Question

Find power series solutions in powers of \(x\) of each of the differential equations in Exercises. $$ y^{\prime \prime}-x y^{\prime}+(3 x-2) y=0 $$

Short answer

The general power series solution for the given differential equation is: \[ y(x) = a_0 + a_1x + \sum_{n=2}^{\infty} \frac{2a_{n-2}}{(n-2)(n-1)} x^n \]

Step-by-step solution

Rewrite the differential equation in standard form

The given differential equation is already in standard form: \[ y'' - xy' +(3x-2)y = 0 \]

Assume a power series solution for y

Assume that the solution y can be expressed as a power series: \[ y(x) = \sum_{n=0}^{\infty} a_nx^n \] Now, find the first and second derivatives of y(x) and plug these into the given equation.

Compute derivatives and plug in the power series

First, find the first derivative, y'(x): \[ y'(x) = \sum_{n=1}^{\infty} na_nx^{n-1} \] Next, find the second derivative, y''(x): \[ y''(x) = \sum_{n=2}^{\infty} n(n-1)a_nx^{n-2} \] Now, plug y(x), y'(x), and y''(x) into the differential equation: \[ \left(\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2} \right) -x \left(\sum_{n=1}^{\infty} na_nx^{n-1} \right) +(3x-2) \left(\sum_{n=0}^{\infty} a_nx^n\right)=0 \]

Equate coefficients of like powers

Multiply and combine the power series, then equate coefficients on the left-hand side with those on the right-hand side (which is always 0). \[ \sum_{n=2}^{\infty} n(n-1)a_nx^{n-2} -\sum_{n=1}^{\infty} n a_nx^n +3\sum_{n=1}^{\infty} a_nx^n -2\sum_{n=0}^{\infty} a_nx^n = 0 \] Equate coefficients for each power of x from the left and right side: \[ n(n-1)a_n - na_n + 3a_n - 2a_{n-2} = 0 \] Simplify the equation by solving for a_n: \[ a_n = \frac{2a_{n-2}}{(n-2)(n-1)} \]

Obtain the general power series solution

Using the recursive formula for the coefficients, we can find the general power series solution for y: \[ y(x) = a_0 + a_1x + \sum_{n=2}^{\infty} a_nx^n = a_0 + a_1x + \sum_{n=2}^{\infty} \frac{2a_{n-2}}{(n-2)(n-1)} x^n \] This is the general power series solution for y in powers of x for the given differential equation.

Key concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations that involve some ordinary derivatives (as opposed to partial derivatives) of a function. They are fundamental in expressing the laws and phenomena of varied disciplines like physics, engineering, and economics. An ODE is typically composed of a function and its derivatives, relating how the function changes with respect to one independent variable. For instance, the equation
\[ y'' - xy' + (3x-2)y = 0 \]
is a second-order linear homogeneous ODE with variable coefficients. It’s second-order because of the presence of the second derivative \(y''\), and it’s homogeneous because all terms involve the function \(y\) or its derivatives, equating to zero. In this scenario, the solution we seek is not just a single function, but a whole family of functions that satisfy the given equation under particular initial conditions.

Series Solution Method

The series solution method is a powerful technique for solving differential equations that might not have straightforward solutions. In this approach, the solution to the ODE is presumed as an infinite sum of powers of the independent variable—commonly known as a power series. In mathematical terms, we express the solution \(y(x)\) as
\[ y(x) = \sum_{n=0}^{\infty} a_nx^n \]
where \( a_n \) are the coefficients of the series to be determined. By plugging in this assumed solution and its derivatives into the original ODE, we can obtain a new equation involving the power series. This step is crucial since it allows us to use the properties of power series to simplify and equate coefficients, leading us toward finding the unknown coefficients \( a_n \) and eventually the power series solution of the ODE.

Coefficient Equating

Coefficient equating is a process used in the series solution method where we match the coefficients of corresponding powers of \(x\) on both sides of the differential equation. After plugging the series representations of \(y\), \(y'\), and \(y''\), we get a new series where each term is equal to zero, since the original equation is equated to zero. This step is instrumental as it gives rise to a recursive relation that helps determine the coefficients \(a_n\) of the power series.
For example, through the equating process we can derive
\[ n(n-1)a_n - na_n + 3a_n - 2a_{n-2} = 0 \]
which further simplifies to
\[ a_n = \frac{2a_{n-2}}{(n-2)(n-1)} \]
This recursive formula allows us to calculate all the \( a_n \) coefficients in terms of \( a_0 \) and \( a_1 \) (the first two terms of the series, which are usually determined by initial conditions). Successive application of the recursive formula results in the building of the power series solution, revealing the full structure of the solution to the differential equation.