91Ó°ÊÓ

In solving first-order linear differential equations like the one given, the integrating factor is a crucial tool. The integrating factor is a function, often denoted as \( \mu(t) \), used to simplify and solve differential equations. For an equation of the form \( \frac{dr}{dt} + a(t)r = b(t) \), the integrating factor is calculated as \( \mu(t) = e^{\int a(t) \, dt} \).

In our problem, we find that \( a(t) = \frac{1}{t \ln t} \). So, we calculate:

• \( \int \frac{1}{t \ln t} \, dt = \ln(\ln t) \)
• Thus, \( \mu(t) = e^{\ln(\ln t)} = \ln t \)

By multiplying the entire differential equation by this integrating factor, we transform the equation into a simpler form. This allows us to express the left-hand side as the derivative of a product, making it straightforward to integrate both sides of the equation. This method essentially "factors" the differential equation into a form that is easier to integrate directly.

A first-order linear differential equation typically involves only the first derivative of the unknown function, which in this example problem is \( r(t) \). These equations have a general structure:

• \( \frac{dr}{dt} + a(t)r = b(t) \)

The challenge is to find the function \( r(t) \) that satisfies this equation.

Our original equation, \( t \ln t \frac{dr}{dt} + r = t e^t \), fits this pattern once we rearrange it:

• Rewritten as: \( \frac{dr}{dt} + \frac{1}{t \ln t} r = e^t \)

This standard linear form allows us to apply techniques such as the integrating factor method to find the solution. Identifying the coefficient functions \( a(t) \) and \( b(t) \) is the key initial step for applying the integrating factor and solving the equation.

Integration by parts is a technique used to simplify the integration of products of functions. It is based on the formula:

• \( \int u \, dv = uv - \int v \, du \)

This method is particularly useful when one function in the product becomes simpler when differentiated or when its integral is easily calculable.

In our equation, we reached a point where we needed to integrate \( \int \ln t \cdot e^t \, dt \). Here's how we apply integration by parts:

• Choose \( u = \ln t \), so \( du = \frac{1}{t} \, dt \)
• Let \( dv = e^t \, dt \), so \( v = e^t \)
• Apply the formula: \( \int \ln t \cdot e^t \, dt = (\ln t \cdot e^t) - \int e^t \frac{1}{t} \, dt \)

This results in an equation that can be solved step-by-step, enabling us to express the integral as a result in terms that incorporate the original functions \( t \) and \( e^t \). This method is often used in solving differential equations where simple antiderivatives are not readily available.