Question

Find a general solution for the differential equation \(x y^{\prime}+3 y=4 x^{2}-3 x .\) Assume \(x>0\).

Short answer

The general solution is \( y = \frac{2}{3}x^2 - \frac{3}{5}x + \frac{C}{x^4} \), where \( x > 0 \).

Step-by-step solution

Identify the Type of Differential Equation

The given differential equation is \( x y' + 3y = 4x^2 - 3x \). This is a first-order linear differential equation of the form \( a(x)y' + b(x)y = f(x) \), where \( a(x) = x \), \( b(x) = 3 \), and \( f(x) = 4x^2 - 3x \).

Solve for the Integrating Factor

The integrating factor, \( \mu(x) \), is given by:\[\mu(x) = e^{\int \frac{b(x)}{a(x)} \, dx} = e^{\int \frac{3}{x} \, dx} = e^{3 \ln |x|} = |x|^3.\] Since \( x > 0 \), we have \( \mu(x) = x^3 \).

Multiply the Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( x^3 \):\[x^3(x y' + 3y) = x^3(4x^2 - 3x).\]This simplifies to:\[x^4 y' + 3x^3 y = 4x^5 - 3x^4.\]

Recognize the Exact Equation

Notice that the left side is the derivative of \( x^4 y \), making the equation:\[\frac{d}{dx}(x^4 y) = 4x^5 - 3x^4.\]

Integrate Both Sides

Integrate both sides with respect to \( x \):\[\int \frac{d}{dx}(x^4 y) \, dx = \int (4x^5 - 3x^4) \, dx.\]The left side integrates to \( x^4 y \). The right side evaluates to:\[\frac{4}{6}x^6 - \frac{3}{5}x^5 + C = \frac{2}{3}x^6 - \frac{3}{5}x^5 + C.\]

Solve for y

Solve for \( y \):\[x^4 y = \frac{2}{3}x^6 - \frac{3}{5}x^5 + C.\]Dividing through by \( x^4 \) gives:\[y = \frac{2}{3}x^2 - \frac{3}{5}x + \frac{C}{x^4}.\]

Final Answer

The general solution of the differential equation is:\[y = \frac{2}{3}x^2 - \frac{3}{5}x + \frac{C}{x^4},\quad x > 0.\]

Key concepts

Integrating Factor

In solving first-order linear differential equations, the integrating factor is a pivotal concept. An integrating factor is used to transform such equations into a format that can be easily solved. For a differential equation of the form \(a(x)y' + b(x)y = f(x)\), we find the integrating factor \(\mu(x)\) using the formula:

• \(\mu(x) = e^{\int \frac{b(x)}{a(x)} \, dx}\)

This expression is derived from the concept that multiplying through by the integrating factor allows us to convert the left side of the equation into the derivative of a product.
In our exercise, \(a(x) = x\) and \(b(x) = 3\), so the integrating factor becomes \(e^{\int \frac{3}{x} \, dx}\). Simplifying this gives \(|x|^3\), and for \(x > 0\), it simplifies further to \(x^3\).
Applying this integrating factor, \(x^3\), to the differential equation enables the transformation of the equation into one that is more straightforward to solve.

General Solution

The concept of a general solution in differential equations entails finding a solution that contains all possible particular solutions of the equation. This typically involves solving an equation with one or more arbitrary constants, usually denoted by \(C\).
After applying the integrating factor, we turned our original differential equation into an equation that could be integrated as a whole derivative:

• \(\frac{d}{dx}(x^4 y) = 4x^5 - 3x^4\)

By integrating both sides with respect to \(x\), and solving for \(y\), we find:

• \(y = \frac{2}{3}x^2 - \frac{3}{5}x + \frac{C}{x^4}\)

This is the general solution for the differential equation, meaning it accounts for any \(x > 0\) and includes an arbitrary constant \(C\) which can be adjusted based on additional conditions, if provided.

Exact Equation

The term "exact equation" refers to a differential equation derived from the derivative of a function such that the equation can be seen as a perfect derivative of two variables. In our problem, after multiplying by the integrating factor \(x^3\), we obtained:

• \(x^4 y' + 3x^3 y = 4x^5 - 3x^4\)

Recognizing that the left side is the derivative of \(x^4 y\), we confirmed that the equation reached a form:

• \(\frac{d}{dx}(x^4 y) = 4x^5 - 3x^4\)

This realization was crucial, as it allowed us to integrate directly. An equation of this type can be efficiently solved, ensuring that the method applied correctly captures the essence of the original problem. Thus, recognizing and working with exact equations simplifies the process of finding the general solution.