Question

Solve the initial-value problems. $$ x \frac{d y}{d x}-2 y=2 x^{4}, \quad y(2)=8 $$

Short answer

The particular solution to the initial-value problem is \(y(x) = \frac{x}{x^2 + 14}\).

Step-by-step solution

Separate Variables

We start by rewriting the given differential equation into a form where all terms with y are on one side and all terms with x are on the other side. This means we want to get the dy/dx term by itself as follows: \(x \frac{dy}{dx} - 2y = 2x^4\) \(\frac{dy}{dx} - \frac{2}{x}y = 2x^3\) Now we have separated the variables.

Integrate Both Sides

First, we introduce an integrating factor to simplify the left side of the equation. The integrating factor I(x) is defined as \(\exp{\left(\int -\frac{2}{x} dx \right)}\),which is equal to: \(I(x) = x^{-2}\) Now, we multiply both sides of the equation by the integrating factor: \(x^{-2} \frac{dy}{dx} - 2x^{-3}y = 2x\) \(x^{-2} \frac{d(xy^{-2})}{dx} = 2x\) Next, we integrate both sides with respect to x: \(\int x^{-2} \frac{d(xy^{-2})}{dx} dx = \int 2x dx\)

Evaluate the Integrals

Evaluate the left side integral using the substitution method. Let u = xy^{-2}, then \(du = x^{-2} d(xy^{-2})\). \(\int du = \int 2x dx\) \(xy^{-2} = x^2 + C\)

Solve for y(x) and Apply the Initial Condition

Now, we solve for y(x): \(y(x) = \frac{x}{x^2 + C}\) Finally, apply the initial condition to find the value of the constant C: \(8 = \frac{2}{2^2 + C}\) Using the initial condition, we get: \(C = 14\) Thus, our particular solution is: \(y(x) = \frac{x}{x^2 + 14}\)

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) involves solving a differential equation together with a given condition, often specified at a particular point. This given condition helps to determine the unique solution to the differential equation.

• The problem specifies an equation: \( x \frac{d y}{d x} - 2 y = 2 x^4 \)
• And an initial condition: \( y(2) = 8 \)

This initial condition can be visualized as a dot on a graph, where the solution curve must pass through. Solving the IVP means finding the right curve which fits both the differential equation and the initial condition.
Without the initial condition, there would be infinitely many possible solution curves. Hence, this helps pin down exactly one solution that satisfies all conditions.

Integrating Factor

When dealing with a linear differential equation, an integrating factor is a strategic function we use to simplify the solving process. The integrating factor, \( I(x) \), is often employed to transform the differential equation into a more manageable form.

• For our equation: \( x \frac{d y}{d x} - 2 y = 2 x^4 \)
• We rearrange to: \( \frac{dy}{dx} - \frac{2}{x}y = 2x^3 \)

The integrating factor is computed as \( \exp{\left(\int -\frac{2}{x}\, dx \right)} = x^{-2} \).
This factor simplifies the differential equation into something easier to integrate. It's like finding the perfect tool that makes the work of integration seamless by transforming the left side into an exact derivative of a product.

Variable Separation

Separating variables is a technique used to rearrange differential equations so that different variables reside on separate sides of the equation. This technique often aids simple integration.

• Initially, the equation \( x \frac{d y}{d x} - 2 y = 2 x^4 \) needs transformation.
• Rewriting gives us: \( \frac{dy}{dx} - \frac{2}{x}y = 2x^3 \)

The aim is to get all terms involving \( y \) and \( dy \) on one side, while all terms involving \( x \) and \( dx \) on the other side.
In some cases, complete separation is not feasible, thus an integrating factor is deployed. Once variables are separated or prepared for integration, the path is clear to solve through integrating both sides.

Integration

Integration is the process of finding the integral of a function, which is a critical step in solving differential equations. In this context, integration helps unravel solutions from equations where derivatives are given.

• We have \( x^{-2} \frac{d(xy^{-2})}{dx} = 2x \)
• By integrating both sides, \( \int x^{-2} \frac{d(xy^{-2})}{dx} \, dx = \int 2x \, dx \)

This step brings us closer to finding \( y(x) \), which represents the relationship we are after.
The technique may involve simple integrations or use of substitution, depending on the complexity of the expression.

Particular Solution

The quest for a particular solution means finding the one solution to a differential equation that also meets the initial conditions laid out in the problem.

• After integration, the general solution idea becomes: \( y(x) = \frac{x}{x^2 + C} \)
• Using the initial condition \( y(2)=8 \) helps isolate the constant \( C \).

Solving \( 8 = \frac{2}{2^2 + C} \) gives us \( C = 14 \).
The final particular solution is \( y(x) = \frac{x}{x^2 + 14} \), which is the sole solution that satisfies both the differential equation and the initial condition.