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. $$x^{2} y^{\prime \prime}-6 y=0$$

Short answer

General solution is a combination of power series: \[ y = C_{1} \sum_{n=0}^{\infty} a_{n} x^{n+3} + C_{2} \sum_{n=0}^{\infty} a_{n} x^{n-2} \]

Step-by-step solution

Identify the type of differential equation

The given differential equation is of the form: \[ x^{2} y^{\text{''}} - 6 y = 0 \] Notice that this is a linear second-order differential equation.

Assume a power series solution

Assume a solution of the form: \[ y = \sum_{n=0}^{\infty} a_{n} x^{n+r} \] Then, the first derivative is: \[ y' = \sum_{n=0}^{\infty} a_{n} (n+r) x^{n+r-1} \] And the second derivative is: \[ y'' = \sum_{n=0}^{\infty} a_{n} (n+r)(n+r-1) x^{n+r-2} \]

Substitute the series into the differential equation

Substitute the series and its derivatives into the differential equation: \[ x^{2} \sum_{n=0}^{\infty} a_{n} (n+r)(n+r-1) x^{n+r-2} - 6 \sum_{n=0}^{\infty} a_{n} x^{n+r} = 0 \] Simplifying gives: \[ \sum_{n=0}^{\infty} a_{n} (n+r)(n+r-1) x^{n+r} - 6 \sum_{n=0}^{\infty} a_{n} x^{n+r} = 0 \]

Combine and simplify

Combine terms with the same power of \( x \): \[ \sum_{n=0}^{\infty} \left[ a_{n} (n+r)(n+r-1) - 6 a_{n} \right] x^{n+r} = 0 \]

Set up the indicial equation

The coefficients must satisfy the following equation for each power of \( x \): \[ a_{n} (n+r)(n+r-1) - 6 a_{n} = 0 \]

Solve the indicial equation

Factoring out \( a_{n} \), we get: \[ (n+r)(n+r-1) - 6 = 0 \] Expanding and simplifying gives: \[ n^{2} + (2r-1)n + r(r-1) - 6 = 0 \]

Solve for r

This simplifies to: \[ n^{2} + (2r-1)n + r^{2} - r - 6 = 0 \] For the series to be valid, we must solve for \( r \). For this specific problem: Setting \( n = 0 \), the characteristic equation becomes: \[ r(r-1) - 6 = 0 \] Solving for \( r \), we get the roots: \[ r = 3, \ r = -2 \]

Form the general solution

Using the roots found, the general solution is a combination of the two series solutions: For \( r = 3 \): \[ y_{1} = \sum_{n=0}^{\infty} a_{n} x^{n+3} \] For \( r = -2 \): \[ y_{2} = \sum_{n=0}^{\infty} a_{n} x^{n-2} \] Thus, the general solution is: \[ y = C_{1} y_{1} + C_{2} y_{2} \]

Key concepts

differential equations

Differential equations involve functions and their derivatives. These equations describe a relationship between a function and its rates of change. In this exercise, we deal with a second-order linear differential equation expressed as: \[ x^{2} y'' - 6y = 0 \] Second-order indicates that the highest derivative is the second derivative. Understanding these equations is crucial as they model many physical and engineering systems. The goal is to find the function y that satisfies this relationship.

power series solutions

A power series is a series of the form \( \sum_{n=0}^{\infty} a_{n} x^{n} \). For our differential equation, we assume a solution in the generalized power series form: \[ y = \sum_{n=0}^{\infty} a_{n} x^{n+r} \] Here, \( a_n \) are the coefficients and r are fixed numbers determined by the indicial equation. This method lets us express solutions in a series to solve differential equations when standard methods fail or are too complex.

indicial equation

The indicial equation helps find possible values for r that allow the series solution to work. It's derived by equating the lowest power coefficient of x in the series solution to zero. From our example, the indicial equation was obtained from: \[ (n+r)(n+r-1)-6=0 \] Solving this gives: \[ r=3, \ r=-2 \] These values of r are crucial as they help determine the form of the basic solutions and ensure the correctness of our power series.

characteristic equations

The characteristic equation is a polynomial derived from a linear differential equation. Specifically, for the indicial equation, it facilitates finding the roots of r in the power series solution method. Expanding: \[ \sum_{n=0}^{\infty} a_{n} [(n+r)(n+r-1)-6] x^{n+r} = 0 \] Results in the characteristic equation by equating the factor of x to zero: \[ r(r-1)-6=0 \] It simplifies solving the linear second-order differential equations by providing the roots directly related to the differential equation's behavior.

general solutions

The general solution combines all independent solutions derived from the roots of the indicial equation. For our example, the roots r=3 and r=-2 give two basic series solutions:

• For \( r=3 \): \( y_1 = \sum_{n=0}^{\infty} a_{n} x^{n+3} \)
• For \( r=-2 \): \( y_2 = \sum_{n=0}^{\infty} a_{n} x^{n-2} \)

Thus, the general solution is a linear combination of these as: \[ y = C_1 y_1 + C_2 y_2 \] where \( C_1 \) and \( C_2 \) are arbitrary constants. This form covers all possible solutions, making it fundamental in understanding the complete behavior of the differential equation.