Question

Give the first 5 terms of the series that is a solution to the given differential equation. $$y^{\prime \prime}=-y, \quad y(0)=0, y^{\prime}(0)=1$$

Short answer

The first 5 terms are 0, 1, 0, -1/6, 0.

Step-by-step solution

Assume a Series Solution

We start by assuming a power series representation for the solution of the differential equation. Let \( y(x) = \sum_{n=0}^{\infty} a_n x^n \), where \( a_n \) are coefficients to be determined.

Compute Derivatives of the Series

The first derivative is \( y'(x) = \sum_{n=1}^{\infty} a_n n x^{n-1} \) and the second derivative is \( y''(x) = \sum_{n=2}^{\infty} a_n n(n-1) x^{n-2} \).

Substitute into the Differential Equation

Substitute \( y''(x) = -y(x) \) into the equation, resulting in \( \sum_{n=2}^{\infty} a_n n(n-1) x^{n-2} = -\sum_{n=0}^{\infty} a_n x^n \).

Equate Coefficients

Shift indices to equate powers of \( x \). For the right-hand side, rewrite as \( \sum_{n=0}^{\infty} a_{n} x^n \) and change to \( \sum_{n=2}^{\infty} a_{n-2} x^n \) to match the powers of left-hand side. By equating, we get \( a_n n(n-1) = -a_{n-2} \) for \( n \geq 2 \).

Apply Initial Conditions

Use the initial conditions to find the coefficients: \( y(0)=0 \Rightarrow a_0 = 0 \) and \( y'(0)=1 \Rightarrow a_1 = 1 \).

Solve for Coefficients Recursively

From equating coefficients, \( a_n = -\frac{a_{n-2}}{n(n-1)} \). Start from known coefficients to find others: \( a_2 = 0, \; a_3 = -\frac{1}{6}, \; a_4 = 0, \; a_5 = \frac{1}{120} \).

Write the Series and First 5 Terms

Using the coefficients, write the series: \( y(x) = x - \frac{x^3}{6} + \frac{x^5}{120} + \cdots \). The first 5 terms are \( 0, 1, 0, -\frac{1}{6}, 0, \frac{1}{120} \).

Key concepts

Initial Conditions in Differential Equations

Initial conditions in differential equations provide specific values that help define a unique solution for the equation. These conditions are essential because, for a given differential equation, solutions can vary greatly. Initial conditions guide us in finding the specific solution that applies to a real-world scenario or mathematical problem. Consider the initial conditions given in the exercise:

• \( y(0) = 0 \) sets the starting point for the function \( y \), determining that when \( x = 0 \), \( y \) also equals zero.
• \( y'(0) = 1 \) informs us about the slope of \( y \) at \( x = 0 \), meaning that initially, the function increases as \( x \) does.

Employing these conditions allows us to pinpoint the coefficients in our power series solution, ensuring it is not just any solution, but the right one that works with the given constraints.

Recursive Determination of Coefficients

When solving differential equations using power series, we often determine coefficients in a recursive manner. This means each coefficient depends on the previous coefficients, following a specific pattern or rule. Let's explain how this works with our example.

• After arranging and equating powers of \( x \), we discovered a relationship: \( a_n n(n-1) = -a_{n-2} \) for \( n \geq 2 \).
• This recursive formula implies that each \( a_n \) connects directly to the coefficient that comes two steps before it, \( a_{n-2} \).
• Given our initial conditions, we start our series with \( a_0 = 0 \) and \( a_1 = 1 \). The rest follows:
  • \( a_2 = 0 \)
  • Using \( a_3 = -\frac{1}{6} \), found by substituting into the recursive formula with \( a_1 \).
  • \( a_4 = 0 \)
  • Next, \( a_5 = \frac{1}{120} \), calculated similarly.

This approach offers a methodical way to build the series solution step-by-step, tracing through the patterns logically and systematically.

Second-Order Differential Equations

Second-order differential equations involve the second derivative denoted as \( y'' \). These types of equations frequently occur in physics and engineering as they describe various systems' dynamics, such as motion or heat distribution. Our example uses such a differential equation: \( y'' = -y \). Let's break down how we handle this.

• "Second-order" means that the highest derivative involved is the second. Thus, we have not only \( y \) but also its first and second derivatives in the equation.
• In our exercise, converting this second-order equation into a power series allowed us to find the solution by determining each coefficient through substitution and equating like terms.
• These types of problems require finding two initial conditions to ascertain a unique solution, which we discussed earlier. Knowing both the value of \( y \) and its derivative at a starting point is critical.

By transforming and solving second-order differential equations like this, we can better understand and predict a system's behavior under various conditions.