Problem 21 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

\(\frac{d y}{d x}=y \ln \sqrt{x} ; y=-4\) when \(x=1\)

Step-by-Step Solution

Answer

The solution to the differential equation is \(y = -4e^{\frac{x}{2}(x \ln x - x)}\).

Step 1: Recognize the Differential Equation Type

Identify that the equation \(\frac{dy}{dx} = y \ln \sqrt{x}\) is a separable differential equation.

Step 2: Rewrite the Equation

Rewrite the given differential equation in a separable form: \(\frac{1}{y} dy = \ln \sqrt{x} dx\).

Step 3: Simplify the Logarithmic Term

Simplify the term \(\ln \sqrt{x}\) to \(\frac{1}{2} \ln x\). This makes the equation: \(\frac{1}{y} dy = \frac{1}{2} \ln x dx\).

Step 4: Integrate Both Sides

Integrate both sides of the equation: \(\int \frac{1}{y} dy = \int \frac{1}{2} \ln x dx\).

Step 5: Solve the Integrals

Solve the integrals to get: \(\ln |y| = \frac{1}{2} (x \ln x - x) + C\).

Step 6: Apply the Initial Condition

Use the initial condition \(y = -4\) when \(x = 1\) to find the constant \(C\): \(\ln |-4| = \frac{0}{2} + C\), which gives \(\ln 4 = C\).

Step 7: Solve for y

Plug \(C\) back into the solution: \(\ln |y| = \frac{1}{2}(x \ln x - x) + \ln 4\), and solve for \(y\).

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. In other words, they describe how a function changes over time or space.
In our example, \(\frac{dy}{dx} = y \ln \sqrt{x}\), this is a first-order differential equation because it involves the first derivative of y. The primary goal is to find the function y(x) that satisfies this relationship.

Initial Conditions

Initial conditions are values that are given at the beginning of a problem to help solve a differential equation. These conditions help us find a specific solution from the general solution.
In our case, we are given that y = -4 when x = 1. This information will be used to find the specific constant C once we integrate both sides of the differential equation. Without initial conditions, we would only have a family of solutions.

Integration Techniques

Integration techniques are methods used to find the integral, or the 'antiderivative,' of a function. For separable differential equations like ours, we separate the variables first and then integrate both sides individually.

Here’s a brief rundown:

• We rewrite the equation: \(\frac{1}{y} dy = \frac{1}{2} \ln x dx\).
• Simplify each side and integrate: \(\int \frac{1}{y} dy\) and \(\int \frac{1}{2} \ln x dx\).

The integration of \(\frac{1}{y} dy\) is \(\ln |y|\) and using integration by parts for \(\frac{1}{2} \ln x dx\), we get \(\frac{1}{2}(x \ln x - x) + C\).

Logarithmic Differentiation

Logarithmic differentiation is a technique used to differentiate functions that are in an inconvenient form for direct differentiation or integration. By taking the natural logarithm of a function, you can simplify it using the properties of logarithms.
For example, \(\ln \sqrt{x}\) can be simplified to \(\frac{1}{2} \ln x\). By doing this, we make the integral more manageable in the integration process.
By simplifying and differentiating, we can integrate equations more effectively and apply initial conditions to find specific constants.