Problem 5 - Thomas Calculus

Question

Solve the differential equations in Exercises \(1-14\) $$ x \frac{d y}{d x}+2 y=1-\frac{1}{x}, \quad x>0 $$

Step-by-Step Solution

Answer

The solution is \( y = \frac{1}{2} - \frac{\ln|x|}{x^2} + \frac{C}{x^2} \).

Step 1: Rearrange the Equation

Start by expressing the given differential equation in a more convenient form:\[ x \frac{dy}{dx} + 2y = 1 - \frac{1}{x} \]To simplify, divide through by \(x\):\[ \frac{dy}{dx} + \frac{2}{x}y = \frac{1}{x} - \frac{1}{x^2} \]

Step 2: Identify Integrating Factor

The equation is now in standard linear form:\[ \frac{dy}{dx} + P(x)y = Q(x) \]Where \(P(x) = \frac{2}{x}\) and \(Q(x) = \frac{1}{x} - \frac{1}{x^2}\). The integrating factor \(\mu(x)\) is given by:\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{2}{x} \, dx} = e^{2 \ln|x|} = x^2 \]

Step 3: Multiply the Entire Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor \(x^2\):\[ x^2 \frac{dy}{dx} + 2xy = x - \frac{1}{x} \]

Step 4: Recognize the Left Side as a Derivative

The left side can be recognized as the derivative of \(x^2 y\):\[ \frac{d}{dx}(x^2 y) = x - \frac{1}{x} \]

Step 5: Integrate Both Sides

Integrate both sides with respect to \(x\):\[ \int \frac{d}{dx}(x^2 y) \, dx = \int \left( x - \frac{1}{x} \right) \, dx \]This yields:\[ x^2 y = \frac{1}{2}x^2 - \ln|x| + C \]

Step 6: Solve for \(y\)

After integration, solve for \(y\):\[ y = \frac{1}{2} - \frac{\ln|x|}{x^2} + \frac{C}{x^2} \]

Key Concepts

Integrating Factor

In solving linear differential equations, the integrating factor is a crucial tool. The integrating factor will convert a non-exact differential equation into an exact one, simplifying the solution process. It is generally denoted as \( \mu(x) \).
\( \mu(x) \) helps eliminate parts of the differential equation that are difficult to solve directly by involving the exponential function.

• The formula for calculating the integrating factor \( \mu(x) \) is:\[ \mu(x) = e^{\int P(x) \, dx} \]
• In our example, we're using \( P(x) = \frac{2}{x} \), resulting in:\[ \mu(x) = e^{\int \frac{2}{x} \, dx} = e^{2 \ln|x|} = x^2 \]

The integrating factor transforms the left-hand side of the differential equation into a product of derivatives, making integration straightforward. By making sure you use the correct integrating factor, you can ensure that the equation becomes much easier to handle.

Linear Differential Equations

Linear differential equations play a vital role in modeling various real-world problems. They consist of derivatives, linear terms in variable \( y \), and varying coefficients that can be functions of \( x \). These equations generally fit the form:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
The solution process involves using the integrating factor to simplify the equation.

• First, arrange the equation in its standard linear form to identify \( P(x) \) and \( Q(x) \).
• Next, compute the integrating factor.
• Multiply the entire equation by the integrating factor to transform the left side into the derivative of a product.

Once these steps are correctly followed, you can integrate both sides to find the solution. Understanding the structure of linear differential equations is fundamental in advanced mathematics.

Integration Techniques

Integration is the final step to finding the solution to a differential equation once it is in a simpler form. Integration techniques vary and depend on the nature of the functions involved.

For our example, after applying the integrating factor, we achieved:
\[ \frac{d}{dx}(x^2 y) = x - \frac{1}{x} \]
Integrating both sides with respect to \( x \) involves using simple integral rules:

• The left side integrates easily as it simplifies to \( x^2 y \).
• For the right side, integrate each term separately:\( \int x \, dx = \frac{1}{2}x^2 \) and \( \int -\frac{1}{x} \, dx = -\ln|x| \).

The results are combined to form the solution:
\[ x^2 y = \frac{1}{2}x^2 - \ln|x| + C \]
Solving for \( y \) involves rearranging terms to isolate \( y \). Integration is critical for turning a differential equation into an explicit form, thus completing the solution process.