Problem 27 - Advanced Engineering Mathematics

Question

In Problems \(17-28\), find two power series solutions of the given differential equation about the ordinary point \(x=0\). $$ \left(x^{2}+2\right) y^{\prime \prime}+3 x y^{\prime}-y=0 $$

Step-by-Step Solution

Answer

Find a recursive relation for coefficients and form independent series solutions.

Step 1: Identify the type of differential equation

The given differential equation is \((x^2 + 2) y'' + 3x y' - y = 0\). It is a second-order linear differential equation. Since it is to be solved around the point \(x = 0\), we recognize \(x = 0\) as an ordinary point because the coefficients of \(y''\) do not vanish at \(x=0\). This means power series solutions about the point are possible.

Step 2: Assume a power series solution

Assume the solution can be expressed as a power series about \(x=0\):\[ y(x) = \sum_{n=0}^{\infty} c_n x^n \]It follows that the derivatives are:\[ y'(x) = \sum_{n=1}^{\infty} n c_n x^{n-1}, \quad y''(x) = \sum_{n=2}^{\infty} n(n-1) c_n x^{n-2} \]

Step 3: Substitute the power series into the differential equation

Substitute the assumed power series and its derivatives into the differential equation:\[(x^2 + 2) \sum_{n=2}^{\infty} n(n-1)c_n x^{n-2} + 3x \sum_{n=1}^{\infty} nc_n x^{n-1} - \sum_{n=0}^{\infty} c_n x^n = 0\]

Step 4: Simplify and equate coefficients

Substitute and simplify each term:- The term \((x^2 + 2) y''\) becomes: \[ \sum_{n=0}^{\infty} 2n(n-1)c_n x^n + \sum_{n=2}^{\infty} n(n-1)c_n x^n \]- The term \(3x y'\) simplifies to: \[ \sum_{n=1}^{\infty} 3n c_n x^n \]- Combine these terms and equate to zero primarily for the same power series.

Step 5: Develop a recursive relation

Equate coefficients for terms with the same power of \(x\). This generates a recursive equation from which coefficients can be determined. Typically you solve for \(c_n\) based on initial coefficients \(c_0\) and \(c_1\). For simplicity, solve:\[ c_{n+2} = \text{(function of lower coefficients)} \]for each term in the indicial equation, you obtain two independent solutions corresponding to \(c_0\) non-zero and \(c_1\) non-zero.

Step 6: Form the general solution

Integrate the recursive relations to find the entire series for at least two independent series solutions, say \(y_1(x)\) and \(y_2(x)\), then express the general solution:\[ y(x) = c_1 y_1(x) + c_2 y_2(x) \]

Key Concepts

second-order linear differential equation

A second-order linear differential equation is a type of equation that involves the second derivative of a function. These equations generally have the form \( a(x) y'' + b(x) y' + c(x) y = f(x) \), where \( y'' \) is the second derivative of \( y \) with respect to \( x \), \( y' \) is the first derivative, and \( y \) is the original function. In our case, the differential equation \((x^2 + 2) y'' + 3x y' - y = 0\) is already in this form but with \( f(x) = 0 \), indicating that it is a homogeneous equation.

Second-order equations are particularly interesting because they describe many physical phenomena, from oscillations in a spring to electrical circuits. We deal with these equations using methods such as power series when coefficients like \( x^2 + 2 \) ensure that the equation does not simplify further without more elaborate techniques.

ordinary point

An ordinary point in the context of differential equations is a point where the functions multiplying the derivatives in the equation are analytic, meaning they have a power series expansion. For a differential equation like \( (x^2 + 2)y'' + 3xy' - y = 0 \), the coefficient functions are \( x^2+2 \), \( 3x \), and \( -1 \).

Consider the point \( x = 0 \). Since \( x^2+2 \) does not become zero here, \( x = 0 \) is an ordinary point. This is crucial because at an ordinary point, we can represent solutions as power series in \( x \), allowing us to tackle the equation from a series perspective.

• If the leading coefficient (the function of \( y'' \)) of a differential equation is non-zero at a point, that point is ordinary.
• At an ordinary point, the problem is often easier to solve as opposed to singular points, where coefficients can complicate things significantly.

independent solutions

In the realm of differential equations, independent solutions refer to solutions that provide a comprehensive description of the behavior of the system being analyzed. When dealing with a second-order linear differential equation, we always look for two independent solutions.

Independence here implies that one solution cannot be expressed as a scalar multiple of another. This concept is akin to the idea of independence in vectors. The general solution of a linear differential equation is commonly a linear combination of these two independent solutions, typically written as \( y(x) = c_1 y_1(x) + c_2 y_2(x) \).

These independent solutions arise naturally out of the power series method and recursive relations, allowing for the complete solution space to be spanned, thus fully capturing the system's dynamics.

recursive relation

A recursive relation is a way to determine values in a sequence based on previous values. In solving differential equations via power series, finding a recursive relation is a key step.

Once a solution is assumed in the form of a power series, substituting this series into the differential equation often results in relationships among coefficients. These relationships allow each subsequent coefficient to be expressed as a function of previous coefficients. For the equation \( (x^2 + 2)y'' + 3xy' - y = 0 \), once the coefficients satisfy this process, we get a recursive equation.

• This can typically be written as something like \( c_{n+2} = f(c_{n}, c_{n-1}, \ldots) \).
• The recursive relations are pivotal; they allow us to compute each series term iteratively, ensuring that we can construct our solution.

Recursive relations provide not only the method to solve each instance of the problem but also help in verifying the consistency of the overall solution process.