Problem 21 - Calculus

Question

In \(21-32\) find a function \(y(x)\) that solves the differential equation. \(d y / d x=x^{2}+\sqrt{x}\)

Step-by-Step Solution

Answer

The solution is \(y(x) = \frac{x^3}{3} + \frac{2}{3}x^{3/2} + C\).

Step 1: Identify the Differential Equation

We start with the given differential equation which is: \(\frac{dy}{dx} = x^{2} + \sqrt{x}\). We need to find the function \(y(x)\) that satisfies this equation.

Step 2: Integrate the Differential Equation

To solve for \(y(x)\), we integrate both sides of the equation with respect to \(x\). The integral of the left side is \(\int \frac{dy}{dx} \; dx = y(x)\), and the integral of the right side is \(\int (x^2 + \sqrt{x}) \; dx\).

Step 3: Integrate Each Term Separately

First, integrate \(x^2\):\[\int x^2 \; dx = \frac{x^3}{3} + C_1\]Now integrate \(\sqrt{x}\):\[\int \sqrt{x} \; dx = \int x^{1/2} \; dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} + C_2\]Where \(C_1\) and \(C_2\) are integration constants.

Step 4: Combine the Results

Combine the results of the integrals to get:\[y(x) = \frac{x^3}{3} + \frac{2}{3} x^{3/2} + C\]Here, \(C\) is a single constant of integration that encompasses \(C_1\) and \(C_2\).

Step 5: State the Solution

The function \(y(x)\) that solves the differential equation \(\frac{dy}{dx} = x^2 + \sqrt{x}\) is:\[y(x) = \frac{x^3}{3} + \frac{2}{3} x^{3/2} + C\]

Key Concepts

Integration

Integration is a fundamental concept in calculus that allows us to find the antiderivative or the integral of a function. In the context of our differential equation, \( \frac{dy}{dx} = x^2 + \sqrt{x} \), we aim to find a function \( y(x) \) whose derivative matches the right-hand side of the equation.

• Indefinite Integration: This refers to finding a function whose derivative gives the original function. For example, integrating \( x^2 \) yields \( \frac{x^3}{3} + C_1 \), where \( C_1 \) is the constant of integration.
• Integration of Power Functions: Basic power functions integrate to another power function. The integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} + C \). Thus, \( \sqrt{x} = x^{1/2} \) integrates to \( \frac{2}{3}x^{3/2} + C_2 \).

After integrating each term individually, you combine them: \( y(x) = \frac{x^3}{3} + \frac{2}{3}x^{3/2} + C \), where \( C \) is a new, combined constant of integration.
Each step in the process of integration helps us reconstruct the function that the differential equation describes, capturing both the essence of the relationship and the accumulated sum of small areas under the curve.

Calculus

Calculus is the vast branch of mathematics concerned with change and motion, utilizing derivatives and integrals as its core tools. When solving differential equations, calculus allows us to move between understanding instantaneous rates of change (derivatives) and accumulated quantities (integrals).

• Differential Equations: These equations express a relationship involving rates of change with respect to a variable. Our task was to resolve one such equation, \( \frac{dy}{dx} = x^2 + \sqrt{x} \), illustrating the process and principles of calculus.
• Fundamental Theorem of Calculus: This theorem bridges derivatives and integrals, signifying that differentiation and integration are inverse operations. In solving our equation, we despatched both integration and differentiation to procure \( y(x) \) from \( \frac{dy}{dx} \).

Using Calculus, we link together the concepts of motion, change, and accumulation, allowing for deep insights and solutions to a vast range of mathematical and real-world problems.

Initial Value Problem

An initial value problem (IVP) is a type of differential equation accompanied by specific conditions provided at some point in the domain of the solution. While this specific exercise did not specify an initial value, understanding IVPs is crucial when working with differential equations.

• Initial Conditions: These are specified values which the solution of a differential equation must satisfy at a particular value of the independent variable, commonly denoted as \( y(x_0) = y_0 \).
• Determining Constants: In the presence of initial conditions, the arbitrary constant \( C \) appearing in the general solution of a differential equation is determined. For instance, if we had \( y(x_0) = y_0 \), we would substitute \( x_0 \) and \( y_0 \) into our solution to solve for \( C \).

While our current example does not incorporate an initial value, the logic follows that an initial condition gives a particular solution among the infinitely many possible solutions provided by adding the integration constant \( C \). Addressing IVPs involves integrating with awareness of any specific values set forth at the onset, thus tailoring the solution to match given criteria.