Question

Solve the following differential equations by power series and also by an elementary method. Verify that the series solution is the power series expansion of your other solution. $$ y^{\prime \prime}-2 y^{\prime}+y=0 $$

Short answer

Power series solution matches with \[ y = C_1 e^x + C_2 x e^x. \]

Step-by-step solution

- Assume a power series solution

Assume a power series solution of the form \[ y = \sum_{n=0}^{\infty} a_n x^n \] where \[ a_n \] are constants to be determined.

- Find the derivatives

Find the first and second derivatives of the power series. \[ y' = \sum_{n=1}^{\infty} n a_n x^{n-1} \] \[ y'' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} \]

- Substitute into the differential equation

Substitute the power series and its derivatives into the given differential equation \[ y'' - 2y' + y = 0 \]. This yields the equation \[ \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - 2\sum_{n=1}^{\infty} n a_n x^{n-1} + \sum_{n=0}^{\infty} a_n x^n = 0 \].

- Combine powers of x

Reindex the sums to combine the powers of \[ x \] and simplify. This gives the combined series \[ a_0 + (2a_2 - 2a_1)x + \sum_{n=2}^{\infty} [(n(n-1)a_n - 2na_{n-1} + a_{n-2}) x^n = 0]. \]

- Solve the recurrence relation

Equate coefficients of like powers of \[ x \] to zero to find the recurrence relation \[ a_n = \frac{2a_{n-1} - a_{n-2}}{n(n-1)}. \] Solving this recurrence, find the series coefficients \[ a_n. \].

- General solution by an elementary method

Solve the differential equation using an elementary method (e.g., characteristic equation). The characteristic equation is \[ r^2 - 2r + 1 = 0 \] which has a repeated root \[ r = 1 \]. Thus, the general solution is \[ y = C_1 e^x + C_2 x e^x.\]

- Verify the power series expansion

Expand \[ y = C_1 e^x + C_2 x e^x \] using a Taylor series to verify that it matches the power series solution found earlier.

Key concepts

Power Series Solution

A power series solution is a technique where we assume the solution to a differential equation can be represented as an infinite sum of powers of the variable. For the equation provided, we assume a solution of the form \ \[ y = \sum_{n=0}^{\begin{infinity}} a_n x^n \], where \[a_n\] are constants to be determined. We then substitute this assumed solution into the differential equation and equate coefficients of like powers of \( x \) to find a recurrence relation for the series coefficients \( a_n \). This provides us with a method to find all the terms in the series, allowing us to express the solution as an infinite series.

Elementary Methods

Elementary methods for solving differential equations involve techniques such as separation of variables, integrating factors, or the characteristic equation. For the given differential equation y'' - 2y' + y = 0, we use the characteristic equation approach. We transform the differential equation into an algebraic equation by assuming solutions of the form \(y = e^{rt}\). Substituting y and its derivatives into the original differential equation, we get the characteristic equation \( r^2 - 2r + 1 = 0 \). Solving this quadratic equation gives us the roots, and hence the general solution is derived. For this particular problem, the roots are repeated \( r = 1 \), resulting in the general solution \( y = C_1 e^x + C_2 x e^x \).

Recurrence Relation

In solving differential equations using a power series, we often derive a recurrence relation. This is an equation that relates each term in the power series to previous terms. For example, after substituting the power series into our differential equation and simplifying, we obtained the recurrence relation \[ a_n = \frac{2a_{n-1} - a_{n-2}}{n(n-1)} \]. This relation allows us to compute the coefficients \( a_n \) in terms of \( a_{n-1} \) and \( a_{n-2} \). By iterating this relation, starting with initial conditions (e.g., \(a_0\) and \(a_1\)), we can find all coefficients of the power series.

Taylor Series

Taylor series is a representation of a function as an infinite sum of terms, calculated from the values of its derivatives at a single point. When we solve differential equations by series, we often verify our series solution by expressing a known solution as a Taylor series. For instance, if we solve \(y'' - 2y' + y = 0 \) by elementary methods and find \( y = C_1 e^x + C_2 x e^x \), we can express \( e^x \) as a Taylor series \[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]. Similarly, \( x e^x \) can be expanded and added. Comparing the coefficients of the Taylor series expansion with our power series solution confirms their equivalence.

Characteristic Equation

Another core concept is the characteristic equation, used in elementary methods to solve linear differential equations with constant coefficients. For our differential equation \( y'' - 2y' + y = 0 \), we assume a solution \( y = e^{rt} \) and substitute it into the differential equation. This substitution transforms our differential equation into an algebraic one: \( r^2 - 2r + 1 = 0 \). Solving this characteristic equation gives the roots, which determine the form of the general solution. For repeated roots, as in this example where \( r = 1 \), the general solution takes the form \( y = C_1 e^x + C_2 x e^x \).