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The concept of an integrating factor is crucial for solving first-order linear differential equations. These equations are typically structured as \( \frac{du}{dt} + P(t)u = Q(t) \). The integrating factor is specifically designed to simplify these equations into an easily solvable form.

To find the integrating factor, the formula \( \mu(t) = e^{\int P(t) \, dt} \) is used. For the given equation \( \frac{du}{dt} + \frac{k}{m} u = 0 \), we identified \( P(t) = \frac{k}{m} \). By integrating \( P(t) \), the integrating factor is calculated as \( e^{\frac{k}{m}t} \).

Once you have the integrating factor, multiply it through the whole differential equation. This transforms the equation into a perfect derivative: \[ \frac{d}{dt}[u \, e^{\frac{k}{m}t}] = 0 \].
This form is straightforward to integrate because the equation now implies that \( u \, e^{\frac{k}{m}t} \) is a constant with respect to \( t \). Solving it gives you the general solution for \( u(t) \).

An initial value problem involves a differential equation alongside an initial condition, which determines a unique solution.

In this problem, the initial condition is \( u(0) = u_0 \), meaning when \( t = 0 \), the value of \( u \) is \( u_0 \).
This allows us to solve for the unknown constant in the general solution obtained using the integrating factor method.

When substituting \( t = 0 \) in the general solution \( u(t) = C e^{-\frac{k}{m}t} \), we get \( u_0 = C e^{0} \). Simplifying, this gives \( C = u_0 \), allowing us to replace \( C \) in our solution. Thus, the unique solution to the initial value problem is \( u(t) = u_0 e^{-\frac{k}{m}t} \).

Without this initial condition, \( C \) could be any value, and we would have infinite possible solutions to the differential equation.

Separable differential equations are a class of differential equations where the variables can be separated on either side of the equation. They are expressed in a form that allows the separation of variables for integration.

In the given problem, we rewrite the equation from \( \frac{du}{dt} + \frac{k}{m} u = 0 \) to \( \frac{du}{dt} = -\frac{k}{m} u \). Now, it's clear how we can segregate the variables:

- Place all terms involving \( u \) on one side: \( \frac{1}{u} \, du \)
- Place all terms involving \( t \) on the other side: \( -\frac{k}{m} \, dt \)

Integrating both sides results in \( \ln |u| = -\frac{k}{m}t + C \).
To resolve \( u \), exponentiate to remove the natural log:
\[ |u| = e^{-\frac{k}{m}t + C} = e^C e^{-\frac{k}{m}t} \]
The solution becomes \( u = C' e^{-\frac{k}{m}t} \), where \( C' = e^C \).

By applying the initial condition \( u(0) = u_0 \), substituting back solves the equation for \( C' \), giving \( C' = u_0 \). Thus, the solution becomes \( u(t) = u_0 e^{-\frac{k}{m}t} \). This demonstrates how separable equations are effective for solving differential equations by utilizing integrable forms.