Solving a Singular Differential Equation with the Method of Frobenius

When faced with a differential equation that has a singular point, such as the one given below, the Method of Frobenius can provide a powerful approach to finding a solution. This article will guide you through the process of solving the differential equation 2x^2y'' - xy' y 0, which possesses a singular point at x 0, using this method.

Understanding the Problem

The differential equation in question is:

2x^2 y'' - xy' y 0

It is important to recognize that this equation has a singular point at x 0. For such cases, the Method of Frobenius is particularly useful in finding a solution. This method is designed for solving linear differential equations around a regular singular point.

Step 1: Identify the Form of the Solution

The first step is to assume a power series solution of the form:

y(x) sum_{n0}^{infty} a_n x^n

Step 2: Compute Derivatives

Next, we need to compute the first and second derivatives of the assumed solution:

y'(x) sum_{n1}^{infty} n a_n x^{n-1}

y''(x) sum_{n2}^{infty} n(n-1) a_n x^{n-2}

Step 3: Substitute into the Differential Equation

Substituting the expressions for y, y', and y'' into the differential equation, we get:

2x^2 left(sum_{n2}^{infty} n(n-1) a_n x^{n-2}right) - x left(sum_{n1}^{infty} n a_n x^{n-1}right) left(sum_{n0}^{infty} a_n x^nright) 0

Step 4: Align the Series

Now, we need to align the indices of the series:

Leaving the first term as is: Changing the index n to n-1 in the second term to align with x^n: Summing the third term directly as it is: Changing the index n to n-1 in the fourth term to align with x^n:

This aligns the series as follows:

2 sum_{n2}^{infty} n(n-1) a_n x^n - sum_{n1}^{infty} n a_n x^n sum_{n0}^{infty} a_n x^n sum_{n1}^{infty} a_{n-1} x^n 0

Step 5: Combine All Terms

Combining all the terms, we have:

sum_{n0}^{infty} [2n(n-1) a_n - n a_n a_n a_{n-1}] x^n 0

Step 6: Set the Coefficients to Zero

For the series to equal zero, each coefficient must equal zero:

For n 0, we have: For n geq 1, we have:

2a_0 0 implies a_0 0

2n(n-1)a_n - n a_n a_n a_{n-1} 0

Step 7: Develop a Recurrence Relation

Rearranging the recurrence relation, we get:

a_{n-1} left[frac{2n(n-1) - n}{1}right] a_n (2n - 2) a_n

Step 8: Solve for Coefficients

Starting with a_0 0, we can compute a few coefficients:

n 1: a_1 frac{a_{-1}}{2} (since a_{-1} is not defined, a_1 0) n 2: a_2 frac{2 cdot 1 cdot 0}{2} - frac{0}{2} 0 Continuing this process, all other coefficients will be zero.

This indicates that the trivial solution y(x) 0 is the only solution to the differential equation.

Conclusion

The solution to the differential equation 2x^2 y'' - xy' y 0, which has a singular point at x 0, is the trivial solution:

y(x) 0