Question

Solve the 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 second solution. $$x^{2}(x-1) y^{\prime \prime}-x(5 x-4) y^{\prime}+(9 x-6) y=0$$

Short answer

The solution is \( y_1 = x^3 \sum_{n=0}^{\infty} a_n x^n \). The second solution is \( y_2 = y_1 \ln(x) + x^3 \sum_{n=0}^{\infty} b_n x^n\).

Step-by-step solution

Identifying the nature of the singularity

Determine if the singular points of the differential equation are regular singular points. The equation given is \[ x^{2}(x-1) y^{\u0018prime}prime}-x(5x-4) y^{\u0018prime}+(9x-6) y=0. \]The singular points are at \(x=0\) and \(x=1\).

Transforming the differential equation

Divide the differential equation by \(x^2(x-1)\) to transform it into the standard form:\[ y^{\u0018prime\text{2}}+\frac{4x-5}{x(x-1)} y^{\text{1}}+\frac{9x-6}{x^2(x-1)}y=0.\]

Applying Frobenius method

Assume a solution of the form \( y = \sum_{n=0}^{\infty} a_n x^{n+s} \). Substitute the series into the transformed differential equation and equate coefficients of like powers of \(x\) to zero.

Determining the indicial equation

From the lowest power of \(x\), derive the indicial equation by setting the coefficient of \(x^s\) to zero. This gives the indicial equation \( s(s-1) -5s +9=0 \).Solve the indicial equation for \(s\).

Solving the indicial equation

The indicial equation \( s^2-6s+9=0 \) simplifies to \( (s-3)^2=0 \). Therefore, \(s=3\) is a double root.

Finding the recurrence relation

For \(s=3\), find the recurrence relation for \(a_n\). Substitute \(y = x^3 \sum_{n=0}^{\infty} a_n x^n\) into the differential equation and simplify to find the coefficients \(a_n\).

Constructing the first solution

Using the recurrence relation, determine the coefficients \(a_n\) for the first solution. The first solution is of the form \( y_1 = x^3 \sum_{n=0}^{\infty} a_n x^n.\)

Verifying Fuchs's theorem conditions

Check that the conditions of Fuchs's theorem are satisfied, confirming that the singular points are regular and the solutions form a Frobenius series.

Finding the second solution

When \(s_1 - s_2\) is an integer and \(s_1\) gives the only solution, the second solution \(y_2\) is given by \( y_2 = y_1 \ln(x) + x^s \sum_{n=0}^{\infty} b_n x^n\). Substitute this into the differential equation to determine \(b_n\).

Constructing the second solution

Combine the logarithmic term with the Frobenius series to construct the second solution: \( y_2 = y_1 \ln(x) + x^3 \sum_{n=0}^{\infty} b_n x^n.\)

Key concepts

Differential Equations

A differential equation is an equation involving an unknown function and its derivatives. In this exercise, we are dealing with a second-order differential equation of the form: \[ x^{2}(x-1) y^{\text{''}} - x(5x-4) y^{\text{'}} + (9x-6) y = 0 \]Here, the unknown function is \( y \), and \( y' \) and \( y'' \) are its first and second derivatives, respectively. Differential equations come in various orders, and their solutions can be functions that describe physical, biological, or even economic systems. This particular equation is a linear homogeneous equation as it involves a linear combination of \( y \), \( y' \), and \( y'' \) equating to zero without any added constants or functions. To find the solutions, specific methods need to be applied. The Frobenius method is one such technique, particularly effective for solving differential equations around singular points.

Fuchs's Theorem

Fuchs's theorem is crucial for understanding solutions to differential equations around singular points. In simple terms, it states that if the singular points of a linear differential equation are regular, you can express the solutions in terms of a Frobenius series. A singular point is where the coefficients of the highest derivative (in our case, \( y'' \)) are either undefined or become infinite. Regular singular points allow for such series solutions, making the Frobenius method applicable. In the given differential equation, the singular points are at \( x = 0 \) and \( x = 1 \). Verifying that these points meet Fuchs's theorem conditions ensures we can proceed with the Frobenius series representation of the solution.

Indicial Equation

The indicial equation is derived from the lowest power of \( x \) in the Frobenius series substitution. It's a critical step in the Frobenius method, as it helps determine the possible values for \( s \), which are essential for constructing the series solution. Given the equation:\[ x^{2}(x-1) y^{\text{''}} - x(5x-4) y^{\text{'}} + (9x-6) y = 0 \]And substituting \( y = \sum_{n=0}^{\infty} a_n x^{n+s} \), we derive the indicial equation:\[ s(s-1) - 5s + 9 = 0 \]Solving this indicial equation results in the roots \( s = 3 \) which occurs as a double root. These values signify the exponents in our Frobenius series and are crucial for solving the differential equation around the singular point.

Recurrence Relation

Once we have the value of \( s \) from the indicial equation, the next step is to find the recurrence relation. This relation connects the coefficients \( a_n \) in the Frobenius series: \( y = x^3 \sum_{n=0}^{\infty} a_n x^n \). By substituting this back into our differential equation and simplifying, we get an equation that relates each \( a_n \) to previous coefficients. The recurrence relation helps in determining each coefficient step by step. For the given problem, once we have the recurrence relation, we can construct the first solution:\[ y_1 = x^3 \sum_{n=0}^{\infty} a_n x^n \]This process involves detailed algebraic manipulation, but the ultimate goal is to express the solution as a power series.