Problem 107 - Calculus Volume 3

Question

Find a power series solution for the following differential equations. $$ y^{\prime \prime}-y=0 $$

Step-by-Step Solution

Answer

The power series solution is \( y = a_0 \cosh(x) + a_1 \sinh(x) \).

Step 1: Assume a Power Series Solution

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

Step 2: Differentiate the Power Series

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

Step 3: Substitute into Differential Equation

Substitute the expressions for \( y \) and \( y'' \) into the differential equation:\[ \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - \sum_{n=0}^{\infty} a_n x^n = 0 \].

Step 4: Adjust Index of Summation

Adjust the indices so that both power series have the same power of \( x \):\[ \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^{\infty} a_n x^n = 0 \].

Step 5: Combine and Compare Coefficients

Combine coefficients of like powers of \( x \):\[ \sum_{n=0}^{\infty} ((n+2)(n+1) a_{n+2} - a_n) x^n = 0 \].For the equality to hold for all \( x \), each coefficient must equal zero. Thus, \[ (n+2)(n+1) a_{n+2} = a_n \].

Step 6: Establish Recurrence Relation

This gives a recurrence relation:\[ a_{n+2} = \frac{a_n}{(n+2)(n+1)} \].

Step 7: Find Initial Terms

Find the terms of the series starting with initial conditions \( a_0 \) and \( a_1 \):For \( n = 0 \): \( a_2 = \frac{a_0}{2} \).For \( n = 1 \): \( a_3 = \frac{a_1}{6} \).Continuing this yields terms for the series.

Step 8: Write the General Solution

The power series solution becomes:\[ y = a_0 \left(1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots \right) + a_1 \left(x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots \right) \].This is equivalent to:\[ y = a_0 \cosh(x) + a_1 \sinh(x) \].

Key Concepts

Recurrence Relation

In solving differential equations using power series, one fundamental step is establishing the recurrence relations. Recurrence relations are equations that express each term in the series as a function of preceding terms. This is essential to determine the coefficients of the series solution.

For the given equation, \( y'' - y = 0 \), by substituting the power series representation for \( y \) and its derivatives, we arrive at the equation:

• \( \sum_{n=0}^{\infty} ((n+2)(n+1) a_{n+2} - a_n) x^n = 0 \)

The solution must hold true for all values of \( x \). This implies that all coefficients of the powers of \( x \) must be zero:
\[ (n+2)(n+1) a_{n+2} = a_n \].

This relationship between successive terms in the series is the recurrence relation. It allows us to compute all coefficients from two initial known values \( a_0 \) and \( a_1 \). Again, this highlights how power series translate operations into algebraic manipulations, making complex differential equations more approachable.

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. They are crucial in understanding natural phenomena, where rates of change are interrelated. In our context, we are dealing with a second-order linear differential equation without external forces:

• \( y'' - y = 0 \)

The objective is to find a function \( y(x) \) that satisfies this equation across an interval.

To discover this, we use the power series approach, which allows for an infinite series of terms \( y = \sum_{n=0}^{\infty} a_n x^n \). By plugging this series into our differential equation, we translate the problem into algebra via substitution, obtaining a system that can be solved using recurrence relations. Thus, solving differential equations through power series leverages series expansion to craft solutions that fulfill the required conditions.

Cosh and Sinh Functions

The solution to the differential equation \( y'' - y = 0 \) leads to terms in the power series that mimic those of the hyperbolic functions, \( \cosh(x) \) and \( \sinh(x) \). These functions are analogs of the trigonometric \( \cos(x) \) and \( \sin(x) \), utilized in hyperbolic geometry, describing the geometry of surfaces of negative curvature.

The hyperbolic cosine function \( \cosh(x) \) is defined by its Taylor series:

• \( \cosh(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots \)

Similarly, the hyperbolic sine function \( \sinh(x) \) has the series:

• \( \sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots \)

Our power series solution for \( y \) can thus be expressed in terms of these known functions, revealing that the general solution is \( y = a_0 \cosh(x) + a_1 \sinh(x) \). This succinctly illustrates how complex differential equations can link directly to elegant mathematical functions, making them easier to interpret and utilize in various applications.