Problem 22 - Calculus

Question

In Exercises \(2-28,\) use separation of variables to find the solutions to the differential equations subject to the given initial conditions. \(\frac{d z}{d t}=t e^{z}, \quad\) through the origin

Step-by-Step Solution

Answer

The solution is \(z = -\ln(1 - \frac{t^2}{2})\).

Step 1: Understanding the Problem

We need to solve the differential equation \(\frac{d z}{d t} = t e^{z}\) using the method of separation of variables with the initial condition that the solution passes through the origin, \((0,0)\). This means the solution should satisfy \(z(0) = 0\).

Step 2: Separating Variables

First, we express all terms involving \(z\) on one side and all terms involving \(t\) on the other side by dividing both sides by \(e^{z}\) and multiplying by \(dt\), resulting in \(\frac{1}{e^{z}} \, dz = t \, dt\).

Step 3: Integrating Both Sides

Integrate both sides of the equation: \(\int \frac{1}{e^{z}} \, dz = \int t \, dt\). The left side integrates to \(-e^{-z}\), and the right side integrates to \(\frac{t^2}{2}\).

Step 4: Adding the Constant of Integration

Include the constant of integration \(C\) after integrating: \(-e^{-z} = \frac{t^2}{2} + C\).

Step 5: Applying the Initial Condition

Using the initial condition \(z(0) = 0\), substitute \(t = 0\) and \(z = 0\) into the equation: \(-e^{0} = 0 + C\). This simplifies to \(-1 = C\).

Step 6: Substitute the Constant Back

Substitute \(C = -1\) back into the equation: \(-e^{-z} = \frac{t^2}{2} - 1\).

Step 7: Solving for z

To solve for \(z\), rearrange the equation: \(e^{-z} = 1 - \frac{t^2}{2}\). Taking the natural log on both sides, we have \(-z = \ln(1 - \frac{t^2}{2})\), which yields the solution by multiplying both sides by \(-1\): \(z = -\ln(1 - \frac{t^2}{2})\).

Key Concepts

Differential Equations

Differential equations are mathematical equations involving derivatives of a function. They can describe numerous physical phenomena, such as motion, heat, or waves. In our problem, we deal with a first-order differential equation given by \( \frac{d z}{d t} = t e^{z} \). This equation relates the rates at which the variables \( z \) and \( t \) change. The goal is to find a function \( z(t) \) that satisfies this relationship. Differential equations are powerful tools for modeling because they allow us to frame a dynamic scenario mathematically and find unknown functions that dictate these changing processes.

Initial Conditions

Initial conditions are specific values given for the solution of a differential equation at a certain point, typically to ensure a unique solution. In the problem at hand, the initial condition is \( z(0) = 0 \), indicating that when \( t = 0 \), \( z \) should also be zero. This condition constrains the potential solutions and provides a particular solution from a family of possible functions. By applying the initial condition, we can determine any arbitrary constants that appear when integrating, ensuring that our solution is not just general but specific to the scenario described.

Integration

Integration is a fundamental tool in solving differential equations, particularly when using the method of separation of variables. Once we've separated the terms involving \( z \) and \( t \, \), the next step is to integrate both sides of the equation. For this exercise, we integrate \( \int \frac{1}{e^{z}} \, dz \) and \( \int t \, dt \). The integration process helps to "undo" the differentiation and find the function \( z(t) \). Specifically, for \( \frac{1}{e^{z}} \, dz \), integration gives us \( -e^{-z} \), while \( t \, dt \) integrates to \( \frac{t^2}{2} \). These integrals play crucial roles in reaching a solution that accounts for the change rates described in the original differential equation.

Exponential Functions

Exponential functions are a critical component of this problem, particularly because our differential equation involves \( e^{z} \). Exponential functions have special properties, such as their unique ability to model growth or decay processes. In this exercise, solving \( \frac{d z}{d t} = t e^{z} \) requires handling the exponential term carefully. The integrating factor \( e^{-z} \) comes into play during the solution process, allowing us to simplify the equation into a form that can be solved via integration. Furthermore, when solving for \( z \), we encounter natural logs, which are the inverse operations of exponentials, further showcasing their intertwined nature in calculus and particularly in solving differential equations.