3. Find the frst four terms in each portion of the series solution around x0 = 0 for the following diferential equation.

(x2 + 1)y′′ − 4xy′ + 6xy = 0

We are tasked with finding the first four terms of the series solution around \( x_0 = 0 \) for the differential equation:

\[
(x^2 + 1)y'' - 4xy' + 6xy = 0
\]

### Step 1: Express the solution as a power series
We assume the solution \( y(x) \) can be written as a power series around \( x_0 = 0 \):

\[
y(x) = \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots
\]

Now we compute the derivatives:

\[
y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} = a_1 + 2a_2 x + 3a_3 x^2 + 4a_4 x^3 + \cdots
\]

\[
y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} = 2a_2 + 6a_3 x + 12a_4 x^2 + \cdots
\]

### Step 2: Substitute the series into the differential equation
We substitute \( y(x) \), \( y'(x) \), and \( y''(x) \) into the given differential equation:

\[
(x^2 + 1)y'' - 4xy' + 6xy = 0
\]

#### 1. Term: \( (x^2 + 1)y'' \)
Expanding \( (x^2 + 1)y'' \):

\[
(x^2 + 1)y'' = (x^2 + 1) \left( 2a_2 + 6a_3 x + 12a_4 x^2 + \cdots \right)
\]
\[
= 2a_2 + 6a_3 x + 12a_4 x^2 + \cdots + 2a_2 x^2 + 6a_3 x^3 + 12a_4 x^4 + \cdots
\]

#### 2. Term: \( -4xy' \)
Expanding \( -4xy' \):

\[
-4xy' = -4x \left( a_1 + 2a_2 x + 3a_3 x^2 + 4a_4 x^3 + \cdots \right)
\]
\[
= -4a_1 x - 8a_2 x^2 - 12a_3 x^3 - 16a_4 x^4 + \cdots
\]

#### 3. Term: \( 6xy \)
Expanding \( 6xy \):

\[
6xy = 6x \left( a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots \right)
\]
\[
= 6a_0 x + 6a_1 x^2 + 6a_2 x^3 + 6a_3 x^4 + \cdots
\]

### Step 3: Collect terms
We now combine all the terms from \( (x^2 + 1)y'' \), \( -4xy' \), and \( 6xy \).

For the constant term (\( x^0 \)):

\[
2a_2 = 0 \quad \Rightarrow \quad a_2 = 0
\]

For the \( x^1 \) term:

\[
6a_3 - 4a_1 + 6a_0 = 0 \quad \Rightarrow \quad 6a_3 = 4a_1 - 6a_0
\]
\[
a_3 = \frac{2a_1 - 3a_0}{3}
\]

For the \( x^2 \) term:

\[
12a_4 + 2a_2 - 8a_2 + 6a_1 = 0 \quad \Rightarrow \quad 12a_4 + 6a_1 = 0 \quad \Rightarrow \quad a_4 = -\frac{a_1}{2}
\]

For the \( x^3 \) term:

\[
12a_4 + 6a_1 x^2 + terms
...
 for

Find the first four terms in each portion of the series solution around x0 = 0 for the following differential equation: (x^2 + 1)y′′ − 4xy′ + 6xy = 0

Explore the series solution of the differential equation (x^2 + 1)y′′ − 4xy′ + 6xy = 0 around x0 = 0. This detailed solution includes key steps such as power series expansion, differentiation, and solving for coefficients. Suitable for advanced calculus or differential equations students.

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