We are tasked with combining the given summations into a single summation where the power of $x$ is the same and all indices start from $n = 0$. Let us break this down systematically.

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Given:

We already know: $y = \sum_{n=0}^\infty a_n x^n, \quad y' = \sum_{n=1}^\infty n a_n x^{n-1}, \quad y'' = \sum_{n=2}^\infty n(n-1)a_n x^{n-2}.$

Substitute these derivatives into the differential equation:

$(-5 + x)y'' + (1 + 4x)y' + 3y = 0.$

Expanding term-by-term:

1. For $(-5 + x)y''$:

$(-5 + x)y'' = -5 \sum_{n=2}^\infty n(n-1)a_n x^{n-2} + \sum_{n=2}^\infty n(n-1)a_n x^{n-1}.$

2. For $(1 + 4x)y'$:

$(1 + 4x)y' = \sum_{n=1}^\infty n a_n x^{n-1} + 4\sum_{n=1}^\infty n a_n x^n.$

3. For $3y$:

$3y = \sum_{n=0}^\infty 3a_n x^n.$

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Adjust indices to match powers of $x^n$:

1. $(-5 + x)y''$:

Split into two parts:

• First part: $-5 \sum_{n=2}^\infty n(n-1)a_n x^{n-2}$

  Shift index: Let $m = n - 2$, so $n = m + 2$: $-5 \sum_{n=2}^\infty n(n-1)a_n x^{n-2} = -5 \sum_{m=0}^\infty (m+2)(m+1)a_{m+2} x^m.$

• Second part: $\sum_{n=2}^\infty n(n-1)a_n x^{n-1}$

  Shift index: Let $m = n - 1$, so $n = m + 1$: $\sum_{n=2}^\infty n(n-1)a_n x^{n-1} = \sum_{m=1}^\infty m(m+1)a_{m+1} x^m.$

Combining, we have: $(-5 + x)y'' = -5 \sum_{m=0}^\infty (m+2)(m+1)a_{m+2} x^m + \sum_{m=1}^\infty m(m+1)a_{m+1} x^m.$

2. $(1 + 4x)y'$:

Split into two parts:

• First part: $\sum_{n=1}^\infty n a_n x^{n-1}$

  Shift index: Let $m = n - 1$, so $n = m + 1$: $\sum_{n=1}^\infty n a_n x^{n-1} = \sum_{m=0}^\infty (m+1)a_{m+1} x^m.$

• Second part: $4\sum_{n=1}^\infty n a_n x^n$

  No index shift is needed: $4\sum_{n=1}^\infty n a_n x^n = 4\sum_{m=1}^\infty m a_m x^m.$

Combining, we have: $(1 + 4x)y' = \sum_{m=0}^\infty (m+1)a_{m+1} x^m + 4\sum_{m=1}^\infty m a_m x^m.$

3. $3y$:

No index shift is needed: $3y = \sum_{n=0}^\infty 3a_n x^n.$

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Combine all terms into a single summation:

Now we combine all terms into a single summation. Grouping terms for $x^m$, we get:

Terms for $x^m$:

• From $(-5 + x)y''$: $-5(m+2)(m+1)a_{m+2} + m(m+1)a_{m+1}.$

• From $(1 + 4x)y'$: $(m+1)a_{m+1} + 4m a_m.$

• From $3y$: $3a_m.$

Combined summation:

$\sum_{m=0}^\infty \left[ -5(m+2)(m+1)a_{m+2} + m(m+1)a_{m+1} + (m+1)a_{m+1} + 4m a_m + 3a_m \right]x^m = 0.$

Simplify the coefficient of $x^m$: $\sum_{m=0}^\infty \left[ -5(m+2)(m+1)a_{m+2} + (m^2 + 2m + 1)a_{m+1} + (4m + 3)a_m \right]x^m = 0.$

For this to hold for all $m$, the coefficient of $x^m$ must be zero: $-5(m+2)(m+1)a_{m+2} + (m^2 + 2m + 1)a_{m+1} + (4m + 3)a_m = 0.$

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Recurrence relation:

The recurrence relation is: $a_{m+2} = \frac{(m^2 + 2m + 1)a_{m+1} + (4m + 3)a_m}{5(m+2)(m+1)}.$

This allows us to compute $a_{m+2}$ iteratively.

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Would you like to compute specific terms or need further clarification?

Related Questions:

1. What are the first four terms of the solution using the recurrence relation?
2. How does the power series solution relate to Frobenius method?
3. What is the radius of convergence of the series $y = \sum a_nx^n$?
4. Can this recurrence relation be solved symbolically for $a_n$?
5. What happens if different initial conditions are applied?

Tip:

When shifting indices in summations, always verify that the powers of $x$ are consistent before combining terms.