Let's break down and simplify each term in the given expressions step by step, before combining them and grouping the powers of $x^{n+r}$.

First Term

The first term is: $2x^2 y'' = 2 \sum_{n=0}^{\infty} a_n (n + r)(n + r - 1) x^{n + r}$ This term represents the second derivative of $y$ multiplied by $2x^2$. The summation already expresses the generalized form of the second derivative, where $a_n$ are the coefficients, and $(n+r)(n+r-1)$ comes from the application of derivatives.

Second Term

The second term is: $(2x+1)xy' = \sum_{n=0}^{\infty} a_n (n+r) \left(2x^{n+r+1} + x^{n+r}\right)$ Here, we have a combination of terms that result from multiplying $(2x+1)$ by the first derivative of $y$. Expanding inside the summation gives us two separate terms: $(2x+1)xy' = \sum_{n=0}^{\infty} a_n (n+r) \left( 2x^{n+r+1} + x^{n+r} \right)$

Third Term

The third term is: $-y = - \sum_{n=0}^{\infty} a_n x^{n+r}$ This is just the function $y$ itself, negated and represented as a power series.

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Now, we combine the terms.

We aim to combine these terms and group powers of $x^{n+r}$. Let's rewrite each of them in terms of powers of $x^{n+r}$.

1. First term: $2x^2 y'' = 2 \sum_{n=0}^{\infty} a_n (n+r)(n+r-1) x^{n+r}$

2. Second term: Expand the second summation: $(2x+1)xy' = \sum_{n=0}^{\infty} a_n (n+r) \left( 2x^{n+r+1} + x^{n+r} \right)$ We split it into two sums: $(2x+1)xy' = 2 \sum_{n=0}^{\infty} a_n (n+r) x^{n+r+1} + \sum_{n=0}^{\infty} a_n (n+r) x^{n+r}$

3. Third term: $-y = - \sum_{n=0}^{\infty} a_n x^{n+r}$

Now, combine all terms: $2 \sum_{n=0}^{\infty} a_n (n+r)(n+r-1) x^{n+r} + 2 \sum_{n=0}^{\infty} a_n (n+r) x^{n+r+1} + \sum_{n=0}^{\infty} a_n (n+r) x^{n+r} - \sum_{n=0}^{\infty} a_n x^{n+r}$

Grouping Powers of $x^{n+r}$

We need to group the terms that have the same power of $x^{n+r}$.

Terms with $x^{n+r+1}$:

$2 \sum_{n=0}^{\infty} a_n (n+r) x^{n+r+1}$ This is already in the form $x^{n+r+1}$, so it stays as is.

Terms with $x^{n+r}$:

We combine all the terms with $x^{n+r}$: $2 \sum_{n=0}^{\infty} a_n (n+r)(n+r-1) x^{n+r} + \sum_{n=0}^{\infty} a_n (n+r) x^{n+r} - \sum_{n=0}^{\infty} a_n x^{n+r}$ Factor out $x^{n+r}$: $\sum_{n=0}^{\infty} a_n x^{n+r} \left[ 2(n+r)(n+r-1) + (n+r) - 1 \right]$ Simplify the expression inside the brackets: $2(n+r)(n+r-1) + (n+r) - 1 = 2(n+r)^2 - (n+r) - 1$

So, the final expression for the $x^{n+r}$ terms is: $\sum_{n=0}^{\infty} a_n x^{n+r} \left[ 2(n+r)^2 - (n+r) - 1 \right]$

Final Simplified Expression

The final expression is: $2 \sum_{n=0}^{\infty} a_n (n+r) x^{n+r+1} + \sum_{n=0}^{\infty} a_n x^{n+r} \left[ 2(n+r)^2 - (n+r) - 1 \right]$ This is the simplified form of the given terms, grouped by powers of $x^{n+r}$.

Would you like further explanation on any step, or more details on how to simplify these kinds of series? Here are 5 related questions that could deepen your understanding:

1. How do you identify patterns in power series when simplifying terms?
2. What is the role of summation in simplifying differential equations?
3. Can we apply this simplification method to nonlinear terms in differential equations?
4. How does changing the order of summation affect the final expression in power series?
5. What are common applications of power series in solving differential equations?

Tip: When simplifying series, always group terms by powers of the variable to see patterns and relationships clearly.