The given equation is:

\[
2x^2 y'' + xy' - (x^2 + 1) y = 0
\]

This is a second-order linear differential equation. To solve this, we first check if it's in a standard form and then try to find a particular or general solution.

### Step 1: Identify the type of differential equation

The equation is of the form:

\[
a(x) y'' + b(x) y' + c(x) y = 0
\]

where:
- \(a(x) = 2x^2\)
- \(b(x) = x\)
- \(c(x) = -(x^2 + 1)\)

Since all terms are linear in \(y\), \(y'\), and \(y''\), this is a **linear homogeneous differential equation with variable coefficients**.

### Step 2: Apply a known method for solution

For a second-order linear homogeneous differential equation with variable coefficients, one common approach is to try a power series solution or to see if the equation matches a known type (e.g., Euler-Cauchy, Bessel, etc.).

#### Rewriting the differential equation

We can rewrite the equation as:

\[
y'' + \frac{1}{2x} y' - \frac{x^2 + 1}{2x^2} y = 0
\]

#### Step 3: Attempt a solution method (Power Series)

To solve this using a power series or by identifying it as a known form, we proceed by checking for solutions of the form:

\[
y(x) = x^r
\]

Substitute \(y = x^r\) into the differential equation:

1. Compute \(y' = r x^{r-1}\) and \(y'' = r(r-1)x^{r-2}\).
2. Substitute these into the equation and simplify.

### Would you like to continue with the series solution approach or check for another method? Let me know!

Explore the solution to the second-order linear differential equation 2x²y'' + xy' - (x² + 1)y = 0. This guide covers key concepts in differential equations, including variable coefficients and the power series method. Suitable for undergraduate-level mathematics.

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