Question

Determine the general solution of the given differential equation that is valid in any interval not including the singular point. \(x^{2} y^{\prime \prime}+3 x y^{\prime}+5 y=0\)

Short answer

Short Answer: The general solution of the given second-order differential equation is: \(y(x) = \sum_{n=0}^{\infty} a_n x^n\), where \(a_n\) satisfies the recursive relation \(a_{n+2} = \frac{-3(n + 1) a_{n+1} - 5 a_n}{(n + 2)(n + 1)}\). This power series solution can be determined with initial conditions, providing a unique solution for a particular problem.

Step-by-step solution

Write down the given differential equation.

Here is the given second-order differential equation: \(x^{2} y^{\prime \prime}+3 x y^{\prime}+5 y=0\)

Assume a power series solution and compute its derivatives.

Assume a power series solution of the form: \(y(x) = \sum_{n=0}^{\infty} a_n x^n\) Compute its first and second derivatives: \(y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1}\) \(y''(x) = \sum_{n=2}^{\infty} n (n-1) a_n x^{n-2}\)

Substitute the power series and its derivatives into the differential equation.

Now we will substitute the power series and its derivatives into the differential equation: \(x^{2} y^{\prime \prime}+3 x y^{\prime}+5 y=0\) \(x^{2} \sum_{n=2}^{\infty} n (n-1) a_n x^{n-2}+3 x \sum_{n=1}^{\infty} n a_n x^{n-1}+5 \sum_{n=0}^{\infty} a_n x^n=0\)

Simplify the equation and combine terms with the same power of x.

Simplify and combine terms with the same power of x: \(\sum_{n=2}^{\infty} n (n-1) a_n x^n+\sum_{n=1}^{\infty} 3 n a_n x^n+\sum_{n=0}^{\infty} 5 a_n x^n=0\) Rewrite the equation, this time shifting the index of summation to start at 0 for all series: \(\sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n+\sum_{n=0}^{\infty} 3 (n+1) a_{n+1} x^n+\sum_{n=0}^{\infty} 5 a_n x^n=0\)

The power series equaling zero implies equal coefficients.

If the power series equals zero, the coefficients in front of the powers of x must all equal zero: \((n + 2)(n + 1) a_{n+2} + 3(n + 1) a_{n+1} + 5 a_n = 0\)

Solve the recursive relation for the coefficients and find the general solution.

Solve the recursive relation for \(a_{n+2}\): \(a_{n+2} = \frac{-3(n + 1) a_{n+1} - 5 a_n}{(n + 2)(n + 1)}\) This relation allows us to find the coefficients \(a_n\) and construct the general solution. Since the differential equation is linear and homogeneous, the general solution is in the form of a linear combination of two linearly independent solutions. The coefficients in the power series depend on the initial conditions and can be determined accordingly. The general solution of the given differential equation is: \(y(x) = \sum_{n=0}^{\infty} a_n x^n\), where \(a_n\) satisfies the recursive relation as shown above.

Key concepts

Power Series Solution

The power series solution to a differential equation is a method for solving linear differential equations by expressing the solution as an infinite sum of power functions. The general form of a power series solution is:

\[\begin{equation} y(x) = \/\sum_{n=0}^{\infty} a_n x^n\end{equation}\]

where \( a_n \) are the coefficients that must be determined. This approach is particularly useful if the differential equation can't be solved via more straightforward methods, or when dealing with differential equations with variable coefficients.

Here's how to proceed:

• First, assume the solution y can be represented as this power series.
• Next, compute the derivatives of the power series term-by-term.
• Then substitute both the power series and its derivatives back into the original differential equation.
• Following this substitution, you should rearrange terms so that each power of x is collected together.
• Considering the fact that this equation must hold for all values of x, equate the coefficients of same power terms to zero to find the relations for the coefficients \( a_n \).

This systematic method provides us with a recursive relationship which can be solved to determine the coefficients, thereon building the solution to the equation step by step. While the computations can be laborious, the clarity and structure offered by the power series solution make it a powerful tool in the analyst's arsenal.

Second-order Differential Equation

The given problem revolves around a second-order differential equation, which is an equation that involves the unknown function y, its first derivative \( y' \), and second derivative \( y'' \). Mathematically, it's expressed as \( F(x, y, y', y'') = 0 \).

These equations are pivotal in physics and engineering, describing systems such as vibrations, oscillations, and dynamics. In our case, we are looking at a linear, homogeneous second-order differential equation with variable coefficients:

\[\begin{equation} x^{2}y'' + 3xy' + 5y = 0\end{equation}\]

The term 'homogeneous' implies that the equation equals zero, and there's no stand-alone function of x on the right-hand side. As such, a key property of these equations is the principle of superposition, which allows us to construct the general solution as a linear combination of two linearly independent solutions.

Approaching such problems with a power series solution is suitable when the coefficients aren't constant and when we are faced with a singularity, a point where the coefficients of the derivatives of y are either zero or undefined, precluding standard solution methods.

Recursive Relation for Coefficients

Discovering the recursive relation for coefficients is a critical step in cementing the power series method. In the context of our differential equation, this recursive relation emerges upon equating coefficients of like powers of x after substituting the series into the equation. The established recursive relation provides a systematic way to compute all subsequent coefficients based on initial values.

In practice, this means:

• We start with an initial set of coefficients, typically \( a_0 \) and \( a_1 \), which might be rooted in initial conditions or boundary values.
• We use the recursive formula to iteratively compute all other coefficients in the series.
• The precise form of the recursive formula depends on the structure of the differential equation.

For our example, the relation is:

\[\begin{equation} a_{n+2} = \frac{-3(n + 1) a_{n+1} - 5 a_n}{(n + 2)(n + 1)}\end{equation}\]

This equation allows us to generate all terms in the power series starting from n = 0 onwards, granting a pathway to construct the series representation of the solution. By piecing together these coefficients with their corresponding powers of x, we unveil the desired general solution to our original problem.