Question

In Exercises \(61-68\) use the method suggested by Exercise 60 to find the general solution on some interval \((0, \rho)\) $$ 4 x^{2}\left(3+2 x+x^{2}\right) y^{\prime \prime}-x\left(3-14 x-15 x^{2}\right) y^{\prime}+\left(3+7 x^{2}\right) y=0 $$

Short answer

Question: Using the method of Frobenius, find the general power series solution to the following differential equation: $$ 4x^2(3+2x+x^2)y'' - x(3-14x-15x^2)y' + (3+7x^2)y = 0 $$ Answer: The general power series solution of the given differential equation on interval (0, ρ) is represented by the series: $$ y(x) = \sum_{n=0}^{\infty} a_n x^{n+r} = a_0 x^r + a_1 x^{r+1} + a_2 x^{r+2} + a_3 x^{r+3} + \cdots $$ where \(a_n\) are the coefficients determined by the following recurrence relation: $$ a_{n+1} = \frac{a_n(7n^2+34nr+34n+14r^2+20r-24)}{(n+1)(n+r+1)(n+r+2)} $$

Step-by-step solution

Determine the general form of the solution

To begin with, we will assume a power series solution of the form: $$ y(x) = \sum_{n=0}^{\infty} a_n x^{n+r} $$ where \(a_n\) are the coefficients and \(r\) is the power exponent to be determined.

Calculate the derivatives

Next, we need to compute the first and second derivatives of the power series solution: First derivative: $$ y'(x) = \sum_{n=0}^{\infty} (n+r)a_n x^{n+r-1} $$ Second derivative: $$ y''(x) = \sum_{n=0}^{\infty} (n+r)(n+r-1)a_n x^{n+r-2} $$

Substitute the derivatives into the given equation

Now, substitute the power series solution \(y(x)\) and its derivatives \(y'(x)\) and \(y''(x)\) into the given differential equation: $$ 4x^2(3+2x+x^2) \left(\sum_{n=0}^{\infty} (n+r)(n+r-1)a_n x^{n+r-2}\right) - x(3-14x-15x^2) \left(\sum_{n=0}^{\infty} (n+r)a_n x^{n+r-1}\right) + (3+7x^2) \left(\sum_{n=0}^{\infty} a_n x^{n+r}\right) = 0 $$

Simplify the equation

To simplify the above equation, we should first distribute the functions inside the sums: $$ \sum_{n=0}^{\infty} (n+r)(n+r-1)a_n x^{n+r} [4(3+2x+x^2)] - \sum_{n=0}^{\infty} (n+r)a_n x^{n+r}[x(3-14x-15x^2)] + \sum_{n=0}^{\infty} a_n x^{n+r}(3+7x^2) = 0 $$ Next, combine all the sums into a single sum: $$ \sum_{n=0}^{\infty} a_n x^{n+r} \left[ (n+r)(n+r-1)4(3+2x+x^2) - (n+r)(3-14x-15x^2) + (3+7x^2) \right] = 0 $$

Form a recurrence relation

Now, we need to find the recurrence relation by setting coefficients of like powers equal to zero, starting with the term not containing \(x\) (constant term \(a_0\)). We can rewrite the equation as: $$ \sum_{n=0}^{\infty} a_n x^{n+r} \left[ (n+r)(n+r-1)12 -(n+r)(3-14x-15x^2) + 3+7x^2 \right] = 0 $$ Setting the coefficients of like powers equal to zero, we obtain the recurrence relation: $$ a_{n+1} = \frac{a_n(7n^2+34nr+34n+14r^2+20r-24)}{(n+1)(n+r+1)(n+r+2)} $$

Find the general solution

Using the recurrence relation, we can now calculate the first few coefficients to determine the general series solution: $$ y(x) = \sum_{n=0}^{\infty} a_n x^{n+r} = a_0 x^r + a_1 x^{r+1} + a_2 x^{r+2} + a_3 x^{r+3} + \cdots $$ In conclusion, the general solution for the given differential equation on interval \((0, \rho)\) is the series \(y(x)\) in the form of the power series involving \(a_n x^{n+r}\).

Key concepts

Power Series Solution

A power series solution is an essential method used to tackle differential equations, particularly when solutions cannot be easily expressed using elementary functions. This method involves representing the solution as an infinite sum, known as a power series, given by:\[y(x) = \sum_{n=0}^{\infty} a_n x^{n+r}\]Here, \(a_n\) are the coefficients of the series, and \(r\) is an exponent that can be determined depending on the equation's nature.
The objective of using a power series is to find a series of simple, calculable terms.
This way, instead of solving complicated expressions directly, we break them down into sum of terms that are easier to handle.
We start by assuming a form for \(y(x)\), and this leads to deriving recurrence relations—which are foundational for finding solutions to differential equations.
Power series solutions offer flexibility and can handle various differentials, especially those that cannot be solved with straightforward methods.

Recurrence Relation

Recurrence relations play a critical role in power series solutions. They describe how to calculate successive terms in a sequence using previous terms. Think of it as a recipe telling you how to compute each term based on the ones before it.
These relations are established by ensuring the coefficients from the power series satisfy the differential equation when substituted back.
In the context of the given problem, upon substituting the power series into the differential equation, we collect all like power terms of \(x\) and set them to zero. This gives us a system from which we extract the recurrence relation:
\[a_{n+1} = \frac{a_n(7n^2+34nr+34n+14r^2+20r-24)}{(n+1)(n+r+1)(n+r+2)}\]This equation lays out how any coefficient \(a_{n+1}\) in the series is related to its predecessor \(a_n\). By computing these coefficients iteratively, we can construct the entire power series solution.

General Solution

The ultimate goal in solving a differential equation is to find the general solution. For the given second order differential equation, this solution is represented by a series:\[y(x) = \sum_{n=0}^{\infty} a_n x^{n+r}\]The terms in this series are calculated using the recurrence relation, providing a systematic approach to finding all possible solutions.
This general solution accounts for all potential solutions to the differential equation by expressing it as an infinite sum of power terms.
Each term is accompanied by a coefficient \(a_n\), which is computed based on the recurrence relation.
Ultimately, the series represents solutions valid within a specific interval \((0, \rho)\), ensuring the most comprehensive answer fitting the original differential equation.

Second Order Differential Equation

A second order differential equation is an equation involving the second derivative of a function, meaning it describes how a rate of change changes over time. Such equations are common in physics and engineering, modeling phenomena like oscillations and waves.
In our exercise, we were dealing with a second order differential equation of the form:\[4 x^{2}(3+2 x+x^{2}) y^{\prime\prime}-x(3-14 x-15 x^{2}) y^{\prime}+ (3+7 x^{2}) y=0\]Understanding these equations involves handling both the function and its derivatives. Solving them often requires advanced techniques like power series methods, as traditional formula-based solutions might not apply.
The reduced complexity of power series methods and recurrence relations offers a robust framework to tackle these types of equations, enabling us to find comprehensive solutions across specified intervals. This example demonstrated how second-order equations can be approached systematically by breaking them down into manageable segments.