Problem 25 - Differential Equations and Linear Algebra

Question

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

Step-by-Step Solution

Answer

Two linearly independent solutions to the given differential equation on $(0, \infty)$ are: $$y_1(x) = a_0\left(1 + \frac{1}{2}x^2 + \frac{3}{20}x^4 + ...\right)$$ $$y_2(x) = a_1\left(x + \frac{2}{12}x^3 +\frac{2}{15}x^5 + ...\right)$$ These solutions are obtained by converting the differential equation into a power series, obtaining a recurrence relation for the coefficients, and then constructing two series solutions from the coefficients.

Step 1: Convert Differential Equation to a Power Series Equation

We can convert our differential equation into a power series equation by assuming that our solution y(x) takes the form of a power series given by: $$y(x) = \sum_{n=0}^{\infty} a_nx^n$$ Then the first and second derivatives of y(x) can be calculated as follows: $$y'(x) = \sum_{n=1}^{\infty} n a_nx^{n-1} \quad \text{and} \quad y''(x) = \sum_{n=2}^{\infty} n (n-1) a_nx^{n-2}$$ Now insert these expressions into the given differential equation and simplify: \(x^{2}(\sum_{n=2}^{\infty} n (n-1) a_nx^{n-2}) - x(1-x)(\sum_{n=1}^{\infty} n a_nx^{n-1}) + (1-x)(\sum_{n=0}^{\infty} a_nx^n) = 0\)

Step 2: Write down the recurrence relation

To be able to equate the power series coefficients on both sides, we first need to match the power of x. We can do this by changing the summation index in the second term of the equation: \(\sum_{n=0}^{\infty} n (n+1) a_{n+2}x^{n} - (1-x)\sum_{n=0}^{\infty} n a_{n+1}x^{n} + (1-x)\sum_{n=0}^{\infty} a_nx^n = 0\) Now, observe that all three series are multiplications of \((1-x)\) term in the second and third series. We can remove (1-x) terms by shifting the index appropriately. Compare the coefficients of x^n on both sides, we can get the following recurrence relation: $$ a_{n+2} = \frac{n-1}{n(n+1)} a_{n}$$

Step 3: Find the first few coefficients of the power series

Using the recurrence relation obtained, we can find the first few coefficients of the power series: $$a_2 = \frac{1}{2} a_0$$ $$a_3 = \frac{2}{3(3+1)} a_1$$ $$a_4 = \frac{3}{4(4+1)} a_2 = \frac{3}{20} a_0$$ $$a_5 = \frac{4}{5(5+1)} a_3 = \frac{2}{15} a_1$$ $$...$$ Notice the first few coefficients suggest there might be two series solutions where we obtain two sequences, one with even powered terms, and one with odd powered terms.

Step 4: Write the final form of the two linearly independent solutions

Since we have found two sequences of coefficients corresponding to odd and even powers of x, we can construct two linearly independent solutions by separating these two sequences: $$y_1(x) = a_0\left(1 + \frac{1}{2}x^2 + \frac{3}{20}x^4 + ...\right)$$ $$y_2(x) = a_1\left(x + \frac{2}{12}x^3 +\frac{2}{15}x^5 + ...\right)$$ These two solutions, \(y_1(x)\) and \(y_2(x)\), are linearly independent and form the solution set of the given differential equation on the interval \((0, \infty)\).

Key Concepts

Power Series Solutions

Turning a differential equation into a power series solution allows us to express our function as an infinite sum of its terms. This method is particularly useful when solving differential equations where traditional techniques may not work. In our problem, we express the solution to the differential equation as a series:

• Begin with the assumption that our solution, \(y(x)\), can be written as a power series: \[ y(x) = \sum_{n=0}^{\infty} a_nx^n \]
• This assumption lets us differentiate term by term to find the derivatives required for the differential equation.

By substituting these series expressions for \(y(x)\), its first derivative \(y'(x)\), and its second derivative \(y''(x)\) into the differential equation, we can simplify the equation to a series form. This method enables us to systematically find the recursion relations needed to solve the equation.

Linear Independence

Linearly independent solutions are crucial in differential equations because they span the solution space for our problem. For our equation, we derived two series solutions, \( y_1(x) \) and \( y_2(x) \), one constructed from coefficients related to even powers of \(x\) and the other from odd powers. These solutions are considered linearly independent if no scalar multiple of one is equal to the other:

• Two functions \(f(x)\) and \(g(x)\) are linearly independent if the equation \( c_1f(x) + c_2g(x) = 0 \) implies \( c_1 = c_2 = 0 \).
• In terms of series, this means that the series home different coefficients and terms to prevent one from being a scalar multiple of the other.

Hence, the solutions \( y_1(x) \) and \( y_2(x) \) derive two independent directions in the solution space for the differential equation, covering all possible solutions on the given interval \((0, \infty)\).

Recurrence Relations

Recurrence relations are equations that recursively define a sequence, once a few initial terms are given. In power series solutions, they guide us in finding coefficients of the series term by term. For our example:

• We translate the differential equation into a condition for each coefficient.
• This serves as a "recipe" for finding subsequent coefficients using initial coefficients, often starting with arbitrary constants, like \(a_0\) and \(a_1\).

The recurrence relation for our equation was found to be: \[ a_{n+2} = \frac{n-1}{n(n+1)} a_{n} \] This framework generates all coefficients of our power series, providing us with the requisite tools to complete the power series solutions.

Second Order Differential Equations

Second order differential equations involve the second derivative of a function and are fundamental in mathematical modeling of physical systems. They typically take the form:

• \( a(x) y'' + b(x) y' + c(x) y = 0 \)

In our specific problem, we have: \[ x^{2} y^{\prime \prime}-x(1-x) y^{\prime}+(1-x) y=0 \] These equations often require at least two initial or boundary conditions to find a unique solution. However, finding two linearly independent solutions spans the space of all potential solutions.

• To solve these, we usually rely on a mixture of analytic and numerical methods, tailored to the characteristics of the coefficients \(a(x)\), \(b(x)\), and \(c(x)\).

By converting to a power series, we re-imagine the solution method suitable for variable coefficients not easily approachable by simple algebraic techniques.