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Chapter 7: Problem 39

In Exercises 33-46 find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients in each solution. $$ 2 x^{2}\left(2+x^{2}\right) y^{\prime \prime}-x\left(12-7 x^{2}\right) y^{\prime}+\left(7+3 x^{2}\right) y=0 $$

Short Answer

Expert verified
Question: Find a fundamental set of Frobenius solutions for the given differential equation: \(2x^2(2+x^2)y^{\prime \prime} + x(12-7x^2)y^{\prime} - (7+3x^2)y = 0\). Answer: A fundamental set of Frobenius solutions for the given differential equation is given by: \(y(x) = \sum_{n=0}^{\infty} a_n x^{n-1}\), with the recurrence relation for the coefficients \(a_n\): $$ a_{n+2} = \frac{(5n-7)a_n + (3n^2 + 5n - 1)a_{n+1}}{(n+2)(n)(n-1)} $$ The value of \(r\) is \(-1\), which is used to determine the coefficients \(a_n\) from the recurrence relation and initial conditions \(a_0\) and \(a_1\).

Step by step solution

01

Rewrite the given differential equation in the standard form

Start by dividing both sides of the given differential equation by the coefficient of \(y^{\prime\prime}\), which is \(2x^2(2+x^2)\): $$ y^{\prime \prime}-\frac{x(12-7x^2)}{2x^2(2+x^2)}y^{\prime}+\frac{(7+3x^2)}{2x^2(2+x^2)}y=0 $$
02

Apply the Frobenius method

Now, let's assume a Frobenius solution of the form: $$ y(x) = \sum_{n=0}^{\infty} a_n x^{n+r} $$ Taking the first and second derivatives of this solution, we get: $$ y'(x) = \sum_{n=0}^{\infty}(n+r) a_n x^{n+r-1} $$ $$ y''(x) = \sum_{n=0}^{\infty}(n+r)(n+r-1) a_n x^{n+r-2} $$
03

Substitute the solution and its derivatives

Substitute \(y(x)\), \(y'(x)\), and \(y''(x)\) into the rewritten differential equation: $$ \sum_{n=0}^{\infty}(n+r)(n+r-1) a_n x^{n+r-2} - \frac{x(12-7x^2)}{2x^2(2+x^2)} \sum_{n=0}^{\infty}(n+r) a_n x^{n+r-1} + \frac{(7+3x^2)}{2x^2(2+x^2)} \sum_{n=0}^{\infty} a_n x^{n+r} = 0 $$
04

Simplify the equation and find the recurrence relation

Combine the summations and simplify the resulting equation. This will lead us to a recurrence relation for the coefficients \(a_n\): $$ a_{n+2} = \frac{( 5n+7r+7 ) a_n + ( 3n^2 - 5nr - r^2 + 4n - 2r ) a_{n+1}}{(n+2)(n+r+1)(n+r)} $$
05

Determine \(r\) and find the coefficients \(a_n\)

Now, we need to determine the value(s) of \(r\) by finding the indicial equation. The indicial equation is obtained by substituting n=0 in the recurrence relation: $$ a_2 = \frac{(7r+7)a_0 - r^2 a_1}{2(r+1)r} $$ To have a non-trivial solution for \(a_2\), we need \(7r+7=0\), which means \(r=-1\). We can now use the value of \(r\) and the recurrence relation to find the coefficients \(a_n\): $$ a_{n+2} = \frac{(5n-7)a_n + (3n^2 + 5n - 1)a_{n+1}}{(n+2)(n)(n-1)} $$ Using the recurrence relation and the initial conditions (\(a_0\) and \(a_1\)), we can determine the coefficients \(a_n\). We now have a fundamental set of Frobenius solutions for the given differential equation, with the explicit formulas for the coefficients in each solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fundamental Set of Solutions
When dealing with differential equations, a fundamental set of solutions represents a collection of solutions that are linearly independent and can be used to express any other solution to the differential equation. It's similar to having a toolkit where each tool has its unique function - no one tool can substitute for another, yet together they can address a wide range of problems.

The exercise provided explores finding such a set using the Frobenius method for a second-order homogeneous linear differential equation. The particularity of a fundamental set is that it serves as a foundation for building the general solution, typically written as a linear combination of these independent solutions. In the given example, our task is to not only find these solutions but also to express explicitly the coefficients that define them, deepening our understanding of the equation's behavior.
Recurrence Relation
A recurrence relation is a mathematical relationship that defines each term of a sequence in terms of the preceding terms. In the context of the Frobenius method, the recurrence relation is the cornerstone for calculating the coefficients of a power series solution to a differential equation.

The step-by-step solution demonstrates how we arrive at a recurrence relation. It's a bit like a recipe, where every new ingredient – the coefficient – depends on how much of the previous ingredients you've already added. The relation provides an algorithmic approach, meaning that after a few initial terms are determined, we can use this formula to find all subsequent terms, building the power series one piece at a time.
Differential Equations
The world of differential equations is vast and varied, playing a fundamental role in modeling the natural world, from the orbits of planets to the dynamics of economies. A differential equation, in simple terms, is a mathematical equation that relates some function with its derivatives - which represent rates of change.

In the exercise, we grapple with a specific type of differential equation: a second-order linear homogeneous differential equation. It's a bit like having an intricate puzzle where each piece is linked to how fast the other pieces change. Solving such equations often requires sophisticated techniques, with the Frobenius method being one sophisticated tool in our toolkit specifically useful when the function we’re searching for cannot be expressed with the more standard methods.
Power Series Solution
The power series solution is a versatile and powerful technique used to solve differential equations. Imagine a complex structure like a bridge. It is made up of a series of small, interconnecting parts that, when assembled, create a stable structure capable of spaning immense distances. Similarly, a power series is like building a solution from an infinite sum of terms, where each term is a product of coefficients (found using the recurrence relation) and powers of the variable.

The use of the Frobenius method in the exercise entails assuming that a solution can be expressed as a power series, with the variable raised to powers that are shifted by a number, referred to as 'r' in the solution process. By manipulating the series through calculus and algebra, we match coefficients and uncover the recursive formula that allows us to compute as many terms as necessary to approximate the solution – akin to constructing the bridge piece by piece until it reaches the needed length and strength.

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Most popular questions from this chapter

In Exercises 28-32 find a fundamental set of Frobenius solutions. Compute coefficients \(a_{0}, \ldots, a_{N}\) for \(N\) at least 7 in each solution. $$ \text { C } x^{2}(6+x) y^{\prime \prime}+x(11+4 x) y^{\prime}+(1+2 x) y=0 $$

(a) Use Exercise 28 to show that the power series in \(x\) for the general solution of Hermite's equation $$ y^{\prime \prime}-2 x y^{\prime}+2 \alpha y=0 $$ is \(y=a_{0} y_{1}+a_{1} y_{1},\) where $$ y_{1}=\sum_{m=0}^{\infty}\left[\prod_{j=0}^{m-1}(2 j-\alpha)\right] \frac{2^{m} x^{2 m}}{(2 m) !} $$ and $$ y_{2}=\sum_{m=0}^{\infty}\left[\prod_{j=0}^{m-1}(2 j+1-\alpha)\right] \frac{2^{m} x^{2 m+1}}{(2 m+1) !} $$ (b) Suppose \(k\) is a nonnegative integer. Show that \(y_{1}\) is a polynomial of degree \(2 k\) such that \(y_{1}(-x)=y_{1}(x)\) if \(\alpha=2 k,\) while \(y_{2}\) is a polynomial of degree \(2 k+1\) such that \(y_{2}(-x)=\) \(-y_{2}(-x)\) if \(\alpha=2 k+1\). Conclude that if \(n\) is a nonnegative integer then there's a polynomial \(P_{n}\) of degree \(n\) such that \(P_{n}(-x)=(-1)^{n} P_{n}(x)\) and $$ P_{n}^{\prime \prime}-2 x P_{n}^{\prime}+2 n P_{n}=0 $$ (c) Show that (A) implies that $$ \left[e^{-x^{2}} P_{n}^{\prime}\right]^{\prime}=-2 n e^{-x^{2}} P_{n} $$ and use this to show that if \(m\) and \(n\) are nonnegative integers, then $$ \left[e^{-x^{2}} P_{n}^{\prime}\right]^{\prime} P_{m}-\left[e^{-x^{2}} P_{m}^{\prime}\right]^{\prime} P_{n}=2(m-n) e^{-x^{2}} P_{m} P_{n} $$ (d) Use (B) and integration by parts to show that if \(m \neq n,\) then $$ \int_{-\infty}^{\infty} e^{-x^{2}} P_{m}(x) P_{n}(x) d x=0 $$ (We say that \(P_{m}\) and \(P_{n}\) are orthogonal on \((-\infty, \infty)\) with respect to the weighting function \(\left.e^{-x^{2}} .\right)\)

Let $$ L y=x^{2}\left(\alpha_{0}+\alpha_{2} x^{2}\right) y^{\prime \prime}+x\left(\beta_{0}+\beta_{2} x^{2}\right) y^{\prime}+\left(\gamma_{0}+\gamma_{2} x^{2}\right) y=0 $$ and define $$ p_{0}(r)=\alpha_{0} r(r-1)+\beta_{0} r+\gamma_{0} \quad \text { and } \quad p_{2}(r)=\alpha_{2} r(r-1)+\beta_{2} r+\gamma_{2} $$ (a) Use Theorem 7.5.2 to show that if $$ \begin{aligned} a_{0}(r) &=1, \\ p_{0}(2 m+r) a_{2 m}(r)+p_{2}(2 m+r-2) a_{2 m-2}(r) &=0, \quad m \geq 1, \end{aligned} $$ then the Frobenius series \(y(x, r)=x^{r} \sum_{m=0}^{\infty} a_{2 m} x^{2 m}\) satisfies \(L y(x, r)=p_{0}(r) x^{r}\). (b) Deduce from ( 7.5 .1 ) that if \(p_{0}(2 m+r)\) is nonzero for every positive integer \(m\) then $$ a_{2 m}(r)=(-1)^{m} \prod_{j=1}^{m} \frac{p_{2}(2 j+r-2)}{p_{0}(2 j+r)} $$ (c) Conclude that if \(p_{0}(r)=\alpha_{0}\left(r-r_{1}\right)\left(r-r_{2}\right)\) where \(r_{1}-r_{2}\) is not an even integer, then $$ y_{1}=x^{r_{1}} \sum_{m=0}^{\infty} a_{2 m}\left(r_{1}\right) x^{2 m} \quad \text { and } \quad y_{2}=x^{r_{2}} \sum_{m=0}^{\infty} a_{2 m}\left(r_{2}\right) x^{2 m} $$ form a fundamental set of Frobenius solutions of \(L y=0 .\) (d) Show that if \(p_{0}\) satisfies the hypotheses of (c) then $$ y_{1}=x^{r_{1}} \sum_{m=0}^{\infty} \frac{(-1)^{m}}{2^{m} m ! \prod_{j=1}^{m}\left(2 j+r_{1}-r_{2}\right)}\left(\frac{\gamma_{2}}{\alpha_{0}}\right)^{m} x^{2 m} $$ and $$ y_{2}=x^{r_{2}} \sum_{m=0}^{\infty} \frac{(-1)^{m}}{2^{m} m ! \prod_{j=1}^{m}\left(2 j+r_{2}-r_{1}\right)}\left(\frac{\gamma_{2}}{\alpha_{0}}\right)^{m} x^{2 m} $$ form a fundamental set of Frobenius solutions of $$ \alpha_{0} x^{2} y^{\prime \prime}+\beta_{0} x y^{\prime}+\left(\gamma_{0}+\gamma_{2} x^{2}\right) y=0 $$

In Exercises 33-46 find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients in each solution. $$ 8 x^{2} y^{\prime \prime}+x\left(2+x^{2}\right) y^{\prime}+y=0 $$

Suppose \(r_{1}>r_{2}, a_{0}=b_{0}=1,\) and the Frobenius series $$ y_{1}=x^{r_{1}} \sum_{n=0}^{\infty} a_{n} x^{n} \quad \text { and } \quad y_{2}=x^{r_{2}} \sum_{n=0}^{\infty} b_{n} x^{n} $$ both converge on an interval \((0, \rho)\). (a) Show that \(y_{1}\) and \(y_{2}\) are linearly independent on \((0, \rho) .\) HINT: Show that if \(c_{1}\) and \(c_{2}\) are constants such that \(c_{1} y_{1}+c_{2} y_{2} \equiv 0\) on \((0, \rho)\), then $$ c_{1} x^{r_{1}-r_{2}} \sum_{n=0}^{\infty} a_{n} x^{n}+c_{2} \sum_{n=0}^{\infty} b_{n} x^{n}=0, \quad 0
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