Question

Solve the following differential equations by using integrating factors. $$ x y^{\prime}=3 y+x^{2} $$

Short answer

The solution to the differential equation is \( y = -x^2 + Cx^3 \).

Step-by-step solution

Simplify the Given Equation

First, we need to rearrange the given differential equation, \( x y^{\prime}=3 y+x^{2} \) to get it into a standard form. Dividing through by \( x \) gives \( y^{\prime} - \frac{3}{x}y = x \). This is now in the form of a linear first-order differential equation.

Identify the Integrating Factor

To solve the differential equation using an integrating factor, we need to identify it. The standard linear equation form is \( y' + P(x)y = Q(x) \). Here, \( P(x) = -\frac{3}{x} \). The integrating factor, \( \mu(x) \), is given by \( e^{\int P(x) \, dx} \), which becomes \( e^{-\int \frac{3}{x} \, dx} = e^{-3 \ln |x|} = |x|^{-3} \). We simplify this to \( \frac{1}{x^3} \) assuming \( x \) is positive.

Multiply Through by the Integrating Factor

Multiply the entire differential equation \( y^{\prime} - \frac{3}{x}y = x \) by the integrating factor \( \frac{1}{x^3} \). This transforms the equation to: \( \frac{1}{x^3}y^{\prime} - \frac{3}{x^4}y = \frac{x}{x^3} \), which becomes \( \frac{1}{x^3}y' - \frac{3}{x^4}y = \frac{1}{x^2} \).

Recognize the Left-Hand Side as a Derivative

The left-hand side of the equation \( \frac{1}{x^3}y' - \frac{3}{x^4}y \) can be identified as the derivative of a product: \( \frac{d}{dx}(\frac{1}{x^3}y) \). Thus, our equation now reads: \( \frac{d}{dx}(\frac{1}{x^3}y) = \frac{1}{x^2} \).

Integrate Both Sides

To solve for \( y \), integrate both sides of the equation with respect to \( x \): \[ \int \frac{d}{dx}(\frac{1}{x^3}y) \, dx = \int \frac{1}{x^2} \, dx \]The left side simplifies to \( \frac{1}{x^3}y \), and the right side integrates to \( -\frac{1}{x} + C \), where \( C \) is the constant of integration.

Solve for y

Now that we have \( \frac{1}{x^3}y = -\frac{1}{x} + C \), multiply both sides by \( x^3 \) to solve for \( y \): \[ y = -x^2 + Cx^3 \] This is the general solution to the differential equation.

Key concepts

First-Order Differential Equations

A first-order differential equation is an equation that involves the derivatives of a function and the function itself. The term "first-order" means that the highest derivative in the equation is a first derivative. This type of differential equation takes the form \( y' = f(x, y) \). Here, \( y' \) indicates the derivative of \( y \) with respect to \( x \). First-order differential equations come in various forms, including separable, exact, and linear.

• Separable: These can be solved by isolating \( y \) and \( x \) on different sides of the equation.
• Exact: These require a specific condition to be met, and if so, can be solved through integration.
• Linear: This type is a focus in many introductory courses because it provides a more structured solution method.

First-order equations appear in various applications such as population dynamics, carbon dating, and simply describing certain physical processes like cooling or heating. Understanding how to approach these equations is foundational for tackling more complex types. By mastering these basics, students can effectively apply the concepts to real-world problems and more intricate mathematical equations.

Linear Differential Equations

Linear differential equations are a subset of first-order differential equations and have a very specific form: \( y' + P(x)y = Q(x) \). The key characteristic of these equations is the presence of terms involving only the first derivative \( y' \) and the function \( y \) itself, without products or powers of these terms. Solutions to linear differential equations often involve integrating factors, a powerful technique used to simplify and solve the equation.

The standard method to solve a linear first-order differential equation involves the following steps:

• Rearrange the equation into the standard linear form, ensuring all terms involving \( y \) and \( y' \) are on one side.
• Calculate the integrating factor, which is \( \mu(x) = e^{\int P(x) \, dx} \).
• Multiply the entire differential equation by this integrating factor. This transforms the left-hand side into a derivative of a product.
• Finally, integrate both sides to find the solution.

Utilizing the concept of linear differential equations allows students to solve a wide array of problems efficiently. The process helps highlight the elegance and utility of integration in differential equations.

Methods of Integration

Integration is a fundamental tool in solving differential equations. It involves finding a function whose derivative is equal to the given function. In the context of differential equations, integration is not just about reversing derivatives but is crucial in identifying relationships among different quantities.

Various methods of integration might be applied depending on the complexity of the differential equation:

• Basic Integration: In straightforward cases, simple antiderivatives are enough.
• Substitution: Useful when the integrand is amenable to substitution, simplifying the expression into a basic form.
• Partial Fractions: Often used when integrating rational functions, breaking them into simpler components.
• Integration by Parts: Particularly valuable when dealing with products of functions.

By understanding these various integration methods, students can choose the most suitable strategy for solving complex differential equations. This flexibility is essential for adapting to different forms and ensuring accurate solutions with correct application of calculus principles.