91Ó°ÊÓ

First-order linear differential equations can often be simplified using an integrating factor. This method helps transform a non-exact equation into an exact one. For a differential equation of the form \[ y' + P(x) y = Q(x) \], you can find the integrating factor \( \mu(x) \) using the formula: \( \mu(x) = e^{\int P(x) \dx} \).

In our problem, we identified \( P(x) = \frac{3}{x} \). Integrating this yields the integrating factor:
\[ \mu(x) = e^{\int \frac{3}{x} \dx} = e^{3 \ln x} = x^3 \]

Multiplying both sides of the equation by this integrating factor enables us to recognize the left-hand side as a derivative. This process simplifies the equation significantly, making it easier to solve.

Integration by parts is a useful technique for solving integrals involving the product of two functions. The formula for integration by parts is:
\[ \int u \, dv = uv - \int v \, du \]

For our integral \( \int x^3 \ln x \, dx \), we set:

• \( u = \ln x \)
• \( dv = x^3 \ dx \)

The corresponding derivatives are:

• \( du = \frac{1}{x} \ dx \)
• \( v = \frac{x^4}{4} \)

Applying the integration by parts formula leads to:
\[ \int x^3 \ln x \, dx = \ln x \cdot \frac{x^4}{4} - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx = \frac{x^4 \ln x}{4} - \int \frac{x^3}{4} \, dx\]

Simplifying further:
\[ \int x^3 \ln x \, dx = \frac{x^4 \ln x}{4} - \frac{x^4}{16} + C \]

This result simplifies our solution process by breaking down the problem into manageable steps.

Solving first-order linear differential equations involves several steps:

• Rewriting the equation in standard form \( y' + P(x) y = Q(x) \)
• Finding the integrating factor \( \mu(x) \)
• Multiplying through by this factor
• Recognizing the left-hand side as a derivative
• Integrating both sides

For our example \( x y^{\prime}+3 y=x \ln x \), after steps 1-3, we get:
\[ \frac{d}{dx} (y x^3) = x^3 \ln x \]
Integrating both sides, we apply integration by parts to solve:
\[ y x^3 = \frac{x^4 \ln x}{4} - \frac{x^4}{16} + C \]

Finally, dividing by \( x^3 \) isolates \( y \):
\[ y = \frac{x \ln x}{4} - \frac{x}{16} + \frac{C}{x^3} \]

By following these steps, we can systematically tackle and solve first-order linear differential equations, making the process less daunting.