In Exercises 23–28, solve the initial value problem.
x dy/dx + 2y = x² + 1, x > 0, y(1) = 1

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Rewrite the given differential equation in the standard linear form. Start by dividing the entire equation by \(x\) (since \(x > 0\)) to isolate \(\frac{dy}{dx}\): \(\frac{dy}{dx} + \frac{2}{x} y = x + \frac{1}{x}\)

Identify the integrating factor \(\mu(x)\), which is given by \(\mu(x) = e^{\int P(x) \, dx}\), where \(P(x) = \frac{2}{x}\). Calculate the integral: \(\int \frac{2}{x} \, dx = 2 \ln|x| = \ln|x|^2\) So, \(\mu(x) = e^{\ln|x|^2} = x^2\) (since \(x > 0\)).

Multiply both sides of the differential equation by the integrating factor \(x^2\): \(x^2 \frac{dy}{dx} + 2x y = x^3 + x\) Notice that the left side is the derivative of the product \(x^2 y\), so rewrite it as: \(\frac{d}{dx} (x^2 y) = x^3 + x\)

Integrate both sides with respect to \(x\): \(\int \frac{d}{dx} (x^2 y) \, dx = \int (x^3 + x) \, dx\) This simplifies to: \(x^2 y = \int x^3 \, dx + \int x \, dx + C\) Calculate the integrals on the right side.

Solve for \(y\) by dividing both sides by \(x^2\): \(y = \frac{1}{x^2} \left( \text{integrated expression} + C \right)\) Finally, use the initial condition \(y(1) = 1\) to find the constant \(C\) by substituting \(x=1\) and \(y=1\) into the equation.

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First-Order Linear Differential Equations

A first-order linear differential equation has the form dy/dx + P(x)y = Q(x). It can be solved using an integrating factor, which simplifies the equation into an exact derivative, allowing integration to find the solution. Recognizing this form is essential for solving the given initial value problem.

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Classifying Differential Equations

Integrating Factor Method

The integrating factor is a function, usually denoted μ(x) = e^(∫P(x)dx), used to multiply both sides of a linear differential equation. This transforms the left side into the derivative of (μ(x)y), enabling straightforward integration. Applying this method helps solve the differential equation efficiently.

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Euler's Method

Initial Value Problems (IVP)

An initial value problem specifies the value of the unknown function at a particular point, such as y(1) = 1. After finding the general solution to the differential equation, the initial condition is used to determine the unique particular solution that satisfies the problem.

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