Question

Solve the given differential equations. $$ \frac{d y}{d x}-\frac{y}{x}=-\frac{y^{2}}{x} $$

Short answer

The solution to the given differential equation is \(y(x) = \frac{1}{x^{-1} + C}\), where \(C\) is the integration constant.

Step-by-step solution

Rewrite the equation to identify a possible integrating factor

First, let's rewrite the given differential equation: \[ \frac{dy}{dx} - \frac{y}{x} + \frac{y^2}{x} = 0. \] We can recognize that this is a Bernoulli differential equation of the form: \[ \frac{dy}{dx} + P(x)y = Q(x)y^n, \] where \(P(x) = -\frac{1}{x}\), \(Q(x) = \frac{1}{x}\), and \(n=2\). To solve this equation, we can use a substitution of the form \(u(x) = y^{1-n}\). In our case, this gives: \[ u(x) = y^{-1}. \]

Multiply the equation by the integrating factor

Now, we will differentiate u(x) with respect to x and solve for the derivative of y(x). The chain rule gives us: \[ \frac{du}{dx} = -y^{-2} \frac{dy}{dx}. \] To solve for \(\frac{dy}{dx}\), we have: \[ \frac{dy}{dx} = -y^2 \frac{du}{dx}. \] We will replace the derivative of y(x) in the original differential equation with this expression and the substitution \(u(x) = y^{-1}\): \[ -y^2 \frac{du}{dx} - \frac{y}{x} + \frac{1}{u} = 0. \] Now we can see that this is a linear first-order differential equation in u(x). We can rewrite this equation as follows: \[ \frac{du}{dx} - \frac{1}{xu} = -\frac{1}{x}. \] To solve this equation, we will use an integrating factor \(\mu(x) = e^{\int -\frac{1}{x} dx}\), which simplifies to: \[ \mu(x) = x^{-1}. \] Multiplying both sides of the equation by the integrating factor gives: \[ x^{-1}\frac{du}{dx} - x^{-2}u = -x^{-2}. \]

Integrate both sides of the equation

Now, the left-hand side of this equation is an exact derivative of the product of the integrating factor and u(x): \[ \frac{d(u/x)}{dx} = -x^{-2}. \] Integrating both sides with respect to x gives: \[ \int \frac{d(u/x)}{dx}dx = \int -x^{-2} dx. \] On the left-hand side, the derivative and integral cancel out: \[ u/x = \int -x^{-2} dx. \] We can now perform the integration on the right-hand side: \[ u/x = -\int x^{-2}dx = x^{-1} + C, \] where C is the integration constant.

Solve for the function y(x)

Finally, we will solve for y(x). We recall the substitution: \[ u(x) = y^{-1}. \] So, \[ y^{-1} = x^{-1} + C, \] or \[ y(x) = \frac{1}{x^{-1} + C}. \] Thus, the solution to the given differential equation is: \[ y(x) = \frac{1}{x^{-1} + C}. \]

Key concepts

First-Order Differential Equation

A first-order differential equation involves an unknown function and its derivative. Such equations take the form \( \frac{dy}{dx} = f(x, y) \). Here, the rate of change \( \frac{dy}{dx} \) is expressed in terms of the independent variable \( x \) and the dependent variable \( y \). In our exercise, the equation \( \frac{dy}{dx} - \frac{y}{x} = -\frac{y^2}{x} \) is a first-order differential equation because it includes \( \frac{dy}{dx} \), which is the derivative of \( y \) with respect to \( x \).
To solve such equations, we may need different methods based on the structure of the equation. For Bernoulli-type equations, like the one given, identifying its specific structure helps us select a suitable solving method, such as the substitution method used here.

Integrating Factor

The integrating factor is a strategic multiplier applied in differential equations to simplify and solve them. The goal of finding an integrating factor is to transform a non-exact differential equation into an exact one, making it easier to integrate.
In our exercise, an integrating factor was used to solve the resulting linear first-order differential equation labeled in terms of \( u(x) \). We determined the integrating factor as \( \mu(x) = e^{\int -1/x\, dx} = x^{-1} \).

• Firstly, express the given differential equation in standard form.
• Calculate the integrating factor, \( \mu(x) \), which typically involves integration.
• Multiply the whole differential equation by the integrating factor.
• Finally, integrate both sides to find the solution.

This approach essentially 'smoothens out' the equation, making it more digestible for integration.

Substitution Method

The substitution method simplifies differential equations by transforming them into a different variable, making them easier to solve. It often involves introducing a new function to substitute either directly into the equation or through the variable's relationships, such as transforming \( y \) into \( u \) as \( u(x) = y^{1-n} \).
In Bernoulli equations, like the one in this exercise, we utilize the substitution of \( u(x) = y^{-1} \). This changes the nature of the equation from Bernoulli to linear, which is more straightforward to solve. Here is the substitution process breakdown:

• Identify a suitable substitution: \( u(x) = y^{1-n} \).
• Differentiate the substitution with respect to \( x \).
• Replace terms in the original equation with expressions in \( u \) and \( \frac{du}{dx} \).

This method significantly reduces the complexity of the task, enabling easier manipulation and integration.

Ordinary Differential Equations

Ordinary differential equations (ODEs) differ from partial differential equations as they involve one independent variable. ODEs like those studied here have solutions that are functions of a single variable, contrasting with partial differential equations, which involve multiple variables.
In the context of our exercise, the ordinary differential equation is governed by the equation \( \frac{dy}{dx} + P(x)y = Q(x)y^n \), which is identified as a Bernoulli equation. This classification guides us to an appropriate solving strategy, utilizing either integrating factors, substitutions, or other methods.
Ordinary differential equations are classified further based on their characteristics (e.g., linear, non-linear), guiding us to specific approaches for finding their solutions. Understanding these classifications is pivotal for determining the most efficient solving techniques. Thus, mastering ODEs involves recognizing their forms and applying the right strategies to find solutions efficiently.