solve y''-2xy'+4y=0 using frobenius method

To solve the differential equation \( y'' - 2xy' + 4y = 0 \) using the Frobenius method, we start by assuming a power series solution of the form:

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

where \( a_n \) are coefficients to be determined.

### Step 1: Compute the derivatives

We need to calculate \( y' \) and \( y'' \):

\[
y' = \sum_{n=1}^{\infty} n a_n x^{n-1}
\]

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

### Step 2: Substitute into the differential equation

Substituting \( y \), \( y' \), and \( y'' \) into the original equation gives:

\[
\sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - 2x \sum_{n=1}^{\infty} n a_n x^{n-1} + 4 \sum_{n=0}^{\infty} a_n x^n = 0
\]

### Step 3: Adjust the indices

Adjust the second term:

\[
-2x \sum_{n=1}^{\infty} n a_n x^{n-1} = -2 \sum_{n=1}^{\infty} n a_n x^n
\]

Now the equation becomes:

\[
\sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - 2 \sum_{n=1}^{\infty} n a_n x^n + 4 \sum_{n=0}^{\infty} a_n x^n = 0
\]

### Step 4: Change the index of the first sum

Change the index of the first sum to match \( x^n \):

\[
\sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n - 2 \sum_{n=0}^{\infty} n a_n x^n + 4 \sum_{n=0}^{\infty} a_n x^n = 0
\]

### Step 5: Combine the series

Now, combine the series:

\[
\sum_{n=0}^{\infty} \left( (n+2)(n+1) a_{n+2} - 2n a_n + 4 a_n \right) x^n = 0
\]

This gives:

\[
(n+2)(n+1) a_{n+2} + (4 - 2n) a_n = 0
\]

### Step 6: Recurrence relation

From the equation, we derive the recurrence relation:

\[
a_{n+2} = \frac{(2n - 4)}{(n+2)(n+1)} a_n
\]

### Step 7: Initial conditions

To find the coefficients, we need initial conditions. Let's assume:

- \( a_0 = a_0 \)
- \( a_1 = a_1 \)

### Step 8: Calculate further coefficients

1. For \( n = 0 \):

\[
a_2 = \frac{(2(0) - 4)}{(0+2)(0+1)} a_0 = -\frac{4}{2} a_0 = -2a_0
\]

2. For \( n = 1 \):

\[
a_3 = \frac{(2(1) - 4)}{(1+2)(1+1)} a_1 = \frac{-2}{6} a_1 = -\frac{1}{3} a_1
\]

3. For \( n = 2 \):

\[
a_4 = \frac{(2(2) - 4)}{(2+2)(2+1)} a_2 = \frac{0}{12} a_2 = 0
\]

This pattern suggests that \( a_4 = 0 \) and every fourth term afterward will also be zero.

### Summary of coefficients

Thus, the series for \( y(x) \) becomes:

\[
y(x) = a_0 \left(1 - 2x^2 + \dots\right) + a_1 \left(x - \frac{1}{3}x^3 + \dots\right)
\]

### Conclusion

The general solution is a combination of two series:

\[
y(x) = c_1 \left( 1 - 2x^2 + 0 \cdots \right) + c_2 \left( x - \frac{1}{3}x^3 + 0 \cdots \right)
\]

where \( c_1 \) and \( c_2 \) are constants determined by initial conditions.

If you need further details or have questions about specific steps, feel free to ask! Here are some related questions:

1. What are the implications of having zero coefficients in the series?
2. How can we determine specific values for \( c_1 \) and \( c_2 \)?
3. What is the convergence radius for the series solution?
4. How does the Frobenius method compare to other solution techniques?
5. Can we apply this method to non-linear differential equations? 

**Tip:** Always check the behavior of the series at different points to ensure convergence and accuracy of the solution.

Discover how to solve the differential equation y'' - 2xy' + 4y = 0 with the Frobenius method. This solution includes detailed steps on setting up a power series, finding recurrence relations, and deriving coefficients for the series solution. Ideal for advanced calculus students exploring differential equations and series methods.

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