91Ó°ÊÓ

Separation of variables is a crucial technique used to solve differential equations, especially when the equation can be expressed as a product of functions, each depending only on one variable. In the given problem, we started with the equation:

• \( \frac{du}{dt} = 2 + 2u + t + tu \)

By factoring the right-hand side, this equation becomes:

• \( \frac{du}{dt} = (2 + t)(1 + u) \)

This factorization was necessary to separate the terms with variables \( u \) and \( t \). The goal is to organize the equation such that all terms involving \( u \) are isolated on one side and terms involving \( t \) are on the other side:

• \( \frac{du}{1 + u} = (2 + t) \, dt \)

This step paves the way for integrating both sides individually according to their respective variables. It transforms the solution of a differential equation into two simpler integrals.

Once the variables are separated, the next step is to integrate both sides of the equation. Integration is the process of finding the integral of a function, which is essentially the reverse of differentiation.

• For the left side, we integrate \( \int \frac{du}{1 + u} \), which results in \( \ln|1 + u| \).
• For the right side, integrate \( \int (2 + t) \, dt \), leading to \( 2t + \frac{t^2}{2} + C \).

Combining these, you get:

• \( \ln|1 + u| = \frac{t^2}{2} + 2t + C \)

The constant \( C \) represents the arbitrary constant of integration that appears because integration is an indefinite process. This step proves how integration transforms separated variables into a form that can be solved for the unknown function.

The exponential function often appears in the solutions of differential equations, especially when dealing with integration involving logarithmic results. After integrating our equation, we arrive at:

• \( \ln|1 + u| = \frac{t^2}{2} + 2t + C \)

To eliminate the natural logarithm and solve for \( u \), we exponentiate both sides:

• \( |1 + u| = e^{\frac{t^2}{2} + 2t + C} \)

This expression utilizes the properties of the exponential function, transforming the equation into a form that allows us to solve explicitly for \( u \). By setting \( A = \pm e^C \), we simplify further to:

• \( u = A e^{\frac{t^2}{2} + 2t} - 1 \)

The exponential function is vital because it helps transform the logarithmic relationship into a solvable expression.

Initial conditions are key to determining specific solutions to differential equations. While differential equations yield a general family of solutions, initial conditions specify a particular solution by settling unknown constants like \( A \) in our case.

• The constant \( A \) in \( u(t) = A e^{\frac{t^2}{2} + 2t} - 1 \) is determined from initial conditions such as \( u(t_0) = u_0 \).

Given a specific initial condition, we substitute the initial values of \( t_0 \) and \( u_0 \) into the general solution to solve for \( A \). This essential technique helps pinpoint the exact solution within the family of solutions that satisfies the conditions at a particular point, making it very practical in scientific and engineering applications.