Question

In Exercises \(61-68\) use the method suggested by Exercise 60 to find the general solution on some interval \((0, \rho)\) $$ 4 x^{2}\left(1+3 x+x^{2}\right) y^{\prime \prime}-4 x\left(1-3 x-3 x^{2}\right) y^{\prime}+3\left(1-x+x^{2}\right) y=0 $$

Short answer

Based on the step-by-step solution provided, find the general solution of the given second-order ODE on the interval (0, ρ). Answer: The general solution on the interval (0, ρ) is given by: $$ y(x) = a_0\sum_{n=0}^{\infty} \frac{x^{n+r}}{4^n\Gamma(n+r+1)} + a_1\sum_{n=0}^{\infty} \frac{x^{n+r}}{4^n\Gamma(n+r+2)} $$

Step-by-step solution

Write down the given ODE

The given ODE is: $$ 4x^{2}(1+3x+x^{2})y^{\prime\prime} - 4x(1-3x-3x^{2})y^{\prime} + 3(1-x+x^{2})y = 0 $$

Rewrite the ODE in a simplified form

It is convenient to rewrite the ODE in a simplified form: $$ (1+3x+x^{2})y^{\prime\prime} - \frac{1-3x-3x^{2}}{x}y^{\prime} + \frac{3}{4x^{2}}(1-x+x^{2})y = 0 $$

Guess the solution using the Frobenius method

We guess the solution in the form of a power series: $$ y(x) = \sum_{n=0}^{\infty} a_nx^{n+r} $$ where \(a_n\) are coefficients to be determined and \(r\) is an unknown exponent.

Compute the first and second derivatives of y(x)

Compute the first and second derivatives of \(y(x)\): $$ y'(x) = \sum_{n=0}^{\infty} (n+r)a_nx^{n+r-1} $$ and $$ y^{\prime\prime}(x) = \sum_{n=0}^{\infty} (n+r)(n+r-1)a_nx^{n+r-2} $$

Substitute the derivatives into the ODE and simplify

Now, substitute \(y(x)\), \(y'(x)\), and \(y^{\prime\prime}(x)\) into the simplified ODE and simplify the result: $$ \sum_{n=0}^{\infty}[(n+r)(n+r-1)a_n(1+3x+x^{2})x^{n+r-2} - (n+r)a_n(1-3x-3x^{2})x^{n+r-1} + (1-x+x^{2})a_nx^{n+r-2}] = 0 $$

Equate coefficients of the same powers of x

To satisfy the equation for all values of \(x\), the coefficients of the same powers of \(x\) must be equal: $$ (n+r)(n + r - 1)a_n(1+3x+x^{2}) - (n+r)a_n(1-3x-3x^{2}) + (1-x+x^{2})a_n = 0 $$ This allows us to write down a three-term recurrence relation: $$ a_{n+2} = \frac{(n+2)(n+1-3x)(n+1)}{4(n+r+2)(n+r+1)}a_n, \quad \text{for} \; n=0, 1, 2,... $$

Solve the recurrence relation

Solve the recurrence relation for \(n=0\) and \(n=1\) respectively to determine the initial coefficients: $$ a_2 = \frac{a_0}{4(2+r)(1+r)} $$ $$ a_3 = \frac{a_1}{4(3+r)(2+r)} $$ Continuing to solve the recurrence relation will provide the general coefficients of \(a_n\).

Obtain the general solution

Use the coefficients obtained from the recurrence relation to find the general solution \(y(x)\): $$ y(x) = a_0\sum_{n=0}^{\infty} \frac{x^{n+r}}{4^n\Gamma(n+r+1)} + a_1\sum_{n=0}^{\infty} \frac{x^{n+r}}{4^n\Gamma(n+r+2)} $$ Here, \(\Gamma(n+1) = n!\) for integer values of n. This is the general solution of the given ODE on the interval \((0, \rho)\).

Key concepts

Frobenius Method

The Frobenius method is a technique used to solve linear ordinary differential equations (ODEs) around singular points. A singular point of a differential equation is a value of the independent variable, in this case, x, where the coefficients of the ODE are undefined or infinite. When an ODE has a regular singular point, this method allows us to find solutions in the form of a power series that involve the variable raised to a power, often with an undetermined exponent r.

For example, given an ODE, the first step is to express the equation in a suitable form highlighting its singular point, usually at x = 0. Then, we assume a solution exists as a power series with coefficients that we need to determine. This series is substituted into the ODE, allowing us to compute derivatives and equate coefficients of corresponding powers of x to derive a recurrence relation, enabling the calculation of the power series coefficients step by step. The Frobenius method systematically builds a solution that can be tailored to a specific ODE, typically producing a series solution that is valid in a certain interval around the singular point.

Recurrence Relation

A recurrence relation in the context of differential equations is a mathematical expression that relates coefficients of different terms in a sequence or series. It is an iterative process, used to determine subsequent terms based on the initial values and the relation itself, often derived from equating coefficients when solving differential equations with methods like Frobenius.

In our exercise, after substituting the assumed power series solution into the original ODE, we arrived at a three-term recurrence relation. This relation connects the coefficient a_n+2 to the coefficient a_n, with specific dependencies on the index n and the undetermined exponent r. The initial coefficients, a₀ and a₁, set the stage for computing the entire series coefficients by applying the relation repeatedly. Recurrence relations not only provide a method to calculate the coefficients but also impose conditions on the solution, ensuring its validity within the interval of interest.

Power Series Solution

A power series solution to an ODE is a series expansion of the unknown function in terms of powers of the independent variable. It is represented as an infinite sum of terms, with each term being a coefficient multiplied by a power of the variable. The power series approach allows for solutions to ODEs that are otherwise difficult or impossible to solve with elementary functions.

In our case, once the coefficients are found through the recurrence relation, the solution is expressed as a power series that is a sum of an infinite number of terms. This power series is centered around the singular point and converges within a certain radius. In the exercise provided, the solution y(x) is expressed as a power series whose coefficients are determined by the recurrence relation. This sort of series solution is especially useful when seeking analytic solutions near points where the ODE's behavior is complex, and it provides a general solution that can be tailored to specific boundary conditions or initial values by choosing appropriate coefficients a₀ and a₁.