Question

Solve each of the following differential equations by the Frobenius method; observe that you get only one solution. (Note, also, that the two values of $s$ are equal or differ by an integer, and in the latter case the larger $s$ gives the one solution.) Show that the conditions of Fuchs's theorem are satisfied. Knowing that the second solution is $\ln x$ times the solution you have, plus another Frobenius series, find the general solution. (It is convenient to note that the value of $s$ in the second Frobenius series is always the same as the second value of $s$ which did not give a solution in the first part of the problem.) $x^2{y}^{\prime \prime }+(x^3-3x)y^{\prime }+(4-2x)y=0$

Short answer

Solve the indicial equation near x=0 using the Frobenius method. Verify the conditions of Fuchs's theorem, and use the larger value of s for the general solution.

Step-by-step solution

Identify the form of the equation

The given differential equation is: $$x^2{y}^{‵‵}+(x^3-3x)y^{‵}+(4-2x)y=0$$ This is a second-order linear homogeneous differential equation.

Find the indicial equation

Rewrite the equation near the regular singular point (here, x = 0). The Frobenius method assumes a solution in the form of a power series: $$y=(x^s{\text{backsum}}_0^{\text{backinfty}}{\text{backalpha}}_n{x}^n)$$. Calculate the indicial equation by substituting this assumed solution into the differential equation and equating the smallest power of x to zero.

Set up the Frobenius series

Given the general form of the Frobenius series, assume: $$y=x^s{\text{backsum}}_{n=0}^{\text{backinfty}}{c}_n{x}^n{\text{y}}^{\prime }={\text{backsum}}_{n=0}^{\text{backinfty}}{c}_n(n+s)x^{n+s-1}{\text{y}}^{″}={\text{backsum}}_{n=0}^{\text{backinfty}}{c}_n(n+s)(n+s-1)x^{n+s-2}$$.

Substitute into the original equation

Substitute the expressions for y, y', and y'' back into the original differential equation:$$x^2{\text{backsum}}_{n=0}^{\text{backinfty}}{c}_n(n+s)(n+s-1)x^{n+s-2}+(x^3-3x){\text{backsum}}_{n=0}^{\text{backinfty}}{c}_n(n+s)x^{n+s-1}+(4-2x){\text{backsum}}_{n=0}^{\text{backinfty}}{c}_n{x}^{n+s}=0$$

Derive the indicial equation

Combine the terms and collect the coefficients of like powers of x. Set the coefficient of the lowest power of x (x=0) to zero to get the indicial equation. This will provide the roots (values of s).

Solve the indicial equation

Solve for s. If there are equal roots or if the roots differ by an integer, there will be only one solution. Find the larger value of s.

Verify Fuchs's theorem

Ensure that the conditions of Fuchs's theorem are satisfied. This theorem guarantees that the solution to the differential equation will consist of a Frobenius series.

Find the general solution

Combine the solution from the larger value of s. For the other value of s, note that the second solution can be written as $$y_2=y_1(\text{backln}x)\text{+}(Frobeniusserieswithsequaltothesecondvalue).$$

Key concepts

Second-Order Linear Differential Equations

In mathematics, differential equations are equations that include derivatives, representing rates of change. A second-order differential equation includes the second derivative, often written as $y^{″}$. These equations are fundamental because they model many physical systems and phenomena, from simple motion to complex electrical circuits.

A linear differential equation means that the function and its derivatives are linear. The given problem is a second-order linear differential equation since it can be written as: $$x^2{y}^{″}+(x^3-3x)y^{\prime }+(4-2x)y=0.$$ Solving such an equation often requires special techniques, especially around points where the equation might behave oddly.
Here is a general form of a second-order linear differential equation: $$a(x)y^{″}+b(x)y^{\prime }+c(x)y=0,$$ where $a(x),b(x),$ and $c(x)$ are functions of $x$. In our problem, $a(x)=x^2,b(x)=x^3-3x,$ and $c(x)=4-2x$.

Indicial Equation

The indicial equation is a tool used in the Frobenius method to find the exponents at a regular singular point. This method assumes a solution in the form of a power series: $$y=x^s(\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^n)$$

By substituting this power series into the differential equation, we collect terms involving different powers of $x$. The indicial equation arises from setting the coefficient of the lowest power of $x$ to zero. This helps determine the possible values of $s$, which are the exponents of the leading term in the solution.

In our sample problem, substituting the power series form into the differential equation and simplifying, we find the indicial equation and solve for $s$. The values of $s$ critically influence the form of the solutions we seek.

Fuchs's Theorem

Fuchs's theorem is crucial for understanding the behavior of solutions to second-order linear differential equations. It states that if a differential equation has a regular singular point at $x=0$, then there exists at least one solution of the form of a Frobenius series. A point is called a regular singular point if the functions $p(x)=\frac{b(x)}{a(x)}$ and $q(x)=\frac{c(x)}{a(x)}$ are analytic (well-behaved) at that point.

In our problem, identifying that $x=0$ is a regular singular point ensures that we can use the Frobenius method for finding solutions. Without this step, we might not be able to guarantee the form or existence of the solutions we need. Fuchs's theorem provides a foundation for working with such differential equations.

Power Series Solutions

The power series solution involves expressing the solution $y$ as a sum of powers of $x$, usually around a point where the function behaves well, often zero (a regular singular point). The form of the power series is: $$y=(x^s{\text{backsum}}_0^{\text{backinfty}}{\text{backalpha}}_n{x}^n).)$$

Substituting this series into the given differential equation allows us to determine the coefficients ${\alpha }_n$. Solving the differential equation in this way can be complex, but it often yields very precise information about the function's behavior near the point of expansion.

In our worked example, after finding the indicial equation and solving for $s$, we use the larger $s$ value to determine one solution. For the second solution (when two $s$ differ by an integer), we need to consider another Frobenius series and potentially logarithmic terms. This approach makes power series solutions a potent tool for solving differential equations around singular points.