91Ó°ÊÓ

Separation of Variables is a powerful technique used to solve differential equations. It involves rearranging the equation so that each variable is on a different side of the equation. This makes it easier to focus on one variable at a time while solving the equation. For the differential equation \( \frac{d z}{d t} = z + z t^2 \), we start by factoring the common term \( z \) from the right-hand side, yielding \( z(1 + t^2) \).

Next, we divide both sides by \( z \) and multiply both sides by \( dt \) to separate the variables completely. This gives us \( \frac{1}{z}dz = (1+t^2)dt \). Now, the left side of the equation depends only on \( z \) and the right side only on \( t \). With the variables separated like this, we can make progress by integrating both sides separately.

Integration is the next step after separating the variables. It involves finding the antiderivative or the integral of each side of the equation. For our separated equation \( \frac{1}{z}dz = (1+t^2)dt \), we integrate both sides. The left side of the equation, \( \int \frac{1}{z} \, dz \), integrates to \( \ln|z| + C_1 \), while the right side, \( \int (1 + t^2) \, dt \), results in \( t + \frac{t^3}{3} + C_2 \).

This integration transforms our separated variables back into a connected equation, \( \ln|z| = t + \frac{t^3}{3} + C \), where \( C = C_2 - C_1 \) is the constant of integration. Integration helps us to find a general solution to the differential equation.

Initial conditions provide specific values for a differential equation at a particular point, helping us determine the arbitrary constants in our general solution. In this exercise, the initial condition given is \( z = 5 \) when \( t = 0 \). By substituting these values into the general solution \( \ln|z| = t + \frac{t^3}{3} + C \), we solve for the constant \( C \).

Rewriting in exponential form, the equation is \( z = e^C e^{t + \frac{t^3}{3}} \), and by using the initial condition, \( z = 5 \), we substitute \( t = 0 \) ensuring \( z = 5 = A e^0 \), which implies that \( A = 5 \). Thus, the specific solution is \( z = 5 e^{t + \frac{t^3}{3}} \). Initial conditions are essential to refining a general solution to meet real-world criteria.

Exponential Functions frequently appear in solutions to differential equations. Once we integrate the variables, we often end up with natural logarithms, such as \( \ln |z| \), that lead to exponential forms upon solving for the variable \( z \).

For the integrated equation \( \ln|z| = t + \frac{t^3}{3} + C \), exponentiating both sides yields \( z = e^C e^{t + \frac{t^3}{3}} \). This equation illustrates how exponentials can model continuous growth or decay processes frequently found in real-world applications.

With the determined constant \( A = 5 \), we see that \( z = 5 e^{t + \frac{t^3}{3}} \) describes how \( z \) changes over time. Thus, exponential functions are crucial for expressing solutions involving growth rates or dynamic changes in such systems.