Question

Give the first 5 terms of the series that is a solution to the given differential equation. $$y^{\prime \prime}=2 y, \quad y(0)=1, y^{\prime}(0)=1$$

Short answer

The first 5 terms are: 1, x, x^2, \(\frac{x^3}{3}\), \(\frac{x^4}{6}\).

Step-by-step solution

Define the power series

Assume the solution is a power series: \[ y(x) = \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots \] Given the initial conditions: \( y(0) = 1 \) implies \( a_0 = 1 \) and \( y'(0) = 1 \) implies \( a_1 = 1 \).

Derive the first and second derivatives

Calculate the derivatives:\[ y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} = a_1 + 2a_2 x + 3a_3 x^2 + 4a_4 x^3 + \cdots \] \[ y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} = 2a_2 + 6a_3 x + 12a_4 x^2 + 20a_5 x^3 + \cdots \]

Use differential equation

According to the differential equation \( y'' = 2y \), set:\[ \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} = 2 \sum_{n=0}^{\infty} a_n x^n \]Equating terms by compensating powers of \(x\), determine relations for coefficients.

Relate coefficients

For \( n=0 \): \( 2a_2 = 2a_0 \), therefore \( a_2 = a_0 = 1 \).For \( n=1 \): \( 6a_3 = 2a_1 \), so \( a_3 = \frac{a_1}{3} = \frac{1}{3} \).For \( n=2 \): \( 12a_4 = 2a_2 \), hence \( a_4 = \frac{a_2}{6} = \frac{1}{6} \).For \( n=3 \): \( 20a_5 = 2a_3 \), thus \( a_5 = \frac{a_3}{10} = \frac{1}{30} \).

Write series and calculate first 5 terms

Substituting the known coefficients \(a_0, a_1, a_2, a_3, a_4\): \[ y(x) = 1 + x + x^2 + \frac{x^3}{3} + \frac{x^4}{6} + \cdots \] The first five terms of the series are: \( 1, x, x^2, \frac{x^3}{3}, \frac{x^4}{6} \).

Key concepts

Power Series Solution

A power series solution is a way of expressing the solution to a differential equation in the form of an infinite sum. This series can be written as:\[ y(x) = \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots \]Here, the solution is approached by assuming that the function can be expressed as a power series. The coefficients \(a_n\) need to be determined. This representation is useful, particularly for solving differential equations that don't have easily accessible closed-form solutions.💡 By breaking the solution into individual terms, students can manage complex equations in smaller, more manageable pieces.

Initial Value Problem

An initial value problem involves finding a function that satisfies a differential equation and meets certain specified conditions at an initial point, often where the variable is zero. For this problem, these initial conditions include:- \( y(0) = 1 \), which implies that the coefficient \( a_0 = 1 \).- \( y'(0) = 1 \), leading to \( a_1 = 1 \).These conditions anchor the solution, ensuring it not only satisfies the differential equation but also fits the specific situation described. Such boundary conditions can provide necessary information to find the particular solution of a differential equation.

Series Expansion

Series expansion is the process of representing a function as a power series. By expanding a function, you can express it as a sum of its terms with ascending power of \(x\). Consider the derivatives of our assumed solution:- The first derivative gives us: \[ y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} \]- The second derivative: \[ y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} \]Using these expansions in the differential equation allows us to equate terms for each power of \(x\) with corresponding terms on both sides of an equation. This method of term-by-term comparison helps to deduce relationships among coefficients.

Coefficients Determination

Determining coefficients involves finding values for \(a_n\) in the series expansion. This often requires solving equations derived from substituting our series and derivatives into the differential equation. In this exercise, equating the series for \(y'' = 2y\) gives us:

• For \(n = 0\), \(2a_2 = 2a_0\) leading to \(a_2 = a_0 = 1\).
• For \(n = 1\), \(6a_3 = 2a_1\) resulting in \(a_3 = \frac{1}{3}\).
• For \(n = 2\), \(12a_4 = 2a_2\) hence \(a_4 = \frac{1}{6}\).
• For \(n = 3\), \(20a_5 = 2a_3\) thus \(a_5 = \frac{1}{30}\).

By successively solving for each coefficient, we obtain enough terms to express the series solution, in this case, up to the fifth term. Understanding and applying these sequential steps clarify how each coefficient impacts the overall series solution.