Problem 45 - Calculus

Question

Solve the differential equations in Problems \(44-47 .\) Assume \(x, y, t > 0\). $$\frac{d x}{d t}=\frac{x \ln x}{t}$$

Step-by-Step Solution

Answer

The solution is \( x = e^{At} \), where \( A \) is a constant.

Step 1: Identify the Type of Differential Equation

We have the differential equation \( \frac{dx}{dt} = \frac{x \ln x}{t} \). This is a separable differential equation because it can be rearranged so that all terms involving \(x\) are on one side and all terms involving \(t\) are on the other side.

Step 2: Separate Variables

Rearrange the equation to separate variables: \( \frac{dx}{x \ln x} = \frac{dt}{t} \). Now we have all \(x\)-related terms on one side and all \(t\)-related terms on the other.

Step 3: Integrate Both Sides

Integrate both sides of the equation: \( \int \frac{dx}{x \ln x} = \int \frac{dt}{t} \). This gives us \( \int \frac{1}{x \ln x} dx = \ln|t| + C \) on the right side.

Step 4: Solve the Integral on the Left

The integral \( \int \frac{1}{x \ln x} dx \) can be solved using substitution. Set \( u = \ln x \), then \( du = \frac{1}{x} dx \). This transforms the integral into \( \int \frac{1}{u} \, du = \ln|u| + C \). Hence, \( \ln|\ln x| = \ln|t| + C \).

Step 5: Simplify the Solution

Exponentiate both sides to simplify, resulting in \( \ln x = e^{\ln|t| + C} = t \cdot e^C \). Letting \( A = e^C \), we find \( \ln x = At \). Solving for \( x \), we obtain \( x = e^{At} \).

Step 6: Express the Final Solution

The general solution to the differential equation is \( x = e^{At} \), where \( A \) is a constant determined by initial conditions.

Key Concepts

Separable Equations

Separable equations are a specific type of differential equations that allow us to rewrite them such that each variable appears on different sides of the equation. This means that all terms involving one variable are grouped together on one side of the equation, and all terms involving the other variable are on the opposite side.
For example, in the equation \( \frac{dx}{dt} = \frac{x \ln x}{t} \), we can rearrange the terms to \( \frac{dx}{x \ln x} = \frac{dt}{t} \).

• Here, the left side contains only terms in \(x\), and the right side includes only terms in \(t\).

By separating the variables, we make it easier to integrate both sides and eventually solve the differential equation.
This method is particularly useful for equations that can be written in this format, making the equation readily solvable by integration.

Integration

Integration is a fundamental step in solving separable differential equations. Once the variables are separated, we integrate both sides with respect to their respective variables.
This means we solve \( \int \frac{dx}{x \ln x} \) for the \(x\)-part, and \( \int \frac{dt}{t} \) for the \(t\)-part.

• Integration of the right side, \( \int \frac{dt}{t} \), yields \( \ln|t| + C \), where \(C\) is the constant of integration.
• The left side, \( \int \frac{1}{x \ln x} dx \), is slightly complex and requires substitution to solve.

Through integration, both sides of the equation are expressed in terms of known functions, which we can later manipulate to find an overall solution to the differential equation.

Substitution Method

The substitution method is used for making complex integrals easier to solve. In our example, solving \( \int \frac{1}{x \ln x} dx \) directly is challenging, so we use substitution to simplify it.
We set \( u = \ln x \) so that \( du = \frac{1}{x} dx \).
This substitution transforms the original integral into something much simpler: \( \int \frac{1}{u} du \).

• Now, integrating \(\frac{1}{u}\) is straightforward, resulting in \( \ln|u| + C \).

After integrating, we substitute back the original variable to maintain consistency.
This step simplifies the process and leads us closer to the general solution.

General Solution

The general solution to a differential equation is the expression that includes an arbitrary constant and satisfies the equation for various initial conditions.
After solving \( \int \frac{1}{x \ln x} dx = \ln|t| + C \), we substitute back and simplify to find \( \ln x = At \), where \( A = e^C \).
This leads to the general solution \( x = e^{At} \).

• The general solution includes a constant \( A \), which can be determined using initial conditions whenever provided.
• The solution describes all potential functions that could satisfy the original differential equation under different initial states.

Generally, understanding and deriving the general solution is the final goal when solving differential equations, providing insight into the behaviors and characteristics of the system described by the equation.