Problem 21 - Calculus Single Variable

Question

In each of Exercises \(17-28,\) solve the given initial value problem.\ $$ \frac{d y}{d x}+\frac{2}{x} y=7 \sqrt{x} \quad y(4)=17 $$

Step-by-Step Solution

Answer

The solution is \( y(x) = 2x^{3/2} + \frac{48}{x^2} \).

Step 1: Identify the Differential Equation and Initial Value

The given problem is a first-order linear differential equation: \( \frac{dy}{dx} + \frac{2}{x}y = 7\sqrt{x} \) with the initial condition \( y(4) = 17 \). Our goal is to find the function \( y(x) \) that satisfies both the differential equation and the initial condition.

Step 2: Find the Integrating Factor

For a first-order linear differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \), the integrating factor \( \mu(x) \) is given by \( e^{\int P(x) \, dx} \) where \( P(x) = \frac{2}{x} \). We calculate the integrating factor: \( \mu(x) = e^{\int \frac{2}{x} \, dx} = e^{2 \ln |x|} = x^2 \).

Step 3: Multiply Through by the Integrating Factor

Multiply the entire differential equation by \( \mu(x) = x^2 \) to make the left side an exact derivative: \( x^2 \frac{dy}{dx} + x \cdot 2y = 7x^{5/2} \). The left-hand side can now be expressed as a derivative: \( \frac{d}{dx}(x^2 y) = 7x^{5/2} \).

Step 4: Integrate Both Sides

Integrate both sides with respect to \( x \): \( \int \frac{d}{dx}(x^2 y) \, dx = \int 7x^{5/2} \, dx \). The left side results in \( x^2 y \), and the right side evaluates to \( \frac{14}{7} x^{7/2} + C \) upon integrating, giving us \( x^{2}y = \frac{14}{7}x^{7/2} + C \).

Step 5: Solve for the Constant Using Initial Condition

Substitute the initial condition \( y(4) = 17 \) into the equation \( x^2 y = \frac{14}{7} x^{7/2} + C \): \( 4^2 * 17 = \frac{14}{7} * 4^{7/2} + C \), which simplifies to \( 272 = 224 + C \), so \( C = 48 \).

Step 6: Write the General Solution

With the constant \( C = 48 \), the particular solution is \( x^2 y = \frac{14}{7} x^{7/2} + 48 \). Dividing both sides by \( x^2 \) gives the solution: \( y(x) = 2x^{3/2} + \frac{48}{x^2} \).

Key Concepts

First-order Linear Differential Equation

A first-order linear differential equation is a mathematical expression that relates a function to its first derivative, often involving both terms like \(y\) and \(\frac{dy}{dx}\). These types of equations are called 'linear' because they do not involve powers or products of the function \(y\) or \(\frac{dy}{dx}\).

In this context, we are dealing with an equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \). Here, \(P(x)\) and \(Q(x)\) are known functions of the independent variable \(x\). The example in our exercise, \( \frac{dy}{dx} + \frac{2}{x} y = 7 \sqrt{x} \,\) exemplifies a first-order linear differential equation. Our task is to find the unknown function \(y(x)\) that would both satisfy this equation and the given initial condition \(y(4)=17\).

Understanding the structure and properties of the differential equation allows us to begin the process to solve it. The initial condition provides more information that helps integrate and find a specific solution for the differential equation.

Integrating Factor

An integrating factor is a mathematical function used to simplify and solve first-order linear differential equations. It is an essential tool that converts an equation into an easier-to-solve form. For an equation given as \( \frac{dy}{dx} + P(x)y = Q(x) \,\) the integrating factor \( \mu(x) \) is computed based on the function \(P(x)\).

The recipe involves calculating \(\mu(x) = e^{\int P(x) \, dx}\).For our equation, \(P(x) = \frac{2}{x}\), which leads to:

• \(\mu(x) = e^{\int \frac{2}{x} \, dx} = e^{2 \ln |x|} = x^2\)

The use of the integrating factor allows us to multiply through the entire equation, transforming the left side into a perfect derivative. This maneuver is pivotal in rewriting the differential equation in a form that is straightforward to integrate, facilitating the construction of the solution.

Exact Derivative

Once the integrating factor has been applied to a differential equation, the transformed left-hand side becomes an exact derivative. An exact derivative is simply the derivative of a particular composite function. For the equation \(x^2 \frac{dy}{dx} + x \cdot 2y = 7x^{5/2}\), the left-hand side can be expressed as:

\[ \frac{d}{dx}(x^2 y) \].

This reformulation means we can now directly integrate both sides with respect to \(x\). The right-hand side of the equation becomes entirely manageable, allowing us to seamlessly find an expression for \(x^2 y\).

The conversion to an exact derivative is critical because it reduces complexity, effectively allowing the differential equation to be expressed in terms of simpler expressions involving integrals. This lays down a simplified path toward finding the function that satisfies both the differential equation and the initial condition.

Particular Solution

Solving differential equations involves finding a particular solution that fits both the equation and any initial condition provided. In this scenario, integrating both sides of the exact derivative form, we are led to:

\(x^2 y = \frac{14}{7}x^{7/2} + C\).

The initial condition \(y(4) = 17\) is employed to solve for the constant \(C\). By substituting \(x = 4\) and \(y = 17\), we find that \(C = 48\).

Thus, the specific function that fits the initial value problem is found by substituting back into the expression, yielding:

\[ y(x) = 2x^{3/2} + \frac{48}{x^2} \].

This expression is the particular solution to the initial value problem. It not only solves the differential equation, but also fulfills the initial condition, showcasing how the underlying theory aligns with practical calculations and applications.