Question

Find power series solutions in powers of \(x-1\) of each of the differential equations in Exercises. Find the power series solution in powers of \(x-1\) of the initial-value problem $$ x y^{\prime \prime}+y^{\prime}+2 y=0, \quad y(1)=2, \quad y^{\prime}(1)=4 $$

Short answer

The power series solution in powers of \(x-1\) of the initial-value problem \(x y'' + y' + 2y = 0, \quad y(1) = 2, \quad y'(1) = 4\) is: \[y(x) = 2 - 2\sum_{n=2}^{\infty} (x-1)^n.\]

Step-by-step solution

Assume Power Series Solution Form

Assume that the solution to the differential equation is of the form: \[y(x) = \sum_{n=0}^{\infty} a_n (x-1)^n \quad \text{where} \: a_n \: \text{are constants} .\]

Differentiate the Power Series to Get y' and y''

Compute the first and second derivatives of \(y(x)\) with respect to \(x\): \[y'(x) = \sum_{n=1}^{\infty} na_n (x-1)^{n-1} \] \[y''(x) = \sum_{n=2}^{\infty} n(n-1)a_n (x-1)^{n-2} \]

Substitute y, y' and y'' into Differential Equation

Substitute the power series expressions for \(y\), \(y'\), and \(y''\) obtained in Step 2 into the given differential equation: \[x\left(\sum_{n=2}^{\infty} n(n-1)a_n (x-1)^{n-2}\right) + \left(\sum_{n=1}^{\infty} na_n (x-1)^{n-1}\right) + 2\left(\sum_{n=0}^{\infty} a_n (x-1)^n\right) = 0.\]

Combine Power Series Terms

Multiply the first power series by \(x\), change indices as required, and combine the three power series into one: \[\sum_{n=0}^{\infty}((n+2)(n+1)a_{n+2}x + na_{n+1} + 2a_n)(x-1)^n = 0 \] For this power series to be zero, the coefficient of each power of \((x-1)\) must be zero. Thus, we have the recursive relation for the coefficients: \[(n+2)(n+1)a_{n+2} + na_{n+1} + 2a_n = 0.\]

Use Initial Conditions to Determine Coefficients

In order to find the power series coefficients, we use the given initial conditions: For \(y(1) = 2\), we know that \[y(x) = 2 = \sum_{n=0}^{\infty} a_n (x-1)^n\] Therefore, the power series becomes: \[y(x) = 2 + \sum_{n=1}^{\infty} a_n (x-1)^n\] This gives us \(a_0 = 2\). For \(y'(1) = 4\), we know that \[y'(x) = 4 = \sum_{n=1}^{\infty} na_n (x-1)^{n-1}\] Therefore, the power series becomes: \[y'(x) = 4 + \sum_{n=2}^{\infty} na_n (x-1)^{n-1}\] This gives us \(a_1 = 4\).

Compute Further Coefficients Using Recursive Relations

Now we use the recursive relation we found in Step 4 to compute further coefficients using the given initial coefficients \(a_0\) and \(a_1\). For n = 0: \[(n+2)(n+1)a_{n+2} + na_{n+1} + 2a_n = 0\] \[(2)(1)a_2 + 0a_1 + 2a_0 = 0\] This gives us \(a_2 = -2\). For n = 1: \[(n+2)(n+1)a_{n+2} + na_{n+1} + 2a_n = 0\] \[(3)(2)a_3 + 1a_1 + 2a_0 = 0\] This gives us \(a_3 = -2\). Continuing this process, we can compute more coefficients as needed.

Write the Final Power Series Solution

Now we express the power series solution of the initial-value problem using the coefficients we have found: \[y(x) = 2 - 2(x-1)^2 - 2(x-1)^3 + \cdots = 2 - 2\sum_{n=2}^{\infty} (x-1)^n.\] The power series solution in powers of \(x-1\) of the initial-value problem is given by the power series we computed in this step.

Key concepts

Ordinary Differential Equations

An ordinary differential equation (ODE) is a relation that contains functions of only one independent variable and their derivatives. ODEs are used to model continuously changing systems such as the motion of planets or the growth of populations. In our exercise, the ODE given is
\[x y^{\textprime \textprime} + y^{\textprime} + 2 y = 0\]
where \(y\) is the unknown function of \(x\), and the prime symbols denote derivatives with respect to \(x\). Solving an ODE typically involves finding a function that satisfies the given relationship. Power series methods are particularly useful when dealing with complex ODEs where standard methods of integration are not applicable. The exercise demonstrates a structured approach to solving ODEs by assuming that the solution can be expressed as a power series centered around a specific point—in this case, \(x-1\).

Initial-Value Problem

An initial-value problem in the context of differential equations is a problem where the solution to the ODE is found under specific starting conditions. These conditions specify the values of the function and possibly its derivatives at a particular point, which are essential in determining a unique solution. For the given ODE, the initial conditions are
\[y(1) = 2, \quad y^{\textprime}(1) = 4.\]
These conditions tell us the value of the function \(y\) and its first derivative \(y^{\textprime}\) at the point \(x=1\). To incorporate these conditions, we first express \(y\) and \(y^{\textprime}\) as power series and then plug \(x=1\) into these series. The given initial values help in determining the coefficients \(a_0\) and \(a_1\), which are the starting points for finding all the other coefficients of the power series.

Recursive Relations for Coefficients

Recursive relations are valuable tools in determining the coefficients of an assumed power series solution of an ordinary differential equation. Once we have a relation that describes how the coefficients relate to each other, we can compute subsequent terms based on initial values.

In our exercise, after substituting the power series into the differential equation, we obtain a relation:
\[(n+2)(n+1)a_{n+2} + na_{n+1} + 2a_n = 0.\]
This recursive formula allows us to calculate the coefficients of the series for all values of \(n\) using just the first two coefficients, \(a_0\) and \(a_1\), which we determined from the initial conditions. Utilizing this recursive relationship effectively converts the complex problem of solving a differential equation into a simpler algebraic process, thus unveiling the power of power series methods in solving initial-value problems for ODEs.