We are tasked with solving the differential equation:

$(2x + x^3)y'' - y' - 6xy = 0$

using a power series solution.

Step 1: Assume a power series solution for $y(x)$

We assume the solution for $y(x)$ in the form of a power series around $x = 0$:

$y(x) = \sum_{n=0}^{\infty} a_n x^n$

Step 2: Derivatives of $y(x)$

To substitute into the given differential equation, we need the first and second derivatives of $y(x)$.

$y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1}$ $y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}$

Step 3: Substitute into the differential equation

Substitute $y(x)$, $y'(x)$, and $y''(x)$ into the given equation $(2x + x^3)y'' - y' - 6xy = 0$.

• The term $(2x + x^3)y''$ becomes: $(2x + x^3) \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} = 2x \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} + x^3 \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}$

  Breaking these terms up: $2x \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} = \sum_{n=2}^{\infty} 2n(n-1) a_n x^{n-1}$ $x^3 \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} = \sum_{n=2}^{\infty} n(n-1) a_n x^{n+1}$

• The term $-y'$ becomes: $- \sum_{n=1}^{\infty} n a_n x^{n-1}$

• The term $-6xy$ becomes: $-6x \sum_{n=0}^{\infty} a_n x^n = - \sum_{n=0}^{\infty} 6 a_n x^{n+1}$

Step 4: Combine like powers of $x$

Now, we combine all the terms together and collect powers of $x$.

For the $x^0$ term: $0 \text{ (no contribution from the other terms)}$

For the $x^1$ term: $2a_2 - a_1 = 0$ which gives $a_2 = \frac{a_1}{2}$.

For the $x^n$ term ($n \geq 2$): $2n(n-1) a_n + a_{n+2} - n a_n - 6a_{n-1} = 0$ This is the recurrence relation that will help us solve for the coefficients $a_n$.

Step 5: Solve the recurrence relation

The recurrence relation is:

$2n(n-1) a_n + a_{n+2} - n a_n - 6a_{n-1} = 0$

For specific values of $n$, this relation will generate the coefficients $a_n$.

Let me know if you want to expand on the steps to find the coefficients in more detail.

---

Do you want to go deeper into this