Question

For the differential equation \(\frac{d y}{d x}-\frac{y}{x}=x^{2}, x>0\), the integrating factor is \(e^{\int(-1 / x) d x} .\) The general antiderivative \(\int\left(-\frac{1}{x}\right) d x\) is equal to \(-\ln x+C .\) (a) Multiply both sides of the differential equation by \(\exp \left(\int\left(-\frac{1}{x}\right) d x\right)=\exp (-\ln x+C), \quad\) and show that \(\exp (-\ln x+C)\) is an integrating factor for every value of \(C .\) (b) Solve the resulting equation for \(y\), and show that the solution agrees with the solution obtained when we assumed that \(C=0\) in the integrating factor.

Short answer

The integrating factor works for any \(C\), and the solution is \(y = \frac{x^3}{2} + C_1x\).

Step-by-step solution

Understanding the Problem

We are given a differential equation \(\frac{dy}{dx} - \frac{y}{x} = x^2\) with an integrating factor \(e^{\int(-1/x) dx}\). We need to show that \(e^{(-\ln x + C)}\) is an integrating factor and solve the equation for \(y\).

Simplifying the Integrating Factor

The integrating factor is given by \(e^{\int(-1/x) dx} = e^{-\ln x + C}\). Simplifying this expression, we have: \[ e^{-\ln x + C} = e^{-\ln x} \cdot e^{C} = \frac{e^C}{x}. \]This shows that for any \(C\), \(\frac{e^C}{x}\) is an integrating factor.

Multiplying the Differential Equation

Multiply both sides of the differential equation by the integrating factor \(\frac{e^C}{x}\):\[ \frac{e^C}{x} \left( \frac{dy}{dx} - \frac{y}{x} \right) = \frac{e^C}{x} \cdot x^2. \]Simplifying both sides gives:\[ \frac{e^C}{x} \cdot \frac{dy}{dx} - \frac{e^Cy}{x^2} = e^C \cdot x. \]

Recognizing the Exact Differential

Notice that the left side is the derivative of \(\frac{e^C}{x} \cdot y\):\[ \frac{d}{dx} \left( \frac{e^C}{x} \cdot y \right) = e^C \cdot x. \]This confirms that the integrating factor makes the equation exact.

Solving the Exact Differential Equation

Integrate both sides with respect to \(x\):\[ \int \frac{d}{dx} \left( \frac{e^C}{x} \cdot y \right) \, dx = \int e^C \cdot x \, dx. \]This gives:\[ \frac{e^C}{x} \cdot y = \frac{e^C}{2}x^2 + C_1. \]

Solving for y

Solve for \(y\) by multiplying both sides by \(x/e^C\):\[ y = \frac{x}{e^C} \left( \frac{e^C}{2}x^2 + C_1 \right) = \frac{x^3}{2} + C_1x. \]

Solution verification when C = 0

Assume \(C = 0\), then the integrating factor becomes \(\frac{1}{x}\) and the solution is:\[ y = \frac{x^3}{2} + C_1x. \]This matches the solution obtained earlier, confirming that the solution is consistent regardless of \(C\).

Key concepts

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives.
They are used to model real-world phenomena where quantities change with respect to one another.
For instance, they can describe how populations grow or how heat distributes across a surface.

In any differential equation, the main goal is to find a function, often denoted by \( y \), that satisfies the relationship defined by the equation.
The process involves understanding how changes in one variable are related to changes in another, which is expressed through derivatives like \( \frac{dy}{dx} \).

There are various types of differential equations, including ordinary differential equations (ODEs), which involve functions of a single variable and their derivatives, and partial differential equations (PDEs), which involve multiple variables.
In this context, we're dealing with an ODE.

Exact Differential Equation

An exact differential equation is a type of differential equation that can be written in a form where one side is the exact derivative of a function.
This means there exists a function \( F(x, y) \) such that the differential \( dF = M(x, y) \, dx + N(x, y) \, dy = 0 \) becomes zero only on differentiating \( F \).

Having an equation in this form allows us to find a solution efficiently.
When we multiply a differential equation by an integrating factor, it can sometimes transform into an exact differential equation.
This transformation simplifies the equation, making it easier to solve.

• An integrating factor is a function that, when multiplied by the equation, allows such a transformation.
• In our example, the integrating factor is \( \exp(-\ln x + C) \), which was used to make the given differential equation exact.

Solving Differential Equations

Solving differential equations involves finding the unknown function that satisfies the equation.
This process can vary greatly depending on the type and complexity of the differential equation.

For an exact differential equation, once the integrating factor is applied, the equation can be rewritten in terms of an exact derivative.
In our example, we transformed the original equation by multiplying with \( \frac{e^C}{x} \).
This showed that the left side was the derivative of \( \frac{e^C}{x} \cdot y \).

By recognizing this, we can easily integrate both sides to find the solution for \( y \).
This technique greatly simplifies the process of solving differential equations, especially when dealing with real-world applications.
The solution helps us uncover the relationship between variables described by the differential equation, aligning with the given conditions.

Integration

Integration is the process of finding the antiderivative or the integral of a function.
In the context of differential equations, integrating allows us to find the function \( y \) from its derivative.

Once we have an exact differential in the form of \( \frac{d}{dx} (\text{expression}) \), we integrate both sides to solve for \( y \).

• The integral of a derivative gives back the function itself, plus an integration constant.
• In our task, after applying the integrating factor, we integrated \( e^C \, x \), which yielded \( \frac{e^C}{2}x^2 + C_1 \).

This process is crucial for solving differential equations as it links the derivative structure provided by the differential equation with a function on \( y \).
Mastering integration techniques is essential when working with differential equations, as it unlocks the function behind observed rates of change.