Problem 15 - Differential Equations and Linear Algebra

Question

Determine two linearly independent solutions to the given differential equation on \((0, \infty)\) $$x^{2} y^{\prime \prime}+x(1-x) y^{\prime}-y=0$$

Step-by-Step Solution

Answer

Using the Frobenius method, we found the first solution \(y_1(x) = x \sum_{n=0}^{\infty} a_n x^n\), where the recursion relation is given by \[a_{n+2}=\frac{a_n}{(n+2)(n+1)}\]. Then, with the method of reduction of order, we obtained the second solution as \(y_2(x) = y_1(x)u(x)\) with \(u(x) = c_1 x + c_2 - \int x v(x) u^{\prime}(x) dx\), where \(v(x)=\frac{y_1^{\prime}(x)}{y_1(x)}\).

Step 1: Frobenius method for the first solution (y1)

We search for a solution of the form \(y_1(x) = x^r \sum_{n=0}^{\infty} a_n x^n\), where r is a constant. Plugging this into the given differential equation and simplifying, we obtain the indicial equation: \[r(r-1)+r(1-r)-1=0\] Solving for r, we have: \[r^2 - 1 = 0 \Rightarrow r_1=1, r_2=-1\] Since the difference between the two roots is an integer, we can obtain only one solution from this method. Choose \(r_1=1\). The recursion relation becomes \[a_{n+2}=\frac{a_n}{(n+2)(n+1)}\] Now we construct the series for \(y_1(x)\) using the recursion relation: \[y_1(x) = x \sum_{n=0}^{\infty} a_n x^n\]

Step 2: Finding the second solution (y2) by the method of reduction of order

To find the second solution, we use the method of reduction of order. We consider a new function \(u(x)\) such that its product with the first solution, \(y_1\), gives a second independent solution: \[y_2(x) = y_1(x)u(x)\] Now, differentiate \(y_2(x)\) twice with respect to x: \[y_2^\prime (x) = y_1^\prime (x) u(x) + y_1(x) u^\prime (x)\] \[y_2^{\prime\prime} (x) = y_1^{\prime\prime}(x)u(x)+2y_1^\prime (x)u^\prime (x)+y_1(x)u^{\prime\prime}(x)\] Plug these expressions for \( y_2(x), y_2^\prime(x), y_2^{\prime\prime}(x)\) into the given differential equation, and group terms by common factors u(x), u'(x), and u''(x). For this equation to hold, the sum of terms in each group must be equal to zero. We focus on the terms involving u'(x): \[0 = x(1-x)(2y_1^\prime u^\prime + y_1u^{\prime\prime})\] Dividing both sides by the non-zero term \(x(1-x)y_1\), we get: \[0 = 2u^{\prime} + u^{\prime\prime}v\] where \(v(x)=\frac{y_1^{\prime}(x)}{y_1(x)}\). Now, we integrate this equation with respect to x once: \[u^\prime (x) = c_1 - \int v(x) u^{\prime\prime}(x) dx\] Integrate once more with respect to x: \[u(x) = c_1 x + c_2 - \int x v(x) u^{\prime}(x) dx\] Now, we can find the second solution, \(y_2(x)\), by multiplying \(u(x)\) with \(y_1(x)\): \[y_2(x) = y_1(x)u(x) = y_1(x) [c_1 x + c_2 - \int x v(x) u^{\prime}(x) dx]\] Thus, we have found two linearly independent solutions, \(y_1(x)\) and \(y_2(x)\), for the given differential equation on \((0, \infty)\).

Key Concepts

Frobenius method

The Frobenius method is a powerful technique for solving linear differential equations, especially near singular points. In this method, we look for a solution in a series form:

• Assume a solution of the form: \(y(x) = x^r \sum_{n=0}^{\infty} a_n x^n\), where \(r\) is a constant.
• Substitute this series into the differential equation.
• Solve for \(r\) using the indicial equation, which determines the possible values for the power series.
• Find a recurrence relation to determine the coefficients \(a_n\).

By considering all terms and coefficients from the series, we ensure that the assumption solves the differential equation at each order of \(x\). In the original solution, applying Frobenius method led us to find one particular solution \(y_1(x)\) of the form \(x \sum_{n=0}^{\infty} a_n x^n\). This first solution fits into the framework provided by the Frobenius approach.

Linear Independence

In the context of solutions to differential equations, "linear independence" refers to the property that one solution cannot be written as a constant multiple or linear combination of another. This is crucial when finding more than one solution, because:

• Two linearly independent solutions can be combined to form the general solution of a second-order homogeneous differential equation.
• These solutions span the solution space, providing the entire set of possible solutions.

For a pair of functions \(y_1(x)\) and \(y_2(x)\) to be linearly independent, their Wronskian determinant must be non-zero over the interval of interest. In simpler terms, if you can show that \(y_1\) and \(y_2\) cannot be made into each other through any constant factor, they are independent. In this problem, the method of Frobenius gives just one solution, and the reduction of order helps us find the second one.

Reduction of Order

"Reduction of Order" is a technique used to find a second linearly independent solution when one solution is already known. Here's how it works:

• Start with a known solution, \(y_1(x)\).
• Assume a second solution of the form \(y_2(x) = y_1(x)u(x)\), where \(u(x)\) is an unknown function to be determined.
• Plug \(y_2(x)\) into the original differential equation, simplify, and separate terms involving \(u(x)\), \(u'(x)\), and \(u''(x)\).

The process simplifies the original equation into a form that allows us to solve for \(u(x)\). The method capitalizes on the existing solution ensuring that by choosing the correct \(u(x)\), the modified solution satisfies the differential equation. This approach was used to find \(y_2(x)\) from the first solution obtained by the Frobenius method.

Indicial Equation

When applying the Frobenius Method to a differential equation, the "indicial equation" arises as a crucial step. It determines the initial values required for solving the differential equation near a singular point. Here's what you need to know:

• The indicial equation is derived by substituting the Frobenius series into the differential equation and analyzing the lowest powers of \(x\).
• Solving this equation gives the possible values for the exponent \(r\) in the power series expansion.
• These values of \(r\) are key as they dictate the form and behavior of the solutions near the singularity.

In the step-by-step solution, solving the indicial equation \(r^2 - 1 = 0\) gives roots \(r_1 = 1\) and \(r_2 = -1\). These roots guide us in constructing solutions, and the difference between them (an integer) tells us how many solutions the Frobenius method initially provides. It also indicates whether additional techniques, such as reduction of order, might be necessary to find all linearly independent solutions.