91Ó°ÊÓ

Separable equations are a special type of differential equations. They can be rearranged in such a way that one side of the equation contains only the variable \( y \) and its derivatives, and the other side only the variable \( x \). This separation allows us to simplify the equation and solve it using integration.
To determine if a differential equation is separable, we look to see if we can manipulate it into the form \( g(y) \, dy = f(x) \, dx \), where \( g(y) \) and \( f(x) \) are functions of \( y \) and \( x \), respectively.
In the given exercise, the equation \( y' = x^m y^n \) is identified as separable because it can be rewritten as \( \frac{1}{y^n} \, dy = x^m \, dx \), effectively separating variables.

Integration is a key mathematical process used to find the antiderivative or integral of a function. It's crucial in solving differential equations, especially separable ones. Once variables are separated, integration is needed to solve for \( y \) in terms of \( x \).

When integrating, remember these key points:

• The integral of \( x^m \) with respect to \( x \) is \( \frac{x^{m+1}}{m+1} + C \), where \( C \) is the constant of integration.
• The integral of \( \frac{1}{y^n} \) with respect to \( y \) is \( \frac{y^{1-n}}{1-n} + C \).

In the solution, each side of the separated equation \( \frac{1}{y^n} \, dy = x^m \, dx \) is integrated independently, yielding the expressions \( \frac{y^{1-n}}{1-n} + C_1 \) and \( \frac{x^{m+1}}{m+1} + C_2 \). Ensuring accurate integration is critical to deriving correct solutions.

The general solution of a differential equation includes all possible solutions, incorporating an arbitrary constant typically denoted as \( C \). This solution arises from integrating the differential equation, as integration does not yield a single answer, but rather a family of solutions that can be tailored by changing \( C \).

For the given problem, after integrating, we find the general solution is:\[y^{1-n} = (1-n) \frac{x^{m+1}}{m+1} + C (1-n)\]This reflects all potential solutions for different values of \( C \). Finally, simplifying this expression gives us:\[y = \left( (1-n) \frac{x^{m+1}}{m+1} + C \right)^{\frac{1}{1-n}}\]It's important to express the general solution clearly, explicitly showing the dependency of \( y \) on \( x \) through the term involving \( C \).

Variables separation is the method used to solve separable differential equations. It involves rearranging the equation to isolate \( y \) terms on one side and \( x \) terms on the other, making it possible to independently address each side with integration.

In practice, you often start with a differential equation like \( y' = x^m y^n \). By dividing both sides by \( y^n \) and multiplying both sides by \( dx \), you're able to rewrite it in a separable form: \( \frac{1}{y^n} \, dy = x^m \, dx \).

This transformation is powerful because it simplifies the differential equation into two simpler integral problems. This focused approach is the foundation for finding solutions to many differential equations, providing a systematic method to handle otherwise complex functions.