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Chapter 5: Problem 7

Find all singular points of the given equation and determine whether each one is regular or irregular. \((x+3) y^{\prime \prime}-2 x y^{\prime}+\left(1-x^{2}\right) y=0\)

Short Answer

Expert verified
Answer: The singular point of the given differential equation is x = -3, and it is a regular singular point.

Step by step solution

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01

Identify the leading coefficient of the highest derivative

The leading coefficient of the highest derivative, \(y''\), is \((x+3)\).
02

Find singular points by setting the leading coefficient to zero

To find the singular points, we set the leading coefficient to zero: \((x+3)=0\). Solving for x, we get: \(x = -3\). So, there is only one singular point at \(x=-3\).
03

Determine if the singular point is regular or irregular

A singular point is regular if all the coefficients of the equation have converging power series expansions around that point. Here, all coefficients are polynomials and they all converge everywhere (including at the singular point \(x=-3\)). Hence, the singular point \(x=-3\) is a regular singular point.

Key Concepts

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

Singular Points
In differential equations, singular points are specific values of the independent variable where the equation might not behave normally. This happens when the coefficient of the highest derivative (in our case, the second derivative, or \(y''\)) becomes zero.
For the equation
  • \((x+3)y'' - 2xy' + (1-x^2)y = 0\)
we see that the leading coefficient is \((x+3)\). To find singular points, we set this coefficient to zero, which gives us \(x = -3\).
Thus, \(x = -3\) is a singular point. Whenever solving similar equations, identifying these points is crucial as they often require special techniques to find solutions.
Regular Singular Point
A singular point is labeled as regular if the behavior of the differential equation around that point remains well-defined and manageable. Mathematically, this means the coefficients involved should have power series expansions that converge at the singular point.
In our exercise, the singular point occurs at \(x = -3\). The step involves examining each coefficient to see if a power series expansion around \(x = -3\) is possible. For
  • \((x+3)y'' - 2xy' + (1-x^2)y = 0\)
all coefficients are polynomials, which naturally converge everywhere, including at \(x = -3\).
Therefore, \(x = -3\) qualifies as a regular singular point. Recognizing regular singular points is important as they often allow the use of series solutions, like Frobenius method, to solve the differential equation.
Irregular Singular Point
Irregular singular points behave differently and more erratically than regular singular points. At an irregular singular point, the differential equation's coefficients do not have a simple, converging power series expansion.
This irregular behavior makes finding solutions more complex and sometimes requires more advanced techniques beyond standard series solutions.
In contrast to our current exercise, where we confirmed \(x = -3\) is regular, an irregular point would involve coefficients that lead to divergence or undefined expansions at that specific point.
Understanding the distinction between regular and irregular points helps in choosing the right approach to handle the unique challenges posed by irregular singular points in solving differential equations.

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Most popular questions from this chapter

Find two linearly independent solutions of the Bessel equation of order \(\frac{3}{2}\), $$ x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-\frac{9}{4}\right) y=0, \quad x>0 $$

Determine the general solution of the given differential equation that is valid in any interval not including the singular point. \(2 x^{2} y^{\prime \prime}-4 x y^{\prime}+6 y=0\)

It can be shown that \(J_{0}\) has infinitely many zeros for \(x>0 .\) In particular, the first three zeros are approximately \(2.405,5.520, \text { and } 8.653 \text { (see figure } 5.8 .1) .\) Let \(\lambda_{j}, j=1,2,3, \ldots,\) denote the zeros of \(J_{0}\) it follows that $$ J_{0}\left(\lambda_{j} x\right)=\left\\{\begin{array}{ll}{1,} & {x=0} \\ {0,} & {x=1}\end{array}\right. $$ Verify that \(y=J_{0}(\lambda, x)\) satisfies the differential equation $$ y^{\prime \prime}+\frac{1}{x} y^{\prime}+\lambda_{j}^{2} y=0, \quad x>0 $$ Ilence show that $$ \int_{0}^{1} x J_{0}\left(\lambda_{i} x\right) J_{0}\left(\lambda_{j} x\right) d x=0 \quad \text { if } \quad \lambda_{i} \neq \lambda_{j} $$ This important property of \(J_{0}\left(\lambda_{i} x\right),\) known as the orthogonality property, is useful in solving boundary value problems. Hint: Write the differential equation for \(J_{0}(\lambda, x)\). Multiply it by \(x J_{0}\left(\lambda_{y} x\right)\) and subtract it from \(x J_{0}\left(\lambda_{t} x\right)\) times the differential equation for \(J_{0}(\lambda, x)\). Then integrate from 0 to \(1 .\)

The Legendre Equation. Problems 22 through 29 deal with the Legendre equation $$ \left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\alpha(\alpha+1) y=0 $$ As indicated in Example \(3,\) the point \(x=0\) is an ordinaty point of this equation, and the distance from the origin to the nearest zero of \(P(x)=1-x^{2}\) is 1 . Hence the radius of convergence of series solutions about \(x=0\) is at least 1 . Also notice that it is necessary to consider only \(\alpha>-1\) because if \(\alpha \leq-1\), then the substitution \(\alpha=-(1+\gamma)\) where \(\gamma \geq 0\) leads to the Legendre equation \(\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\gamma(\gamma+1) y=0\) Show that the I.egendre equation can also be written as $$ \left[\left(1-x^{2}\right) y^{\prime}\right]=-\alpha(\alpha+1) y $$ Then it follows that \(\left[\left(1-x^{2}\right) P_{n}^{\prime}(x)\right]^{\prime}=-n(n+1) P_{n}(x)\) and \(\left[\left(1-x^{2}\right) P_{m}^{\prime}(x)\right]^{\prime}=\) \(-m(m+1) P_{m}(x) .\) By multiplying the first equation by \(P_{m}(x)\) and the second equation by \(P_{n}(x),\) and then integrating by parts, show that $$ \int_{-1}^{1} P_{n}(x) P_{m}(x) d x=0 \quad \text { if } \quad n \neq m $$ This property of the Legendre polynomials is known as the orthogonality property. If \(m=n,\) it can be shown that the value of the preceding integral is \(2 /(2 n+1) .\)

The Legendre Equation. Problems 22 through 29 deal with the Legendre equation $$ \left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\alpha(\alpha+1) y=0 $$ As indicated in Example \(3,\) the point \(x=0\) is an ordinaty point of this equation, and the distance from the origin to the nearest zero of \(P(x)=1-x^{2}\) is 1 . Hence the radius of convergence of series solutions about \(x=0\) is at least 1 . Also notice that it is necessary to consider only \(\alpha>-1\) because if \(\alpha \leq-1\), then the substitution \(\alpha=-(1+\gamma)\) where \(\gamma \geq 0\) leads to the Legendre equation \(\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\gamma(\gamma+1) y=0\) Show that, if \(\alpha\) is zero or a positive even integer \(2 n,\) the series solution \(y_{1}\) reduces to a polynomial of degree \(2 n\) containing only even powers of \(x\). Find the polynomials corresponding to \(\alpha=0,2,\) and \(4 .\) Show that, if \(\alpha\) is a positive odd integer \(2 n+1,\) the series solution \(y_{2}\) reduces to a polynomial of degree \(2 n+1\) containing only odd powers of \(x .\) Find the polynomials corresponding to \(\alpha=1,3,\) and \(5 .\)
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