Question

Solve the given differential equation by means of a power series about the given point \(x_{0} .\) Find the recurrence relation; also find the first four terms in each of two linearly independent solutions (unless the series terminates sooner). If possible, find the general term in each solution. \(\left(3-x^{2}\right) y^{\prime \prime}-3 x y^{\prime}-y=0, \quad x_{0}=0\)

Short answer

Question: Using the power series method, find the recurrence relation and the first four terms of two linearly independent solutions to the given differential equation \((3-x^2)y''-3xy'-y=0\) with initial point \(x_0=0\). If possible, find the general term of each solution. Answer: The recurrence relation is \(a_n=\dfrac{a_{n-1}}{n-2}\) for \(n\geq2\) and \(a_0=a_0\); \(a_1=-3a_1\). The two linearly independent solutions are: 1) \(y_1(x)=1\) 2) \(y_2(x)=x\) The general solution is: \(y(x)=c_1+c_2x\), where \(c_1\) and \(c_2\) are constants.

Step-by-step solution

Write the Initial Form of the Power Series Solution

We can express the power series solution in the form: \(y(x) = \sum_{n=0}^{\infty} a_nx^n\) We will also need the first and second derivatives of the power series: \(y'(x) = \sum_{n=1}^{\infty} na_nx^{n-1}\) \(y''(x) = \sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}\) These will be substituted into the given differential equation.

Substituting the Power Series into the Differential Equation

Now, let's substitute these three expressions into the differential equation \((3-x^2)y''-3xy'-y = 0\): \((3-x^2)\left(\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}\right) -3x\left(\sum_{n=1}^{\infty} na_nx^{n-1}\right) - \sum_{n=0}^{\infty} a_nx^n=0\)

Simplifying the Equation and Finding the Recurrence Relation

Now, let's simplify the equation by distributing the terms: \(\sum_{n=2}^{\infty} n(n-1)a_n(3x^{n-2}-x^{n}) + \sum_{n=0}^{\infty} a_nx^n -3\sum_{n=1}^{\infty} na_nx^n=0\) To find the recurrence relation, we'll match the coefficients of like powers of x: For \(n \geq 2\): \(3n(n-1)a_n- n(n-1)a_n=-na_{n-1}\) \(a_n=\dfrac{a_{n-1}}{n-2}\) For \(n=0\): \(a_0=a_0\) For \(n=1\): \(a_1=-3a_1\) This gives us the recurrence relation that we were looking for.

Calculating Coefficients and Writing the Linearly Independent Solutions

Using the recurrence relation, we can now find the first few coefficients: \( a_1 = -3a_1 \Rightarrow 4a_1=0 \Rightarrow a_1=0\) \(a_2 = \dfrac{a_1}{0} = \text{Undefined}\), so we can't find a general solution with \(a_2\). For a multiple general solution, we can try by choosing another coefficient to have nonzero value, to find two linearly independent solutions: 1) Choose \(a_0=1\) and \(a_1=0\): \(y_1(x)=1+\sum_{n=2}^{\infty}a_nx^n\) 2) Choose \(a_1=1\) and \(a_0=0\): \(y_2(x)=x+\sum_{n=2}^{\infty}a_nx^n\) Then, both solutions are linearly independent.

Finding the First Four Terms in Each Solution

Now that we have the general power series forms of the two linearly independent solutions (\(y_1(x)\) and \(y_2(x)\)), we'll find their first four terms: 1) For \(y_1(x)\): \(a_0=1, a_1=0\) \(a_2 = \dfrac{a_1}{0} = \text{Undefined}\) Thus, the first four terms of \(y_1\) are: \(y_1(x) = 1\) 2) For \(y_2(x)\): \(a_0=0, a_1=1\) \(a_2 = \dfrac{a_1}{0} = \text{Undefined}\) Thus, the first four terms of \(y_2\) are: \(y_2(x) = x\) Since the series terminate at \(n=1\), the general solution can be written as: \(y(x)=c_1y_1(x) + c_2y_2(x)=c_1+c_2x\) Where \(c_1\) and \(c_2\) are constants.

Key concepts

Power Series Solution

When solving differential equations, a power series solution is an excellent method, especially near certain points. This technique involves expressing the unknown function as an infinite sum of powers of a variable, in this case, \(x\). The general form for a power series is \(y(x) = \sum_{n=0}^{\infty} a_nx^n\). Here, \(a_n\) are coefficients that we need to find in order to solve the differential equation.

This method is particularly useful because it transforms a differential equation, which can be complex to handle, into an algebraic problem. The power series form allows us to plug into the differential equation and makes it possible to equate coefficients of like powers of \(x\), simplifying the problem greatly.

In practice, this approach involves writing the function and its derivatives as power series. Then, we substitute these into the differential equation to solve for the coefficients \(a_n\). Once we establish the pattern of coefficients, the solution can be expressed as a series or summed for specific terms.

Recurrence Relation

A recurrence relation is an equation that recursively defines a sequence: each term is a function of its preceding terms. In the context of power series solutions to differential equations, the recurrence relation indicates how each coefficient in the series is related to previous coefficients.

To derive a recurrence relation, we often equate the coefficients of each power of \(x\) after substituting the power series into the differential equation. This step reduces the problem to deciding the relationships between successive coefficients \(a_n\). For example, in our problem, we found that \(a_n = \frac{a_{n-1}}{n-2}\), which tells us how to calculate each new coefficient using the ones before it.

The key is identifying patterns among the coefficients. Once uncovered, these patterns guide us in calculating as many terms as needed, and understanding these relationships can help establish whether the series solution converges and fully satisfies the original equation.

Linearly Independent Solutions

Linearly independent solutions are an important concept when dealing with differential equations. In simple terms, two solutions are linearly independent if one is not just a scalar multiple of the other. This independence is crucial when seeking comprehensive solutions to differential equations.

For our differential equation, after deriving the power series and recurrence relation, we constructed two solutions: \(y_1(x)\) and \(y_2(x)\). Each starts with different initial conditions (meaning we chose different starting coefficients \(a_0\) and \(a_1\)), ensuring they are fundamentally different in behavior and not just multiple versions of the same function.

Typically, we aim to find as many linearly independent solutions as the order of the differential equation dictates. These solutions, when combined, form the general solution to the differential equation, encapsulating all possible ways the function can behave according to the given equation.