Problem 114 - Calculus Volume 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 \( y(x) = -2 + 10x \).

Step 1: Assume a General Power Series Solution

Let's assume that the solution to the differential equation can be expressed as a power series at x = 0: \[ y(x) = \sum_{n=0}^{fty} a_n x^n \]where \(a_n\) are constants to be determined.

Step 2: Derive the First Two Derivatives

We need the first and second derivatives of the power series. Start by finding the first derivative:\[ y'(x) = \sum_{n=1}^{fty} n a_n x^{n-1} \]And the second derivative:\[ y''(x) = \sum_{n=2}^{fty} n(n-1) a_n x^{n-2} \]

Step 3: Substitute into the Differential Equation

Substitute the first and second derivatives into the differential equation:\[ y''(x) - 8y'(x) = \sum_{n=2}^{fty} n(n-1) a_n x^{n-2} - 8 \sum_{n=1}^{fty} n a_n x^{n-1} = 0 \]

Step 4: Align Powers of x

To simplify, align the powers of \(x\) by adjusting the indices of summation. Set each power's coefficient to zero:For \(x^n\) where \(n\geq 2\):\[ n(n-1) a_n - 8na_{n} = 0 \]This simplifies to:\[ a_n(n(n-1) - 8n) = 0 \]

Step 5: Solve for the Coefficients

This reduces to:\[ n(a_n)(n-9) = 0 \]Thus, \(a_n\) must be zero unless \(n = 9\). This means that every coefficient except those determined by initial conditions will be zero.

Step 6: Apply Initial Conditions

Use the initial conditions \(y(0) = -2\) and \(y'(0) = 10\) to find \(a_0\) and \(a_1\):Since \(y(0) = a_0 = -2\).And \(y'(0) = a_1 = 10\).

Step 7: Construct the Solution

The series simplifies because all higher terms vanish (\(a_n = 0\) for \(n\geq 2\)). Thus, the power series solution becomes:\[ y(x) = a_0 + a_1 x = -2 + 10x \]

Step 8: Verify the Solution

Check that the form satisfies the differential equation as well as initial conditions:\( y''(x) = 0 \) and \(8y'(x) = 8 imes 10 = 80\). The equation \(0 - 80 = 0\) confirms that the form satisfies the differential equation.The initial conditions are met: \(y(0) = -2\) and \(y'(0) = 10\).

Key Concepts

Differential Equations

Differential equations are mathematical expressions that link functions with their derivatives. These equations play critical roles in modeling various real-world phenomena, such as growth rates, heat transfer, and other dynamic systems. In our particular exercise, we are dealing with a linear differential equation of the form \( y'' - 8y' = 0 \). This equation involves a second derivative, \( y'' \), and a first derivative, \( y' \), but not the function \( y \) itself. This specific structure suggests a relationship between how a function changes and its rate of change.

• Linear: The equation is linear because each term either involves a constant or the function derivates multiplied by a constant.
• Homogeneous: The equation is homogeneous, meaning it is set equal to zero.

Finding a solution usually entails solving for the function \( y(x) \) that meets these conditions, often requiring assumptions or initial conditions to be completely determined.

Initial Conditions

Initial conditions provide the necessary information to pinpoint a specific solution out of the infinite possibilities that a differential equation might permit. For our example, the initial conditions are given as \( y(0) = -2 \) and \( y'(0) = 10 \). These describe the state of the function and its derivative at \( x = 0 \).

• Initial conditions help determine unique coefficients in the power series representation of the solution.
• They guide us in forming a complete solution that fits the real-world scenario being modeled.

These conditions allow us to determine values for the coefficients \( a_0 \) and \( a_1 \) specifically, making the solution specific to our problem rather than a generalized answer.

Derivatives

Derivatives, a fundamental concept in calculus, express how a function changes at any given point. In this problem, we calculate both the first derivative \( y'(x) \) and the second derivative \( y''(x) \) of the assumed power series solution. These derivatives are essential because they allow us to plug into our differential equation.

• First Derivative \( y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} \).
• Second Derivative \( y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} \).

By substituting these derivatives into the differential equation, you can work toward finding a solution in terms of a power series. Understanding derivatives in this context allows you to comprehend how a function and its rates of change interact in dynamic systems.

Coefficients Determination

The determination of coefficients in a power series solution involves aligning terms by powers of \( x \) and setting each coefficient equal to zero, as seen in our example. For our differential equation, this means arranging terms so that each power of \( x \) has its own equation.

• For each power of \( x^n \), ensure the equation \( n(n-1)a_n - 8na_n = 0 \) holds.
• This leads to the equation \( a_n(n(n-1) - 8n) = 0 \), reducing further.
• Beyond computation, using initial conditions was crucial to concretely determine \( a_0 \) and \( a_1 \).

In this process, meaningful coefficients are extracted, while others become zero, simplifying the overall power series. This technique of coefficients determination is pivotal in obtaining a precise and workable power series solution.