The families of curves represented by the solution of the equation

$\frac{dy}{dx}=-{\left(\frac{x}{y}\right)}^n$

for n = −1 and n = +1, respectively, are

To determine the families of curves represented by the given differential equation, we need to solve the equation for two specific values of $n$: $n=-1$ and $n=+1$.

Differential Equation Analysis

The general form of the differential equation provided is:

$\frac{dy}{dx}=-{\left(\frac{x}{y}\right)}^n$

We will proceed by analyzing this differential equation for each specified value of $n$.

Solution for n = −1

When $n=-1$, the differential equation transforms into:

$\frac{dy}{dx}=-{\left(\frac{x}{y}\right)}^{-1}$

Simplifying the exponent, we utilize the property $a^{-1}=1/a$:

$\frac{dy}{dx}=-\frac{y}{x}$

This is a separable differential equation. To solve it, we rearrange the terms to gather $y$ with $dy$ and $x$ with $dx$:

$\frac{dy}{y}=-\frac{dx}{x}$

Next, we integrate both sides of the equation:

$\int \frac{dy}{y}=\int -\frac{dx}{x}$

Performing the integration yields:

$\ln |y|=-\ln |x|+C$

Where $C$ represents the constant of integration. Using the logarithm property $\ln (a)+\ln (b)=\ln (ab)$, we can combine the logarithmic terms:

$\ln |y|+\ln |x|=C$

$\ln |xy|=C$

To eliminate the natural logarithm, we exponentiate both sides with base $e$:

$e^{\ln |xy|}=e^C$

$|xy|=e^C$

Let $A=\pm e^C$, where $A$ is an arbitrary non-zero constant. This gives us the equation:

$xy=A$

The equation $xy=A$ is the standard form for a family of hyperbolas. These hyperbolas have the coordinate axes as their asymptotes.

Solution for n = +1

When $n=+1$, the differential equation is given as:

$\frac{dy}{dx}=-{\left(\frac{x}{y}\right)}^1$

This simplifies to:

$\frac{dy}{dx}=-\frac{x}{y}$

This is also a separable differential equation. We separate the variables by cross-multiplication:

$ydy=-xdx$

Now, we integrate both sides of the equation:

$\int ydy=\int -xdx$

Performing the integration, we get:

$\frac{y^2}{2}=-\frac{x^2}{2}+C^{\prime }$

Where $C^{\prime }$ is the constant of integration. To clear the denominators, we multiply the entire equation by 2:

$y^2=-x^2+2C^{\prime }$

Rearranging the terms to bring $x^2$ and $y^2$ to one side:

$x^2+y^2=2C^{\prime }$

Let $R^2=2C^{\prime }$, where $R^2$ is a positive constant, since the sum of squares $x^2+y^2$ must be non-negative. The equation then takes the form:

$x^2+y^2=R^2$

The equation $x^2+y^2=R^2$ represents a family of circles centered at the origin with radius $R$.

Families of Curves Summary

To summarize the results from solving the differential equation for the specified values of $n$:

- For $n=-1$, the solution $xy=A$ describes a family of hyperbolas.
- For $n=+1$, the solution $x^2+y^2=R^2$ describes a family of circles.

Thus, the families of curves are Hyperbolas and Circles, respectively.