Question

Find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients. \(x^{2}(1-2 x) y^{\prime \prime}+3 x y^{\prime}+(1+4 x) y=0\)

Short answer

#Answer# The general solution to the given differential equation can be expressed as a linear combination of two linearly independent Frobenius series solutions: \(y(x) = C_1y_1(x) + C_2y_2(x)\) Where \(y_1(x)\) and \(y_2(x)\) are given by: \(y_1(x) = \sum_{n=0}^\infty a_n x^{n+r_1}\) \(y_2(x) = \sum_{n=0}^\infty \tilde{a}_n x^{n+r_2}\) The coefficients \(a_n\) and \(\tilde{a}_n\) satisfy the recurrence relation: \(a_n = \frac{2(n+r-1)(n+r)+3(n+r)}{(n+r)(r+n-1)}a_{n-1} - \frac{4}{(n+r)(r+n-1)}a_{n-1}\) Using different values for \(r_1\) and \(r_2\), we can compute the coefficients for each solution and form a fundamental set of Frobenius solutions with explicit formulas for the coefficients.

Step-by-step solution

Express the solution in Frobenius series

We'll start by expressing the solution as a Frobenius series: \(y(x) = \sum_{n=0}^\infty a_n x^{n+r}\) The first derivative is: \(y'(x) = \sum_{n=0}^\infty (n+r) a_n x^{n+r-1}\) The second derivative is: \(y''(x) = \sum_{n=0}^\infty (n+r)(n+r-1) a_n x^{n+r-2}\)

Substitute the Frobenius series into the differential equation

Plug the Frobenius series for \(y(x)\), \(y'(x)\), and \(y''(x)\) into the given differential equation: \(x^{2}(1-2x)\sum_{n=0}^\infty (n+r)(n+r-1)a_n x^{n+r-2} + 3x\sum_{n=0}^\infty(n+r) a_n x^{n+r-1} + (1 + 4x)\sum_{n=0}^\infty a_n x^{n+r} = 0\)

Simplify the equation

Now, we will simplify the equation by combining the summations and simplifying the powers of x: \(\sum_{n=0}^\infty x^{n+r}(a_n(n+r)(n+r-1)(1-2x)+3(n+r)a_nx+((1+4x)a_n))=0\)

Collect coefficients and remove x terms

To satisfy the equation, the coefficient of each power of x must be equal to zero. We'll now collect coefficients of the same power of x and remove x terms: \(\sum_{n=0}^\infty a_n(n+r)(n+r-1)(1-2x)x^{n+r} + \sum_{n=0}^\infty 3(n+r)a_n x^{n+r} + \sum_{n=0}^\infty (1+4x)a_n x^{n+r} = 0\)

Compare coefficients and express dependence

Comparing coefficients yields the following relationships: \((n+r)(n+r-1)a_n - 2(n+r)(n+r-1)a_{n-1} + 3(n+r)a_n + a_n + 4a_{n-1} = 0\) Re-arranging the equation, we get the recurrence relation for the coefficients: \(a_n = \frac{2(n+r-1)(n+r)+3(n+r)}{(n+r)(r+n-1)}a_{n-1} - \frac{4}{(n+r)(r+n-1)}a_{n-1}\)

Find the two linearly independent solutions

The general solution to the differential equation is a linear combination of two linearly independent solutions. The linearly independent solutions can be expressed in terms of the coefficients of the Frobenius series as follows: \(y_1(x) = \sum_{n=0}^\infty a_n x^{n+r_1}\) \(y_2(x) = \sum_{n=0}^\infty \tilde{a}_n x^{n+r_2}\) Using the recurrence relation for \(a_n\), we can compute the coefficients for each solution using different values of \(r_1\) and \(r_2\). Then, from that recurrence relation, we can find the fundamental set of Frobenius solutions and provide explicit formulas for the coefficients.

Key concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations involving a function and its derivatives. They come in various types and orders. In particular, the order of an ODE refers to the highest derivative present in the equation.

The exercise we're focusing on involves a second-order ODE, which is common in mathematical physics. These equations often model systems where acceleration (the second derivative of displacement) is involved, such as in mechanical vibrations.

Solving ODEs often requires particular techniques depending on their complexity. In this instance, we employ the Frobenius method, which is particularly effective for equations with regular singular points. This method allows the solution to be expressed as a power series, making it easier to handle complex functions that cannot be solved with simpler methods.

Series Solutions

The Frobenius method used in solving the given ODE is a type of series solution. The core idea behind series solutions is to express the unknown function as a series, often a power series or a Frobenius series, to reformulate the original equation.

In this context, a function is expressed as:
\[y(x) = \sum_{n=0}^\infty a_n x^{n+r}\]
where \(a_n\) are the coefficients of the series, and \(r\) is determined through substitution into the differential equation.

By substituting the series into the ODE and aligning terms of equal powers of \(x\), we can create a recurrence relation for the coefficients. This method is incredibly useful for equations with variable coefficients that aren't easily solvable by traditional means.

Linear Independence

Linear independence is a crucial concept in solving differential equations using series solutions. Solutions to differential equations generally involve combinations of functions, and determining their linear independence is key to obtaining a fundamental set of solutions.

For two functions to be linearly independent, there must be no constant \(c\) such that:
\[y_1(x) = c \, y_2(x)\]
In the case of the Frobenius method, we identify two linearly independent solutions \(y_1(x)\) and \(y_2(x)\) that form the fundamental set. This is accomplished by finding different roots \(r_1\) and \(r_2\) in the series solution, leading to distinct solutions. These solutions are not just arbitrary; they ensure that any potential solution to the ODE can be expressed as a combination of \(y_1(x)\) and \(y_2(x)\).

Recurrence Relations

Recurrence relations play a pivotal role in deriving the coefficients in the series solution of an ODE. Derived from the process of substituting a series into the ODE, a recurrence relation provides a way to compute each coefficient \(a_n\) in terms of preceding ones.

In our specific example, this process led to the relation:
\[a_n = \frac{2(n+r-1)(n+r)+3(n+r)}{(n+r)(r+n-1)}a_{n-1} - \frac{4}{(n+r)(r+n-1)}a_{n-1}\]
This relation allows us to systematically calculate all coefficients required for the series, starting from a known coefficient, often \(a_0\).

Recurrence relations simplify the calculation process significantly, ensuring we can construct the series solution accurately for any required degree of precision.