{y}{x} $$ Differentiate both sides with respect to x: $$ \frac{dv}{dx}=\frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} $$

Simplify the Differential Equation

Replacing \(v\) and \(y'=\frac{dy}{dx}\) into the given homogeneous equation, we'll obtain: $$ \frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2}=\frac{x+y}{x-y} $$ Now, substitute \(y = xv\) to simplify the equation further: $$ \frac{1}{x} \frac{d(xv)}{dx} - \frac{xv}{x^2}=\frac{x+xv}{x-xv} $$

Separate Variables and Integrate

Now, our equation is separable. So, let's rewrite it in terms of \(x\) and \(v\). First, cancel the \(x\) from each term: $$ \frac{dv}{dx}+v=\frac{1+v}{1-v} $$ Separating the variables, we obtain: $$ \frac{dv}{1+v^2}=\frac{dx}{x} $$ Now, we integrate both sides: $$ \int\frac{dv}{1+v^2}=\int\frac{dx}{x} $$ The left integral can be solved as: $$ \tan^{-1}(v)=\ln|x|+C $$ where C is the constant of integration.

Find the Solution Implicitly

Now, we must go back to our original variables \(y\) and \(x\). Recall that we had \(v = \frac{y}{x}\). Therefore, the implicit solution to the given homogeneous differential equation is: $$ \tan^{-1}\left(\frac{y}{x}\right)=\ln|x|+C $$

Key concepts

First-Order Differential Equations

First-order differential equations are equations that involve a function and its first derivative. The general form of a first-order differential equation is \( y' = f(x, y) \). Such equations describe how a quantity changes and are fundamental in modeling real-world phenomena. These can be classified into several types, including linear, separable, and homogeneous equations. In our example, we have a homogeneous equation, which is a special type of differential equation. The given equation, \( y' = \frac{x+y}{x-y} \), is homogeneous because it can be rewritten as a function of the ratio \( \frac{y}{x} \). This concept plays a crucial role in finding simpler forms of the equation to solve it more easily.

Separable Differential Equations

Separable differential equations are a subset of differential equations that can be expressed as a product of a function of \( x \) and a function of \( y \). These equations can be solved by separating the variables \( x \) and \( y \) on different sides of the equation, hence the name "separable." In our problem, after the substitution \( v = \frac{y}{x} \), the equation simplifies into a separable form \( \frac{dv}{1+v^2} = \frac{dx}{x} \). This means that both \( v \) and \( x \) can be integrated separately. It simplifies the process, making it easier to find a solution by tackling one integral at a time.

Implicit Solutions

An implicit solution of a differential equation is a relation \( F(x, y) = 0 \) that involves both the dependent and independent variables, rather than solving explicitly for the dependent variable. In our exercise, after integrating, we reach a solution of the form \( \tan^{-1}\left(\frac{y}{x}\right) = \ln|x| + C \). Here, \( C \) is an integration constant, and the solution is expressed implicitly, meaning \( y \) is not isolated. Implicit solutions often naturally arise when the variables are deeply intertwined, making it difficult or impossible to solve explicitly for one of them. They are crucial in expressing more complex relationships in a concise form.