Question

Find the general solution to the differential equations. $$ y^{\prime}=y \ln y $$

Short answer

The general solution is \( y = e^{Ce^x} \).

Step-by-step solution

Identify the Type of Differential Equation

The given differential equation is \( y' = y \ln y \). It's a first-order differential equation, and it can be solved using the method of separation of variables because \( y \) terms can be isolated from \( x \) terms.

Rewrite the Equation for Separation

Rearrange the equation to separate variables. We have \( \frac{dy}{dx} = y \ln y \). Multiply both sides by \( \frac{1}{y \ln y} \) to get \( \frac{1}{y \ln y} \frac{dy}{dx} = 1 \).

Separate Variables

Now, rewrite the equation with separated variables: \( \frac{1}{y \ln y} dy = dx \).

Integrate Both Sides

Integrate both sides to find the general solution. For the left side, use substitution: let \( u = \ln y \), then \( du = \frac{1}{y} dy \). So, \( \int \frac{1}{y \ln y} dy \) becomes \( \int \frac{1}{u} du \), which integrates to \( \ln |u| + C_1 \). For the right side, \( \int dx = x + C_2 \).

Solve the Integral

Substituting back \( u = \ln y \), the left side becomes \( \ln |\ln y| + C_1 = x + C_2 \). Combine constants: \( \ln |\ln y| = x + C \).

Exponentiate to Solve for \( y \)

Exponentiate both sides to solve for \( y \): \( |\ln y| = e^{x+C} = Ce^x \). Thus, \( y = e^{\pm Ce^x} \).

Write the General Solution

Thus, the general solution to the differential equation is \( y = e^{Ce^x} \), where \( C \) is an arbitrary constant.

Key concepts

Separation of Variables

When tackling differential equations, one powerful method is 'separation of variables.' This technique is especially useful for equations where you can isolate different variables on either side of the equation. The goal is to rewrite the equation such that all terms involving one variable, such as \( y \), are on one side and all terms involving the other variable, usually \( x \), are on the other side.

For example, consider the differential equation \( y' = y \ln y \). By rearranging the equation, you can move the \( y \)-dependent terms to one side, turning it into \( \frac{1}{y \ln y} dy = dx \). This clean separation ensures that when you integrate, each side deals with only one variable at a time.

Once separated, integrating both sides becomes straightforward. This method is particularly handy because it transforms a complex problem into two simpler problems of integration. However, not every differential equation allows for such neat separation. It often depends on the structure and relationship of variables within the equation.

First-Order Differential Equation

A first-order differential equation is a type of equation that involves the derivatives of a function. Specifically, it includes the first derivative but not any higher derivatives. Such equations are widespread in mathematical modeling of real-world phenomena, as they describe the rate of change of one variable in relation to another.

In our exercise, the equation \( y' = y \ln y \) is a first-order differential equation because it contains only the first derivative of \( y \) with respect to \( x \). No second or higher derivatives are involved here.

First-order differential equations can often be solved using techniques like separation of variables, as we saw earlier, or other methods like integrating factors or exact equations. Understanding the type of differential equation you're dealing with can guide you towards the appropriate solution method, making it an essential part of the problem-solving process.

General Solution

The concept of a 'general solution' in the context of differential equations refers to the solution that encompasses all possible solutions. For a first-order differential equation, this general solution often contains an arbitrary constant, denoted as \( C \). This constant represents the family of solutions, indicating that there are infinitely many solutions differing by that constant value.

Applying this to our problem, the expression \( y = e^{Ce^x} \) is the general solution to the differential equation \( y' = y \ln y \). The presence of the constant \( C \) means that for any initial condition provided, a specific value of \( C \) will satisfy the condition and yield a particular solution out of this family of solutions.

General solutions are crucial because they offer the complete set of answers to a differential equation. When additional conditions—often initial or boundary conditions—are applied, we can determine a unique solution or specify the value of the constant \( C \). This makes the general solution a fundamental step in understanding the behavior of dynamic systems described by differential equations.