Problem 15 - Calculus: Early Transcendentals

Question

Find the solution of the differential equation that satisfies the given initial condition. \( x \ln x = y(1 + \sqrt {3 + y^2}) y', y(1) = 1 \)

Step-by-Step Solution

Answer

The solution involves implicit integration and satisfies the initial condition.

Step 1: Identify and Simplify the Differential Equation

The given differential equation is \( x \ln x = y(1 + \sqrt{3 + y^2}) y' \). Let's simplify it by solving for \( y' \):\[ y' = \frac{x \ln x}{y(1 + \sqrt{3 + y^2})}. \]

Step 2: Separate Variables

To separate variables, we rearrange the equation as: \[ y(1 + \sqrt{3 + y^2}) \, dy = x \ln x \, dx. \] This enables us to integrate both sides separately.

Step 3: Integrate Both Sides

To solve the separated equation, integrate both sides:- Integrate the left side with respect to \( y \): \[ \int y(1 + \sqrt{3 + y^2}) \, dy. \]- Integrate the right side with respect to \( x \): \[ \int x \ln x \, dx. \]The equations for these integrals, however, necessitate substitution or numerical methods for computation.

Step 4: Apply the Initial Condition

Given the initial condition \( y(1) = 1 \), substitute \( y = 1 \) and \( x = 1 \) into the integrated terms. Use this condition to solve for the constant of integration, which emerges from solving these integrals.

Step 5: Construct the Solution

Using the integrated forms and the constant derived from the initial condition, construct the final solution function \( y(x) \). This solution will be expressed implicitly due to the complexity of the integration.

Key Concepts

Initial Condition

In differential equations, the **initial condition** provides crucial information to find a specific solution from a set of possible solutions. It is a known value of the function or its derivatives at a particular point. For example, in the problem, we have the differential equation alongside the condition \( y(1) = 1 \). Here, \( y(1) = 1 \) is the initial condition, meaning when \( x = 1 \), the value of \( y \) is 1.

This constrains the solution and allows us to find the specific constant when integrating the equation. When solving differential equations, always use the initial condition after integrating to solve for the constant of integration. This ensures that the solution satisfies both the differential equation and the initial condition, offering the exact function that models the scenario described.

Integration

**Integration** is a fundamental concept in calculus dealing with finding the integral of a function. It is used to calculate areas, volumes, central points, and many other important quantities. In the exercise, integration helps us move from the differential equation (which involves derivatives) to an equation that describes \( y \) or \( x \) exactly without derivatives.

When we integrate the separated variables, we are essentially accumulating the areas under the curves defined by these functions. This operation is crucial, as it transforms the equation from one involving \( y' \) into one that reveals the relationships between \( y \) and \( x \). After integrating both sides, constants of integration appear, which must be solved using the initial condition, ensuring that the solution fits the specific problem requirements.

Variable Separation

In solving differential equations, **variable separation** is a technique used to rearrange a differential equation so that functions of each variable are on opposite sides of the equation. This preparatory step is necessary before integrating the equation. In our case, we began with:

• \( y' = \frac{x \ln x}{y(1 + \sqrt{3 + y^2})} \)

We rearranged this to:

• \( y(1 + \sqrt{3 + y^2}) \, dy = x \ln x \, dx \)

This separation allows us to integrate each side with respect to its respective variable \( x \) and \( y \). It's a powerful technique, making it possible to deal with non-linear differential equations by reducing them into simpler, more manageable parts for integration. Always ensure that every term involving \( y \) stays on one side and every term with \( x \) on the other before moving to integration.

Calculus

**Calculus** is the mathematical study of continuous change. It includes two major branches: differentiation and integration, which are the backbone of solving differential equations. The given differential equation is a perfect illustration of calculus in action, attempting to find a function without explicit knowledge of the function's form.

In calculus, solving such equations involves using different techniques of integration, sometimes requiring substitution or numerical estimation when the integral is not straightforward. Calculus gives us the tools to decide the rate of change of one variable with respect to another (differentiation) and to understand the accumulated change from those rates (integration).

It also allows us to tackle real-world problems in physics, engineering, and other fields by modeling the change mathematically. Understanding these concepts thoroughly helps in not only solving textbook problems but also in applying these skills to solve practical problems.