Problem 32 - Advanced Engineering Mathematics

Question

Use the power series method to solve the given initial-value problem. $$ \left(x^{2}+1\right) y^{\prime \prime}+2 x y^{\prime}=0, y(0)=0, y^{\prime}(0)=1 $$

Step-by-Step Solution

Answer

The solution is \(y(x) = \tan^{-1}(x)\).

Step 1: Assume a Power Series Solution

Assume that the solution to the differential equation is a power series of the form \(y(x) = \sum_{n=0}^{fty} a_n x^n\). Consequently, the derivatives are \(y'(x) = \sum_{n=1}^{fty} n a_n x^{n-1}\) and \(y''(x) = \sum_{n=2}^{fty} n(n-1) a_n x^{n-2}\).

Step 2: Substitute into the Differential Equation

Substitute the power series forms of \(y''(x)\) and \(y'(x)\) into the differential equation: \((x^2+1) y''(x) + 2x y'(x) = 0\). This becomes: \(\sum_{n=2}^{fty} n(n-1) a_n (x^{n} + x^{n-2}) + 2x \sum_{n=1}^{fty} n a_n x^{n-1} = 0\).

Step 3: Align the Power Series

Rewrite the series so that all series are expressed as powers of \(x^n\). For the series involving \(\sum_{n=2}^{fty} n(n-1) a_n x^n\), no change is needed. The \(\sum_{n=2}^{fty} n(n-1) a_n x^{n-2}\) becomes \(\sum_{n=0}^{fty} n(n+1) a_{n+2} x^n\). The \(\sum_{n=1}^{fty} n a_n x^n\) is straightforward. This yields: \(\sum_{n=2}^{fty} n(n-1)a_n x^n + \sum_{n=0}^{fty} n(n+1)a_{n+2} x^n + \sum_{n=1}^{fty} 2na_n x^n = 0\).

Step 4: Combine the Power Series

Combine the power series terms into a single series: \(a_0 \cdot 0 + (2a_1) \cdot x^1 + \sum_{n=2}^{fty} [(n(n-1) + 2n)a_n + n(n+1)a_{n+2}] x^n = 0\). For this expression to be zero for all \(x\), the coefficient of each power of \(x\) must be zero.

Step 5: Solve the Recurrence Relation

Solve the recurrence relation obtained from the series by setting the coefficients of each power equal to zero. For \(x^0\): \(0\cdot a_0 = 0\) is trivially satisfied. For \(x^1\): \(2a_1 = 0\), thus \(a_1 = 1\) (satisfying initial condition). For \(x^n, n\geq 2\): \([(n(n-1) + 2n)a_n + n(n+1)a_{n+2}] = 0\). Solve for \(a_{n+2}\) in terms of \(a_n\): \(a_{n+2} = -\frac{(n(n-1) + 2n)}{n(n+1)} a_n\).

Step 6: Use Initial Conditions

The initial conditions \(y(0) = 0\) and \(y'(0) = 1\) were applied when deriving \(a_0\) and \(a_1\). Given \(a_0 = 0\), using the recurrence relation, compute subsequent terms to identify the pattern.

Step 7: Make Conclusions about the Solution

The solution simplifies significantly because many terms become zero due to the initial condition and the recurrence relation, leading to a simplified power series representation for \(y(x)\). Consequently, the essential terms are derived from initial conditions leading to \(y(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots\).

Step 8: Recognize the Series Form

The recognized pattern shows that the solution is the known Taylor series expansion for \(\tan^{-1}(x)\). Therefore, the function \(y(x) = \tan^{-1}(x)\) satisfies the differential equation and initial conditions provided.

Key Concepts

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. In this context, we're working with a second-order linear homogeneous differential equation:

• It has the form \((x^{2}+1)y'' + 2xy' = 0\), where \(y''\) and \(y'\) are the second and first derivatives of \(y\), respectively.
• Such equations are utilized across various fields of science and engineering to model dynamic systems.

In a typical problem, you may be asked to find the function \(y(x)\) that satisfies this differential equation. Solving differential equations can be challenging, especially those which aren't easily solvable by standard methods. For this problem, the power series method is used. This method is particularly useful when the coefficients of the derivatives in the equation are variable, meaning they change with \(x\). By assuming \(y(x)\) as a power series, like \(y(x) = \sum_{n=0}^{\infty} a_n x^n\), the solution is crafted by transforming the differential equation into a series form. This allows us to equate terms of equal powers of \(x\) to determine the coefficients \(a_n\).

Initial Value Problem

An initial value problem (IVP) is a specific type of differential equation problem which includes conditions at a starting point. Here, the differential equation \((x^2 + 1)y'' + 2xy' = 0\)is given initial conditions:

• \(y(0) = 0\)
• \(y'(0) = 1\)

These conditions tell us the value of the solution \(y(x)\) and its first derivative at \(x = 0\), serving as enough information to find the unique solution curve. Using these conditions is crucial, as they provide the means to determine the specific values of the constants \(a_0\) and \(a_1\) in the power series expansion, which represent the solution. Initial values help in tailoring a general solution to meet specific problem criteria. It ensures that even if multiple solutions exist theoretically, we can pinpoint the one solution that exactly fits the provided conditions.

Recurrence Relation

Within the power series method, a recurrence relation is a pivotal concept. It helps us find each term of the series based on previous terms. By substituting the power series expressions into our differential equation, the series have to match terms of equal powers of \(x\). This gives us a recurrence relation to solve for coefficients like this:

• For \(x^1\): we found \(2a_1 = 0\), leading to \(a_1 = 1\).
• For \(x^n, n\geq 2\): the relation was derived as \(a_{n+2} = -\frac{(n(n-1) + 2n)}{n(n+1)} a_n\).

This mathematical relation tells how to compute each subsequent coefficient in terms of the previous ones, ensuring the series form is accurate. In practice, solving this helps to construct the series solution term-by-term. It's a systematic way to obtain a sequence of coefficients ensuring that the series converges to the particular solution of the differential equation that fulfills the initial conditions.

Taylor Series

A Taylor series is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It allows complex or unknown functions to be expressed in a polynomial form. In the context of this problem, once we derive the recurring coefficients through our series solution, a recognizable pattern emerges:

• The power series solution for the equation is \(y(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots\).
• This mirrors the Taylor series expansion for \(\tan^{-1}(x)\), a well-known inverse trigonometric function.

Thus, the power series method not only solves for \(y(x)\) but also connects our solution back to a standard function we've encountered in calculus, \(\tan^{-1}(x)\). This relation indicates that:

• The solution to the differential equation with the specified initial conditions is \(y(x) = \tan^{-1}(x)\).
• This demonstrates how initial conditions and series competence can lead to standard functions, providing insights into the nature of the solution.