Problem 25 - Calculus

Question

In Exercises \(2-28,\) use separation of variables to find the solutions to the differential equations subject to the given initial conditions. $$\frac{d z}{d t}=z+z t^{2}, \quad z=5 \text { when } t=0$$

Step-by-Step Solution

Answer

The solution is \( z = 5e^{t + \frac{t^3}{3}} \).

Step 1: Rewrite the Differential Equation

First, we rewrite the given differential equation \( \frac{d z}{d t} = z + z t^2 \) by factoring the right-hand side. This gives us \( \frac{d z}{d t} = z(1 + t^2) \). This expression shows that the differential equation can be separated.

Step 2: Separate the Variables

We can now separate the variables to isolate \( z \) on one side and \( t \) on the other. Divide both sides by \( z \) and multiply by \( dt \) to get: \( \frac{1}{z} dz = (1 + t^2) dt \).

Step 3: Integrate Both Sides

Integrate both sides of the equation from Step 2. The integral of \( \frac{1}{z} dz \) is \( \ln |z| + C_1 \) and the integral of \( (1 + t^2) dt \) is \( t + \frac{t^3}{3} + C_2 \). Thus, we have \( \ln |z| = t + \frac{t^3}{3} + C \), where \( C = C_2 - C_1 \).

Step 4: Solve for the Constant Using Initial Conditions

Given the initial condition \( z = 5 \) when \( t = 0 \), substitute these values into the equation \( \ln |z| = t + \frac{t^3}{3} + C \) to find \( C \): \( \ln(5) = 0 + \frac{0^3}{3} + C \), so \( C = \ln(5) \).

Step 5: Solve for the Explicit Solution

Substitute the value of \( C \) back into the equation to get \( \ln |z| = t + \frac{t^3}{3} + \ln(5) \). Exponentiate both sides to solve for \( z \): \( |z| = 5e^{t + \frac{t^3}{3}} \). Since \( z = 5 \) when \( t = 0 \), \( z \) is positive, so \( z = 5e^{t + \frac{t^3}{3}} \).

Key Concepts

Understanding Differential Equations

Differential equations are equations that involve a function and its derivatives. These equations are crucial in mathematics, especially when modeling real-world phenomena where rates of change are involved, like population growth or heat conduction.
A differential equation like \( \frac{dz}{dt} = z + zt^2 \) describes how one variable, \( z \), changes with respect to another variable, \( t \). Here, the equation indicates that the change in \( z \) is affected by both its current value and the square of \( t \).

There are several ways to solve differential equations, and one common method is separation of variables. This technique allows us to rewrite the equation in a form where each variable is on one side of the equation. This separation makes it easier to find a solution through integration.

Clarifying Initial Conditions

When solving differential equations, initial conditions provide specific values that the solution must satisfy at a given point. These conditions help us to find a unique solution to the differential equation where multiple general solutions could be possible without them.
In our example, an initial condition is provided: \( z = 5 \) when \( t = 0 \). This tells us the value of \( z \) at the specific time \( t = 0 \). By using this initial condition, you can determine the constant of integration that typically appears when solving a differential equation. Without the initial condition, we can only find a family of solutions rather than a single explicit one.

The Process of Integration

Integration is a fundamental concept in calculus that involves finding the antiderivative of a function. In the context of solving differential equations, integration allows us to go from a rate of change (given by a derivative) to the original function.

For the given problem, once the equation is separated as \( \frac{1}{z} dz = (1 + t^2) dt \), we integrate each side:

• The integral of \( \frac{1}{z} dz \) is \( \ln |z| \) because the antiderivative of \( \frac{1}{z} \) is the natural logarithm.
• The integral of \( 1 + t^2 \) is \( t + \frac{t^3}{3} \) because each term is integrated separately.

These integrals are then equated, forming the basis for finding the solution to the original differential equation.

Exponential Functions and Their Role

Exponential functions are a key part of many differential equations. This is because they frequently describe natural growth or decay processes, as in biological populations or radioactive decay.
When we solve the given differential equation and arrive at the expression \( \ln |z| = t + \frac{t^3}{3} + \ln(5) \), we exponentiate both sides to isolate \( z \).

The exponential function \( e^{x} \) reverses the natural logarithm \( \ln(x) \). As a result, we get:

• \( |z| = 5e^{t + \frac{t^3}{3}} \)
• Since \( z \) is given as positive for \( t = 0 \), we drop the absolute value, resulting in \( z = 5e^{t + \frac{t^3}{3}} \)

This exponential form succinctly captures the behavior of \( z \) over time \( t \), illustrating how it evolves following the initial conditions and the differential equation.