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In Problems, I through 8, determine whether \(x=0\) is an ordinary point, a regular singular point, or an irregular singular point. If it is a regular singular point, find the exponents of the differential equation at \(x=0 .\) $$ x y^{\prime \prime}+\left(x-x^{3}\right) y^{\prime}+(\sin x) y=0 $$

Short Answer

Expert verified
At \( x = 0 \), it is an ordinary point.

Step by step solution

01

Identify Types of Points

We determine the type of point at \( x=0 \) for the differential equation \( xy'' + (x - x^3)y' + \sin(x) y = 0 \). We first rewrite the differential equation in standard form, i.e., \( y'' + P(x) y' + Q(x) y = 0 \), where \( P(x) = \frac{x-x^3}{x} \) and \( Q(x) = \frac{\sin(x)}{x} \).
02

Analyze P(x) and Q(x) at x=0

Next, let's analyze \( P(x) = 1 - x^2 \) and determine if there are any singularities at \( x=0 \). Both \( P(x) \) and \( Q(x) \) can be expanded using Taylor series to check their behavior at zero: \( P(x) = 1 \) and \( Q(x) \) is expanded using \( \sin(x) \approx x - \frac{x^3}{6} + \cdots \) so \( Q(x) = 1 - \frac{x^2}{6} \). Both functions are regular at \( x=0 \).
03

Define Ordinary and Singular Points

An ordinary point is where both \( P(x) \) and \( Q(x) \) are analytic (i.e., can be expressed by a power series). Since \( P(x) \) and \( Q(x) \) are both analytic at \( x=0 \) (simplifying to literals without division by \( x^2 \) or other such constraints), \( x=0 \) is thus classified as an ordinary point.

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Key Concepts

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

Ordinary Point
In the realm of differential equations, understanding the type of point where a function is defined is crucial. An *ordinary point* is where the functions involved in the differential equation are "well-behaved," meaning both P(x) and Q(x) are analytic at this point. Analytic functions can be expressed by power series. In simpler terms, they can be expanded into a sum of terms that make them easy to study and solve.

For the differential equation given in the exercise, \[ y'' + P(x)y' + Q(x)y = 0 \]the coefficients \(P(x)\) and \(Q(x)\) are constructed from dividing the non-homogeneous parts by \(x\). At \(x=0\), an ordinary point, both \(P(x) = 1 - x^2\) and \( Q(x) = 1 - \frac{x^2}{6} \) can be expressed as power series. As they're both clear from singularities (no abrupt jumps or poles), \(x=0\) is confirmed as an ordinary point.

Points like these make differential equations much more accessible to solve, allowing us to use methods that involve power series solutions.
Singular Point
A singular point of a differential equation is where the differential equation does not behave straightforwardly. At these points, the function is not analytic, and the solutions can become complex or unpredictable.

Singular points are divided into two types:
  • Regular Singular Point: The equation has singularities, but they're not too severe. Techniques like Frobenius series can still solve them.
  • Irregular Singular Point: The singularities involve stronger behaviors, leading to tougher solutions, often involving more complex functions.
At \(x=0\) in our specific differential equation, \(P(x)\) and \(Q(x)\) are analytic, meaning there's no division by zero or other undefined expressions. This gives us confidence that there are no singular points at \(x=0\), neither regular nor irregular.

Understanding whether a point is singular or not helps in choosing the right solution techniques, ultimately simplifying complex problems.
Exponents of Differential Equation
When encountering a regular singular point, the exponents of the differential equation reveal the behavior of its solutions. These exponents are critical in the Frobenius method, a technique used to find power series solutions around singular points.

However, in our exercise, since \(x=0\) is not a singular point but an ordinary one, the need to find these exponents does not arise.

If it were a regular singular point, the typical approach involves constructing an equation called the indicial equation to determine possible exponents. These exponents play a key role in shaping the power series solution. For learners, recognizing when and how to find these exponents is a powerful tool for solving differential equations around complicated points.

While this task was deemed unnecessary for this specific exercise, understanding how exponents fit into the broader scope of differential equations helps prepare for tackling more challenging problems.

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Most popular questions from this chapter

(a) Show that the solution of the initial value problem $$ y^{\prime}=1+y^{2}, \quad y(0)=0 $$ is \(y(x)=\tan x .\) (b) Because \(y(x)=\tan x\) is an odd function with \(y^{\prime}(0)=1\), its Taylor series is of the form $$ y=x+c_{3} x^{3}+c_{5} x^{5}+c_{7} x^{7}+\cdots $$ Substitute this series in \(y^{\prime}=1+y^{2}\) and equate like powers of \(x\) to derive the following relations: $$ 3 c_{3}=1, \quad 5 c_{5}=2 c_{3} $$ \(7 c_{7}=2 c_{5}+\left(c_{3}\right)^{2}\) $$ 9 c_{9}=2 c_{7}+2 c_{3} c_{5} $$ \(11 c_{11}=2 c_{9}+2 c_{3} c_{7}+\left(c_{5}\right)^{2}\) (c) Conclude that $$ \begin{aligned} \tan x=& x+\frac{1}{3} x^{3}+\frac{2}{15} x^{5}+\frac{17}{315} x^{7} \\ &+\frac{62}{2835} x^{9}+\frac{1382}{155925} x^{11}+\cdots . \end{aligned} $$ (d) Would you prefer to use the Maclaurin series formula in (13) to derive the tangent series in part (c)? Think about it!

Find two linearly independent Frobenius series solutions (for \(x>0\) ) of each of the differential equations in Problems 17 through \(26 .\) $$ 2 x y^{\prime \prime}-y^{\prime}-y=0 $$

Find a three-term recurrence relation for solutions of the form \(y=\sum c_{n} x^{n} .\) Then find the first three nonzero terms in each of two linearly independent solutions. $$ y^{\prime \prime}+(1+x) y=0 $$

Follow the steps outlined in this problem to establish Rodrigues's formula $$ P_{n}(x)=\frac{1}{n ! 2^{n}} \frac{d^{n}}{d x^{n}}\left(x^{2}-1\right)^{n} $$ for the \(n\) th-degree Legendre polynomial. (a) Show that \(v=\left(x^{2}-1\right)^{n}\) satisfies the differential equation $$ \left(1-x^{2}\right) v^{\prime}+2 n x v=0 $$ Differentiate each side of this equation to obtain $$ \left(1-x^{2}\right) v^{\prime \prime}+2(n-1) x v^{\prime}+2 n v=0 . $$ (b) Differentiate each side of the last equation \(n\) times in succession to obtain $$ \left(1-x^{2}\right) v^{(n+2)}-2 x v^{(n+1)}+n(n+1) v^{(n)}=0 . $$ Thus \(u=v^{(n)}=D^{n}\left(x^{2}-1\right)^{n}\) satisfies Legendre's equation of order \(n . \quad\) (c) Show that the coefficient of \(x^{n}\) in \(u\) is \((2 n) ! / n ! ;\) then state why this proves Rodrigues' formula. (Note that the coefficient of \(x^{n}\) in \(P_{n}(x)\) is \(\left.(2 n) ! /\left[2^{n}(n !)^{2}\right] .\right)\)

Find general solutions in powers of \(x\) of the differential equations. State the recurrence relation and the guaranteed radius of convergence in each case. $$ \left(x^{2}+2\right) y^{\prime \prime}+4 x y^{\prime}+2 y=0 $$

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