Problem 96 - University Calculus: Early Transcendentals

Question

Solve the initial value problems in Exercises. $$\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}, \quad y(4)=0$$

Step-by-Step Solution

Answer

The solution to the initial value problem is \( y = \sqrt{x} - 2 \).

Step 1: Recognize the Differential Equation

The equation given is \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \). This is a first-order differential equation where the derivative of \( y \) with respect to \( x \) is expressed in terms of \( x \).

Step 2: Isolate the Differential

To solve the differential equation, we can rewrite it in a separable form. This takes the form of separating the variables such that: \( dy = \frac{1}{2\sqrt{x}} \, dx \).

Step 3: Integrate Both Sides

Integrate both sides of the equation to solve for \( y \). The integral of \( dy \) is \( \int dy = y \). The integral of the right side is \( \int \frac{1}{2\sqrt{x}} \, dx = \int x^{-1/2} \, dx \).

Step 4: Solve the Integral

Calculate the integral on the right side. The antiderivative of \( x^{-1/2} \) is \( 2x^{1/2} + C \), where \( C \) is the constant of integration. Therefore, \( y = x^{1/2} + C \).

Step 5: Use Initial Condition

Use the initial condition \( y(4) = 0 \) to find \( C \). Substituting \( x = 4 \) in \( y = x^{1/2} + C \), we have \( 0 = 2 + C \). Solve for \( C \) to get \( C = -2 \).

Step 6: Write the Final Solution

Substitute \( C = -2 \) back into the equation for \( y \). Thus, the solution of the differential equation is \( y = x^{1/2} - 2 \).

Key Concepts

First-Order Differential Equation

In mathematics, a first-order differential equation is an equation that relates a function to its first derivative. This type of equation is called 'first-order' because it involves only the first derivative of a function with respect to the variable. Such equations are essential in modeling real-world phenomena where change is continuous, such as velocity in physics or exponential growth in biology. The general form of a first-order differential equation can be written as \( \frac{dy}{dx} = f(x, y) \), where \( f(x, y) \) is a function of \( x \) and \( y \).
First-order differential equations are fundamental since they often serve as the basis for more complex equations. To solve them, you'll typically look to apply techniques such as separation of variables, integrating factors, or substitution, depending on the form. In the context of the given exercise, the differential equation \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \) is expressed with the derivative of \( y \) in terms of \( x \), making it a classic first-order case.

Separable Equations

Separable equations are a special class of first-order differential equations that can be rewritten so that each variable appears on a distinct side of the equation. The given differential equation \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \) can be manipulated into a separable form by isolating \( dy \) on one side and terms involving \( x \) on the other.
To achieve this, separate the variables by multiplying both sides by \( dx \), resulting in \( dy = \frac{1}{2\sqrt{x}} \, dx \). This separation is crucial because it allows us to integrate both sides independently. When you're able to express the equation in a separable form, you empower yourself to utilize basic integration techniques to find the solution.

Initial Value Problem

An initial value problem in the context of differential equations refers to finding a particular solution that not only satisfies a differential equation but also meets a given initial condition. An initial condition provides specific values for the variables at a particular point, allowing us to determine the constant of integration that emerges during the integration process.
In our example, the initial condition is \( y(4) = 0 \). This condition specifies that when \( x = 4 \), the value of \( y \) should be 0. Applying this information helps determine the constant \( C \) in the general solution obtained from integrating both sides. In practical terms, initial value problems are connected to boundary conditions in physical systems, ensuring our mathematical solutions adhere to the constraints of the real world.

Integration

Integration is a fundamental mathematical operation frequently employed to solve differential equations. It involves finding a function whose derivative is the given function. In our exercise, once variables are separated, we integrate both sides of the equation: \( \int dy = y \) and \( \int \frac{1}{2\sqrt{x}} \, dx \).
The integral on the right simplifies to the antiderivative of \( x^{-1/2} \), which is \( 2x^{1/2} \). Don't forget to add the integration constant \( C \) to the result. Applying the initial condition \( y(4) = 0 \) allows us to solve for this constant, ultimately leading to the specific solution \( y = x^{1/2} - 2 \).
Integration is not only a central technique for solving differential equations but also for a variety of applications in physics, engineering, and beyond, wherever determining the amount accumulated by a continuous change is necessary.