Question

In each exercise, an initial value problem is given. Assume that the initial value problem has a solution of the form \(y(t)=\sum_{n=0}^{\infty} a_{n} t^{n}\), where the series has a positive radius of convergence. Determine the first six coefficients, \(a_{0}, a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\). Note that \(y(0)=a_{0}\) and that \(y^{\prime}(0)=a_{1}\). Thus, the initial conditions determine the arbitrary constants. In Exercises 40 and 41 , the exact solution is given in terms of exponential functions. Check your answer by comparing it with the Maclaurin series expansion of the exact solution. $$ y^{\prime \prime}+t y^{\prime}-2 y=0, \quad y(0)=0, \quad y^{\prime}(0)=1 $$

Short answer

Question: Determine the first six coefficients of the Taylor series expansion of the solution to the initial value problem (IVP) given by the differential equation \(y''(t) + ty'(t) - 2y(t) = 0\) with initial conditions \(y(0)=0\) and \(y'(0)=1\). Answer: The first six coefficients of the Taylor series expansion of the solution are: $$a_0 = 0$$ $$a_1 = 1$$ $$a_2$$ $$a_3 = \frac{1}{3}a_2$$ $$a_4$$ $$a_5 = \frac{1}{4}(2a_3 - 3a_2)$$

Step-by-step solution

Calculate derivatives of the series

First, find the first and second derivatives of the power series form of the solution, y(t): $$y'(t) = \sum_{n=1}^{\infty} na_n t^{n-1}$$ $$y''(t) = \sum_{n=2}^{\infty} n(n-1)a_n t^{n-2}$$

Substitute derivatives into the differential equation

Substitute \(y'(t)\), \(y''(t)\), and \(y(t)\) into the given differential equation: $$\left[\sum_{n=2}^{\infty} n(n-1)a_n t^{n-2}\right] + t\left[\sum_{n=1}^{\infty} na_n t^{n-1}\right] - 2\left[\sum_{n=0}^{\infty} a_n t^n\right] = 0$$

Identify the coefficient of each power of t

Write the differential equation as a series of coefficients for powers of t, which will give us separate equations for each a_n: $$\sum_{n=0}^{\infty} [(n+2)(n+1)a_{n+2} + na_na_{n-1} - 2a_n] t^n = 0$$

Solve for coefficients using the initial conditions

Use the initial conditions \(y(0)=0\) and \(y'(0)=1\) to solve for the first two coefficients \(a_0\) and \(a_1\). $$y(0) = a_0 = 0$$ $$y'(0) = a_1 = 1$$

Solve for the next four coefficients

Use the equations derived in step 3 to find the coefficients \(a_2, a_3, a_4\), and \(a_5\). For each n (from 2 to 5), solve the following equation to find the corresponding coefficient: $$[(n+2)(n+1)a_{n+2} + na_na_{n-1} - 2a_n] = 0$$ For \(n=2\): $$6a_4 - 2a_2 = 0 \Rightarrow a_4 = \frac{1}{3}a_2$$ For \(n=3\): $$12a_5 + 3a_2 - 2a_3 = 0 \Rightarrow a_5 = \frac{1}{4}(2a_3 - 3a_2)$$ For \(n=4\): $$20a_6 + 12a_3 - 2a_4 = 0$$ For \(n=5\): $$30a_7 + 20a_4 - 2a_5 = 0$$ Since we have the values for \(a_0\) and \(a_1\), and relationships between the coefficients, we can list the first six coefficients: $$a_0 = 0$$ $$a_1 = 1$$ $$a_2$$ $$a_3 = \frac{1}{3}a_2$$ $$a_4$$ $$a_5 = \frac{1}{4}(2a_3 - 3a_2)$$

Key concepts

Power Series Solution

When tackling differential equations, a power series solution approach can be highly effective, especially when dealing with non-constant coefficients that make standard methods less applicable.

A power series solution involves representing the unknown function as an infinite sum of power terms, often expressed as \(y(t) = \sum_{n=0}^{\infty} a_n t^n\), where \(a_n\) are the coefficients that need to be determined and \(t\) typically represents the independent variable.

This method hinges on the principle that if the power series converges within an interval, it can be differentiated and integrated term-by-term. The convergence itself is indicated by the radius of convergence, which should be positive for the solution to be valid within some interval centered at the initial value point.

By taking derivatives of the power series representation of the function and substituting them back into the differential equation, a new series is obtained. By equating coefficients of like powers of \(t\) from this series, we can establish relations between the coefficients \(a_n\), which can then be solved sequentially using given initial conditions.

Maclaurin Series Expansion

The Maclaurin series is a specific type of power series that is centered at \(t=0\). It is essentially a Taylor series expansion about zero and is given by \(f(t) = \sum_{n=0}^{\text{\infty}} \frac{f^{(n)}(0)}{n!} t^n\), where \(f^{(n)}(0)\) is the \(n^{\text{th}}\) derivative of the function evaluated at zero.

The Maclaurin series is particularly useful when we have an initial value problem with conditions specified at \(t=0\), which is the case in the exercise we are exploring. By comparing the coefficients from the Maclaurin series of a known exact solution with those obtained from the power series method, one can verify the accuracy of the solution derived using power series.

In practice, once the first few derivatives of the function at \(t=0\) are known, the corresponding Maclaurin series can be constructed up to any desired degree, facilitating an efficient check against solutions of differential equations found through other means.

Differential Equations

Differential equations play a central role in mathematics and its applications to the sciences and engineering because they describe how physical quantities change. They come in many forms, from simple separable equations to complex partial differential equations with multiple variables.

The problem presented here involves a second-order ordinary differential equation (ODE) with initial conditions, an initial value problem (IVP). IVPs typically require finding a function \(y(t)\) that not only satisfies the differential equation but also meets the initial conditions given, such as \(y(0)\) and \(y'(0)\) in the exercise.

To solve an IVP, one can apply various analytical or numerical methods. Analytical methods include separation of variables, integrating factors, or series solutions as demonstrated in the power series approach. Numerical methods, on the other hand, may be used when analytical solutions are difficult to obtain, and they provide approximate solutions over discrete points.

The understanding of differential equations and their solutions is critical for students wishing to apply mathematical principles to real-world problems. Each method of solution offers different insights and tools, ensuring that students are well-equipped to tackle a wide range of problems.