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.