Problem 114 - Calculus 3

Question

Find a power series solution for the following differential equations. $$y^{\prime \prime}-8 y^{\prime}=0, \quad y(0)=-2, \quad y^{\prime}(0)=10$$

Step-by-Step Solution

Answer

The power series solution is: \( y(x) = -2 + 10x + 40x^2 + \frac{320}{3}x^3 + \ldots \)

Step 1: Assume a Power Series Solution

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

Step 2: Differentiate the Power Series

Find the first and second derivatives of the power series:- First derivative: \[ y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} = \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n. \] - Second derivative: \[ y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} = \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n. \]

Step 3: Substitute into the Differential Equation

Substitute the derivatives into the differential equation:\[ y'' - 8y' = \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n - 8 \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n = 0. \]

Step 4: Equate Coefficients to Zero

For the derivatives to equal zero, the coefficients of powers of \(x^n\) must also equal zero, giving:\[ (n+2)(n+1) a_{n+2} - 8(n+1) a_{n+1} = 0. \]Solve for \(a_{n+2}\):\[ a_{n+2} = \frac{8(n+1)}{(n+2)(n+1)} a_{n+1} = \frac{8}{n+2} a_{n+1}. \]

Step 5: Identify and Apply Initial Conditions

Using the initial conditions:- For \(y(0) = -2\): \[ y(0) = a_0 = -2. \]- For \(y'(0) = 10\): \[ y'(0) = a_1 = 10. \]

Step 6: Recursively Find Coefficients

Use the recursive formula \(a_{n+2} = \frac{8}{n+2} a_{n+1}\) to find additional coefficients:- \(n = 0\): \[a_2 = \frac{8}{2} a_1 = 4 \times 10 = 40.\]- \(n = 1\): \[a_3 = \frac{8}{3} a_2 = \frac{8}{3} \times 40 = \frac{320}{3}.\] Continue this process for higher-order coefficients if needed.

Step 7: Construct the Power Series Solution

Combine the coefficients found into the power series:\[ y(x) = -2 + 10x + 40x^2 + \frac{320}{3}x^3 + \ldots \]

Key Concepts

Differential Equations

Differential equations are mathematical expressions that relate some function with its derivatives. They play a crucial role in describing how various physical systems change, often portraying dynamic processes. These equations can vary widely, from simple forms like linear, first-order differential equations to more complex versions involving higher-order derivatives. For instance, in the given exercise, we have a second-order linear differential equation: \[ y^{\prime \prime} - 8 y^{\prime} = 0. \]This equation asks us to find a function, \( y(x) \), whose second derivative minus eight times its first derivative equals zero. To solve this kind of differential equation using a power series approach helps us approximate solutions by iteratively refining coefficients of powers of \( x \). The power series method assumes solutions can be expanded into infinite sums, facilitating the exploration of complex equations.

Initial Conditions

Initial conditions are essential for finding specific solutions to differential equations. They represent known values of the function or its derivatives at a certain point, usually \( x = 0 \). For years, differential equations had countless general solutions; they required additional data to pinpoint unique ones. In the exercise at hand, the initial conditions are:

• \( y(0) = -2 \)
• \( y^{\prime}(0) = 10 \)

These conditions help in determining the exact values of the coefficients in our power series, \( a_0 \) and \( a_1 \), which give us:

• \( a_0 = -2 \)
• \( a_1 = 10 \)

By applying these values, we obtain a solution that not only satisfies the differential equation but also meets these particular constraints. Without initial conditions, we would only have a general solution, unable to precisely match specific scenarios.

Recursive Formula

A recursive formula is a way of defining the terms of a sequence using the preceding terms. In solving differential equations via power series, these formulas are invaluable in determining the subsequent coefficients of the series based on the preceding ones. The recursive relationship discovered in the step-by-step solution is:\[ a_{n+2} = \frac{8}{n+2} a_{n+1}. \]This equation allows us to find every subsequent coefficient in the power series given just the first few coefficients, \( a_0 \) and \( a_1 \). For the exercise, using this recursive formula effectively means:

• From \( a_1 = 10 \), we find \( a_2 = 4 \times 10 = 40 \).
• Then from \( a_2 = 40 \), \( a_3 = \frac{8}{3} \times 40 = \frac{320}{3} \), and so on.

This process can continue for as many terms as needed, making recursion a powerful tool for gradually building up the power series solution.

Coefficients in Power Series

In a power series, the coefficients are the constant multipliers of the terms containing the variable raised to increasing powers. These coefficients are key to constructing the series solution to a differential equation, as they determine the behavior of the solution over a range of values. For this exercise, we express \( y(x) \) as \[ y(x) = \sum_{n=0}^{\infty} a_n x^n. \]In this expansion, \( a_n \) are the coefficients determined recursively. Additionally, the initial conditions allow us to find \( a_0 \) and \( a_1 \), which are pivotal in calculating further coefficients. For example:

• \( a_0 = -2 \)
• \( a_1 = 10 \)
• \( a_2 = 40 \)
• \( a_3 = \frac{320}{3} \)

These coefficients form the basis for constructing the series solution, with each additional term providing greater accuracy for the solution. Understanding and calculating coefficients accurately is crucial for obtaining a valid and applicable mathematical solution.