3 Forms Of Frobenius Method With Equal Roots

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In mathematics, the Frobenius method is a powerful tool used to find solutions to ordinary differential equations (ODEs). This method is particularly useful when dealing with second-order linear ODEs with variable coefficients. One of the most interesting cases that arise in the Frobenius method is when the indicial equation has equal roots. In this article, we will delve into the three forms of the Frobenius method that emerge when the roots are equal, exploring each form in detail and providing practical examples to illustrate the concepts.

Importance of the Frobenius Method

Before diving into the specifics of equal roots, it's essential to understand why the Frobenius method is so important. The Frobenius method provides a systematic way to find solutions to ODEs in the form of a power series. This approach is particularly useful when dealing with ODEs that have variable coefficients, which are common in many areas of physics, engineering, and mathematics. By expressing the solution as a power series, the Frobenius method allows us to find solutions that might be difficult or impossible to obtain using other methods.

Equal Roots in the Indicial Equation

The indicial equation is a quadratic equation that arises in the Frobenius method. It is used to determine the values of the roots, which in turn determine the form of the solution. When the roots of the indicial equation are equal, it means that the discriminant of the quadratic equation is zero. This leads to three distinct forms of the Frobenius method, each with its own set of rules and procedures.

Form 1: The First Form

In the first form, the solution is expressed as a single power series with a common root. This form is similar to the standard Frobenius method, but with the added complexity of equal roots. To find the solution, we assume a power series of the form:

y(x) = ∑[n=0 to ∞] a_n x^(n+r)

where r is the common root. We then substitute this series into the ODE and equate coefficients to determine the values of the coefficients a_n.

Example: First Form

Consider the ODE:

x^2 y'' + 3xy' + (x^2 - 2)y = 0

The indicial equation has equal roots r = 1. We assume a power series solution of the form:

y(x) = ∑[n=0 to ∞] a_n x^(n+1)

Substituting this series into the ODE and equating coefficients, we find that the coefficients a_n satisfy the recurrence relation:

a_n = -a_(n-2) / (n+1)

Using this relation, we can compute the coefficients a_n and obtain the solution:

y(x) = a_0 x + a_1 x^2 - a_0 x^3/3! +...

Form 2: The Second Form

In the second form, the solution is expressed as a sum of two power series with different roots. This form is more complex than the first form, as it involves two separate power series. To find the solution, we assume a power series of the form:

y(x) = ∑[n=0 to ∞] a_n x^(n+r) + ∑[n=0 to ∞] b_n x^(n+s)

where r and s are the equal roots. We then substitute this series into the ODE and equate coefficients to determine the values of the coefficients a_n and b_n.

Example: Second Form

Consider the ODE:

x^2 y'' + 2xy' + (x^2 - 4)y = 0

The indicial equation has equal roots r = s = 2. We assume a power series solution of the form:

y(x) = ∑[n=0 to ∞] a_n x^(n+2) + ∑[n=0 to ∞] b_n x^(n+2)

Substituting this series into the ODE and equating coefficients, we find that the coefficients a_n and b_n satisfy the recurrence relations:

a_n = -a_(n-2) / (n+2) b_n = -b_(n-2) / (n+2)

Using these relations, we can compute the coefficients a_n and b_n and obtain the solution:

y(x) = a_0 x^2 + a_1 x^3 - a_0 x^4/4! +... + b_0 x^2 + b_1 x^3 - b_0 x^4/4! +...

Form 3: The Third Form

In the third form, the solution is expressed as a single power series with a logarithmic term. This form is the most complex of the three, as it involves a logarithmic term in addition to the power series. To find the solution, we assume a power series of the form:

y(x) = ∑[n=0 to ∞] a_n x^(n+r) + c x^r log(x)

where r is the equal root. We then substitute this series into the ODE and equate coefficients to determine the values of the coefficients a_n and c.

Example: Third Form

Consider the ODE:

x^2 y'' + 4xy' + (x^2 - 4)y = 0

The indicial equation has equal roots r = 2. We assume a power series solution of the form:

y(x) = ∑[n=0 to ∞] a_n x^(n+2) + c x^2 log(x)

Substituting this series into the ODE and equating coefficients, we find that the coefficients a_n and c satisfy the recurrence relation:

a_n = -a_(n-2) / (n+2) c = a_0 / 2

Using these relations, we can compute the coefficients a_n and c and obtain the solution:

y(x) = a_0 x^2 + a_1 x^3 - a_0 x^4/4! +... + a_0 x^2 log(x) / 2

Gallery of Frobenius Method Equal Roots

FAQ

In conclusion, the Frobenius method with equal roots is a powerful tool for solving ODEs. The three forms of the method provide a systematic way to find solutions in the form of a power series, even when the roots of the indicial equation are equal. By understanding the rules and procedures for each form, we can tackle a wide range of ODEs and find solutions that might be difficult or impossible to obtain using other methods.