Problem 24 - Differential Equations and Linear Algebra

Question

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

Step-by-Step Solution

Answer

The two linearly independent solutions to the given differential equation on \((0, \infty)\) are: \[y_1(x) = a_0 + a_2 x^2 + a_4 x^4 + a_6 x^6 + ...\] \[y_2(x) = a_1 x + a_3 x^3 + a_5 x^5 + a_7 x^7 + ...\]

Step 1: Assume a power series solution

Assume a solution of the form, \[y(x) = \sum_{n=0}^{\infty} a_n x^n\] where \(a_n\) are the unknown coefficients. Now, calculate the first and second derivatives of the assumed power series.

Step 2: Calculate the derivatives

Differentiate the assumed power series solution with respect to x to get the first and second derivatives: First derivative: \[y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1}\] Second derivative: \[y''(x) = \sum_{n=2}^{\infty} n(N-1) a_n x^{n-2}\]

Step 3: Substitute the derivatives into the differential equation

Plug the derivatives back into the differential equation and obtain: \[\sum_{n=2}^{\infty}n(n-1)a_n x^n + 2\sum_{n=1}^{\infty} n a_n x^n (2+x) + 2\sum_{n=0}^{\infty} a_n x^n (1+x) = 0\]

Step 4: Find the recurrence relation for the coefficients

After equating the powers of x and summing the series, we can determine the recurrence relation for the coefficients \(a_0\), \(a_1\), \(a_2\),...: \[a_{n+2} = \frac{(-2n+1)a_n - 4n a_{n-1}}{(n+1)(n+2)}\]

Step 5: Find two linearly independent solutions

Solve the recurrence relation in Step 4 to find the first few terms of the power series, then we have: \[y_1(x) = a_0 + a_2 x^2 + a_4 x^4 + a_6 x^6 + ...\] \[y_2(x) = a_1 x + a_3 x^3 + a_5 x^5 + a_7 x^7 + ...\] Since \(y_1(x)\) and \(y_2(x)\) have different powers of x, they are linearly independent. The desired two linearly independent solutions to the given differential equation on \((0, \infty)\) are: \[y_1(x) = a_0 + a_2 x^2 + a_4 x^4 + a_6 x^6 + ...\] \[y_2(x) = a_1 x + a_3 x^3 + a_5 x^5 + a_7 x^7 + ...\]

Key Concepts

Power Series Solution

A power series solution is a method used to solve differential equations by expressing the solution as an infinite sum of terms. The general idea is to assume a solution in the form of a power series and then determine its coefficients so that it satisfies the differential equation.

For example, for the differential equation given in the problem, we assume the solution to be:
\[ y(x) = \sum_{n=0}^{\infty} a_n x^n \]
Here, \(a_n\) are the unknown coefficients we need to find. Why use a power series? It's particularly useful when the solution cannot be expressed in terms of elementary functions. It builds up the solution incrementally as a sum of polynomial terms.

**Why Power Series?**

• Flexibility: It allows handling more complex differential equations.
• Approximation: You can approximate solutions using a finite number of terms.
• Convergence: Power series solutions can converge over an interval to provide an accurate solution.

Once we express the assumed solution, differentiate it and substitute back into the differential equation, we can then find the coefficients using a technique called the recurrence relation.

Linearly Independent Solutions

Linearly independent solutions are critical in finding a complete set of solutions to differential equations. For linear differential equations, especially those with constant coefficients, the solutions form a vector space. This means finding a complete basis - a set of solutions from which any other solution can be constructed as a combination.

Two functions, like in this exercise, are linearly independent if no constant exists such that one function is a constant multiple of the other. In simpler terms, they do not overlap in terms of direction or scaling. **Importance of Linearly Independent Solutions**:

• Comprehensive Solution: Ensures all possible solutions are found.
• Non-overlapping: Guarantees distinct solutions that cover different aspects of the problem.

In the exercise, the solutions \( y_1(x) \) and \( y_2(x) \) formed from the power series are linearly independent. They provide a full solution to the differential equation. Mathematically, we say \( y_1(x) = a_0 + a_2 x^2 + a_4 x^4 + \ldots \) and \( y_2(x) = a_1 x + a_3 x^3 + a_5 x^5 + \ldots \) because they arise from different powers of \(x\), ensuring they are non-scalable copies of each other.

Recurrence Relation

A recurrence relation is an equation that expresses the coefficients in a power series in terms of preceding coefficients. This concept is key when dealing with power series solutions for differential equations, allowing us to find all terms iteratively once we have an initial set of terms.

The differential equation in our problem is a perfect use case for recurrence relations. After substituting the derivatives into the differential equation, the series are manipulated into a recurrence relation:
\[ a_{n+2} = \frac{(-2n+1)a_n - 4n a_{n-1}}{(n+1)(n+2)} \]
**Why Recurrence Relations?**

• Generates Terms: Provides a method to generate each term in a series using earlier terms.
• Computational Ease: Simplifies finding many terms without solving complex equations directly.

By using a recurrence relation, we can systematically develop the power series, leading to the complete solutions \( y_1(x) \) and \( y_2(x) \) as described. This step-by-step construction helps in getting precise solutions to complex differential equations without directly solving them.