Question

Solve the following differential equations by the method of Frobenius (generalized power series). Remember that the point of doing these problems is to learn about the method (which we will use later), not just to find a solution. You may recognize some series [as we did in (11.6)] or you can check your series by expanding a computer answer. $$x^{2} y^{\prime \prime}+2 x y^{\prime}-6 y=0$$

Short answer

The general solution is \(y(x) = a_0 x^2 + b_0 x^{-3}\).

Step-by-step solution

- Identify Singular Points

First, identify if the point x=0 is a regular singular point of the differential equation. The differential equation is given as:\[x^{2} y^{\backprime \backprime} + 2x y^{\backprime} - 6 y = 0\]Rewrite the equation in the standard form:\[y^{\backprime \backprime} + \frac{2}{x} y^{\backprime} - \frac{6}{x^2} y = 0\]Here, the functions \(P(x)\) and \(Q(x)\) are \(\frac{2}{x}\) and \(-\frac{6}{x^2}\) respectively. The point x=0 is a regular singular point since both \(x P(x)\) and \(x^2 Q(x)\) are analytic at x=0.

- Assume a Series Solution

Assume a Frobenius series solution of the form:\[y = \sum_{n=0}^{\infty} a_n x^{n + r}\]where \(a_0 eq 0\). Next, find the derivatives: \[y^{\backprime} = \sum_{n=0}^{\infty} a_n (n + r)x^{n + r - 1}\]\[y^{\backprime \backprime} = \sum_{n=0}^{\infty} a_n (n + r)(n + r - 1)x^{n + r - 2}\]

- Substitute the Series into the Differential Equation

Substitute these series into the differential equation and combine like terms:\[x^2 \sum_{n=0}^{\infty} a_n (n + r)(n + r - 1)x^{n + r - 2} + 2x \sum_{n=0}^{\infty} a_n (n + r)x^{n + r - 1} - 6 \sum_{n=0}^{\infty} a_n x^{n + r} = 0\]This simplifies to:\[\sum_{n=0}^{\infty} a_n (n + r)(n + r - 1)x^{n + r} + 2 \sum_{n=0}^{\infty} a_n (n + r)x^{n + r} - 6 \sum_{n=0}^{\infty} a_n x^{n + r} = 0\]

- Find the Indicial Equation

Combine terms to find the indicial equation. From the equation:\[\sum_{n=0}^{\infty} a_n [(n + r)(n + r - 1) + 2(n + r) - 6]x^{n + r} = 0\]For the left side to be zero for all \(x\), the coefficient of each power of \(x\) must be zero. Thus, we get the indicial equation by setting the term with the lowest power of \(x\), which is \(a_0\), equal to zero:\[(r)(r - 1) + 2r - 6 = 0\]This simplifies to:\[r^2 + r - 6 = 0\]Factoring gives two roots:\[(r - 2)(r + 3) = 0\]So, \(r = 2\) and \(r = -3\).

- Find the Recurrence Relation

For each root \(r\), we find the recurrence relation for coefficients. Starting with \(r = 2\):\[((n + 2)(n + 1) + 2(n + 2) - 6) a_n = 0\]This simplifies to:\[(n^2 + 3n + 2) a_n = 0\]Therefore, every coefficient \(a_n\) must be zero except for the first term. This corresponds to the series:\[y_{1} = a_0 x^{2} = a_0 x^2\]Similarly, for \(r = -3\), \[((n - 3)(n - 4) - 6 a_n = 0\]which simplifies to:\[(n^2 - 7n + 12) a_n = 0\]This also means every coefficient \(a_n\) must be zero except for the first term, corresponding to the series:\[y_{2} = b_0 x^{-3} = b_0 x^{-3}\]

- Write the General Solution

Combining the two series solutions, assuming the linearity principle flows through,\[y(x) = a_0 x^2 + b_0 x^{-3}\]Hence, this is the general solution to the differential equation.

Key concepts

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. Specifically, a second-order differential equation involves the second derivative of the function. These types of equations can describe various phenomena in physics, engineering, and other sciences. For example, the equation given in the exercise: \[ x^{2} y^{\backprime \backprime} + 2x y^{\backprime} - 6 y = 0 \] is a second-order linear differential equation. This equation represents how a function and its rate of change relate to each other, highlighting the importance of these equations in modeling real-world situations.

Series Solution

The series solution method involves finding an unknown function as an infinite sum of terms. Each term in the series is a function of the independent variable (like x) multiplied by a coefficient. In this exercise, we assume a Frobenius series solution of the form: \[ y = \sum_{n=0}^{\backslas hinfty} a_n x^{n + r} \] where the coefficients \( a_n \) are to be determined. This series, similar to a power series but with a possible extra factor \( x^r \), will help in systematically solving differential equations with variable coefficients, especially near singular points. This method is particularly powerful because it can handle cases where standard methods fail.

Indicial Equation

The indicial equation is crucial in the Frobenius method. It helps determine the initial terms of the series solution. By substituting the series into the differential equation and equating the coefficient of the lowest power of \( x \) to zero, we derive the indicial equation. For the given differential equation, the indicial equation is formed by: \[ (r)(r - 1) + 2r - 6 = 0 \] Simplifying this gives: \[ r^2 + r - 6 = 0 \] Solving this quadratic equation, we find the roots \( r = 2 \) and \( r = -3 \). These roots are essential as they determine the form of the series solutions.

Recurrence Relation

The recurrence relation is derived to find the coefficients \( a_n \) once the indicial equation has given the possible values of \( r \). For each root of the indicial equation, a recurrence relation provides a way to calculate each term of the series based on the previous terms. For \( r = 2 \), the recurrence relation is: \[ (n^2 + 3n + 2) a_n = 0 \] which simplifies to indicate \( a_n = 0 \) for all \( n > 0 \). This results in the series: \[ y_1 = a_0 x^2 \] Similarly, for \( r = -3 \), the recurrence relation is: \[ (n^2 - 7n + 12) a_n = 0 \] leading to the series: \[ y_2 = b_0 x^{-3} \] These recurrence relations provide the structure of the possible solutions to the differential equation.