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. $$ y^{\prime \prime}=-4 y $$

Short answer

The series solution is the power series expansion of \( y = c_1 \cos(2x) + c_2 \sin(2x) \).

Step-by-step solution

- Assume a power series solution

Assume the solution to the differential equation can be expressed as a power series: \[ y = \sum_{n=0}^{\infty} a_n x^n \]

- Determine derivatives of the power series

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

- Substitute derivatives into the differential equation

Substitute the second derivative into the given differential equation \( y'' = -4y \):\[ \sum_{n=2}^{\infty} a_n n (n-1) x^{n-2} = -4 \sum_{n=0}^{\infty} a_n x^n \]

- Align powers of x

Reindex the left series to match the powers of \(x\) on both sides. Let \(m = n-2\) on the left side:\[ \sum_{m=0}^{\infty} a_{m+2} (m+2)(m+1) x^m = -4 \sum_{n=0}^{\infty} a_n x^n \]

- Equating coefficients

Set coefficients of like powers of \(x\) equal to solve for \(a_n\):\[ a_{m+2} (m+2)(m+1) = -4a_m \]From this, obtain a recurrence relation: \[ a_{m+2} = -\frac{4}{(m+2)(m+1)} a_m \]

- Determine the series coefficients

Start with initial conditions \(a_0\) and \(a_1\). Calculate subsequent coefficients using the recurrence relation:\[ a_2 = -\frac{4}{2 \cdot 1} a_0 = -2a_0 \]\[ a_3 = -\frac{4}{3 \cdot 2} a_1 = -\frac{2}{3}a_1 \]\[ a_4 = -\frac{4}{4 \cdot 3} a_2 = \frac{2}{3} a_0 \]Continue this process to obtain more terms.

- Elementary method solution

Solve the differential equation using a standard technique. The characteristic equation is:\[ r^2 + 4 = 0 \]Thus, the roots are \( r = \pm 2i \). The general solution is:\[ y = c_1 \cos(2x) + c_2 \sin(2x) \]

- Compare series and elementary method solutions

Express the elementary method solution as a power series using the Taylor series expansions for sine and cosine:\[ \cos(2x) = \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n}}{(2n)!} \]\[ \sin(2x) = \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n+1}}{(2n+1)!} \]Verify that the series solutions match the power series obtained from the initial method.

Key concepts

Power Series Solution

A **power series solution** is a method used to solve differential equations by assuming the solution is in the form of a power series. For the given differential equation, we assume the solution to be in the series form: \[ y = \sum_{n=0}^{\text{\infty}} a_n x^n \] Here, \( a_n \) are coefficients that we need to determine.
By substituting the power series and its derivatives into the original differential equation, we reformat the differential equation into an algebraic form with respect to these coefficients. This helps in solving for each coefficient step-by-step using a recurrence relation.

Elementary Method

The **elementary method** uses classical techniques to directly solve a differential equation, often employing known solutions of simpler, characteristic equations.
For the differential equation \( y'' = -4y \), we determine the characteristic equation: \[ r^2 + 4 = 0 \] Solving this gives us the roots \( r = \pm 2i \).
This results in the general solution: \[ y = c_1 \cos(2x) + c_2 \sin(2x) \] where \( c_1 \) and \( c_2 \) are constants determined by initial conditions. By considering the typical forms of solutions involving sine and cosine for such differential equations, it connects the solution obtained to recognizable trigonometric functions.

Recurrence Relation

In solving the series, we often derive a **recurrence relation**. It's an equation that relates terms in a sequence or series to previous terms.
From the step-by-step process, we get the recurrence relation: \[ a_{m+2} (m+2)(m+1) = -4a_m \] Simplifying this, we find: \[ a_{m+2} = -\frac{4}{(m+2)(m+1)} a_m \] Starting with initial coefficients \( a_0 \) and \( a_1 \), we calculate higher-order terms. This systematic approach helps us determine more coefficients efficiently and generates the power series recursively.

Taylor Series Expansion

To verify our power series solution, we convert the elementary solution into a Taylor series. The **Taylor series expansion** at a point provides an infinite series representation of a function.
For trigonometric functions involved in our differential solution: \[ \cos(2x) = \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n}}{(2n)!} \] \[ \sin(2x) = \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n+1}}{(2n+1)!} \] By expanding the cosine and sine functions and comparing them to our series solution, we can confirm them. This not only brings clarity and verification to our process but illustrates the inter-connected nature of different solution techniques through mathematical consistency.