Problem 62 - Calculus for Biology and Medicine

Question

In Problems 59-72, solve the initial-value problem. $$ \frac{d y}{d x}=\frac{1}{2 \sqrt{x}}, \text { for } x \geq 1 \text { with } y=3 \text { when } x=4 $$

Step-by-Step Solution

Answer

The solution is \( y = \sqrt{x} + 1 \).

Step 1: Separate Variables

Given the differential equation \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \), we can treat it as \( dy = \frac{1}{2\sqrt{x}} dx \) to separate the differential equation into steps.

Step 2: Integrate Both Sides

To solve for \( y \), integrate both sides: \( \int dy = \int \frac{1}{2\sqrt{x}} dx \). The left side becomes \( y \), and the right side integrates to \( \int \frac{1}{2}x^{-1/2} dx = x^{1/2} + C = \sqrt{x} + C \).

Step 3: Apply the Initial Condition

Use the initial condition to find the constant \( C \). We know that when \( x = 4 \), \( y = 3 \). Substitute into the equation \( y = \sqrt{x} + C \) to get \( 3 = \sqrt{4} + C \), which gives \( 3 = 2 + C \). Solving for \( C \) gives \( C = 1 \).

Step 4: Write the Particular Solution

Substitute the value of \( C \) back into the general solution to find the particular solution. The complete solution to the initial-value problem is \( y = \sqrt{x} + 1 \).

Key Concepts

Separation of Variables

Separation of Variables is a technique used to simplify and solve differential equations. It involves rewriting the equation in a form where all terms involving one variable are on one side and terms involving the other are on the opposite side. This allows us to integrate both sides independently. For the given problem, we have the differential equation \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}}\). We can multiply both sides by \( dx \) to get \( dy = \frac{1}{2\sqrt{x}}dx \). This rearrangement separates the variables: \( dy \) is on one side, and \( dx \) is on the other.
This separation is a crucial step in solving such equations, as it transforms the differential equation into an algebraic equation that can be integrated.

Differential Equation

A Differential Equation is an equation that involves an unknown function and its derivatives. Solving a differential equation means finding a function that satisfies the equation. In this exercise, the given differential equation is \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \). This is a first-order differential equation because it involves only the first derivative \( \frac{dy}{dx} \).
Differential equations are essential in modeling various phenomena in physics, engineering, and other fields. They are powerful tools because they express how a rate of change (the derivative) relates to the state of a system (the function itself).
In this context, solving the differential equation means finding the function \( y \) that relates to \( x \) in such a way that its derivative matches the given expression.

Integration of Functions

Integration of Functions is the process of finding a function, known as the integral, whose derivative is the given function. It is the reverse operation of differentiation. In our exercise, after separating the variables, we arrived at the equation \( dy = \frac{1}{2\sqrt{x}}dx \), where both sides need to be integrated.
The integral of \( dy \) is simply \( y \), as differentiating \( y \) with respect to \( y \) gives 1. The right-hand side \( \int \frac{1}{2\sqrt{x}} dx \) requires using the power rule of integration. With a variable \( x \), the integral becomes \( \sqrt{x} + C \), where \( C \) is the constant of integration.
This constant is particularly important because it helps satisfy any initial conditions given in the problem. Applying such conditions helps us find the particular solution among the family of curves represented by the general solution.