Chapter 39The Power Series Method and Frobenius Theory

Example 39.1: Maclaurin expansion and radius of convergence of $f(x)=\dfrac1{1+x^2}$

Expand $f(x)=\dfrac{1}{1+x^2}$ in a Maclaurin series and find its radius of convergence using d'Alembert's ratio test.

Solution. Viewing it as a geometric series with common ratio $-x^2$ (substitute $t=-x^2$ into $\dfrac1{1-t}=1+t+t^2+\cdots$), we obtain

$$ f(x) = \frac{1}{1+x^2} = \frac1{1-(-x^2)} = 1+(-x^2)+(-x^2)^2+(-x^2)^3+\cdots = 1-x^2+x^4-x^6+\cdots $$

In this series, all coefficients of odd-degree terms are $0$, and the coefficient of the even-degree term $x^{2m}$ is $a_{2m}=(-1)^m$. The ratio of adjacent coefficients that are not $0$ is

$$ \left|\frac{a_{2m}}{a_{2m+2}}\right| = \left|\frac{(-1)^m}{(-1)^{m+1}}\right| = 1 $$

which is always $1$, regardless of $m$. Therefore, by d'Alembert's ratio test (we use the form of the formula with $x^2$ regarded as a single variable), the radius of convergence is $r=1$; that is, the series converges to $f(x)$ for $\abs{x}\lt1$. (We have verified with SymPy's $\mathrm{series}$ that $1-x^2+x^4-x^6$ agrees with the first four terms of the expansion. In fact, substituting $x=1$ gives the finite value $f(1)=1/2$, but the series side becomes $1-1+1-1+\cdots$, whose value is not determined, so we can also confirm that the series does not converge exactly at the boundary $x=\pm1$.)

Example 39.2: The Maclaurin expansion of $f(x)=e^{-x^2}$ and its radius of convergence

Find the first four terms of the Maclaurin expansion of $f(x)=e^{-x^2}$, and determine its radius of convergence.

Solution. Substituting $t=-x^2$ into $e^t=1+t+\dfrac{t^2}{2!}+\dfrac{t^3}{3!}+\cdots$ of Eq. \eqref{eq:39-exp-series}, we obtain

$$ e^{-x^2} = 1+(-x^2)+\frac{(-x^2)^2}{2!}+\frac{(-x^2)^3}{3!}+\cdots = 1-x^2+\frac{x^4}{2}-\frac{x^6}{6}+\cdots $$

(note that $(-x^2)^2=x^4$ and $(-x^2)^3=-x^6$). These are the first four terms. As for the radius of convergence, the expansion of the original $e^t$ converges for all real numbers $t$ ($R=\infty$; see Eq. \eqref{eq:39-exp-series}), and since $t=-x^2$ also takes a finite real value for every real number $x$, the expansion of $e^{-x^2}$ also converges for all real numbers $x$. Therefore the radius of convergence is $R=\infty$. (We verified that SymPy's $\mathrm{series}(e^{-x^2},x,0,8)$ agrees with $1-x^2+\dfrac{x^4}2-\dfrac{x^6}6$. Incidentally, this function is the kernel of the probability density function $\propto e^{-x^2/2}$ of the normal distribution in statistics, and is a very important function in applications.)

Example 39.3: Confirming that $x=0$ is an ordinary point of $y''+y=0$ and solving it by the power series method

For the equation $y''+y=0$, confirm that $x=0$ is an ordinary point and solve it by the power series method (verify that $\sin x,\cos x$, which you learned in Chapters 1 and 2, are indeed solutions).

Solution. Written in the form of Eq. ①, the equation has $P(x)=0,\ Q(x)=1$; both are constant functions (a special case of polynomials) and hence analytic at every real number. In particular, $x=0$ is an ordinary point. Therefore the solution has the form $y=\displaystyle\sum_{k=0}^{\infty}a_kx^k$. Differentiating term by term,

$$ y'=\sum_{k=1}^{\infty}ka_kx^{k-1},\qquad y''=\sum_{k=2}^{\infty}k(k-1)a_kx^{k-2} $$

Substituting into the original equation, we obtain

$$ \sum_{k=2}^{\infty}k(k-1)a_kx^{k-2}+\sum_{k=0}^{\infty}a_kx^k=0 $$

Shifting the index in the left-hand sum by $k\to k+2$ (when we replace $k$ with $k+2$ in $k(k-1)a_k x^{k-2}$, the lower limit of the sum becomes $k=0$ and $x^{k-2}\to x^k$), we get

$$ \sum_{k=0}^{\infty}(k+2)(k+1)a_{k+2}x^{k}+\sum_{k=0}^{\infty}a_kx^k=0 \ \Longrightarrow\ \sum_{k=0}^{\infty}\bigl\{(k+2)(k+1)a_{k+2}+a_k\bigr\}x^k=0 $$

Since this is an identity in $x$, the coefficient of each power must be $0$, and we obtain the recurrence relation

$$ a_{k+2}=-\frac{a_k}{(k+1)(k+2)} \qquad (k=0,1,2,\ldots) $$

The coefficients $a_0,a_1$ can be chosen freely (they correspond to the initial values). Setting $a_0=1,a_1=0$, we get $a_2=-\dfrac1{2\cdot1}=-\dfrac12$, $a_4=-\dfrac{a_2}{4\cdot3}=\dfrac1{24}$, …, and since the coefficients of all odd powers are $0$, we obtain

$$ y=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots=\cos x $$

In the same way, setting $a_0=0,a_1=1$ gives $y=\sin x$. The general solution is the linear combination $y=C_1\cos x+C_2\sin x$ of these two linearly independent solutions. This coincides exactly with the solution of $y''+y=0$ that you know from high school mathematics—the power series method is a way to “discover” even the familiar functions $\sin,\cos$ from the recurrence relation for the coefficients alone, without any knowledge of exponential or trigonometric functions. (Verified with SymPy that the recurrence relation agrees with the expansion of $\cos x$.)

Example 39.4: Classifying the singular point of $y''-\dfrac1xy'=0$ at $x=0$

For the equation $y''-\dfrac1xy'=0$, determine what kind of point $x=0$ is (an ordinary point, a regular singular point, or something else).

Solution. Putting the equation in the form of Eq. ①, we have $P(x)=-\dfrac1x,\ Q(x)=0$. The function $P(x)=-1/x$ is not even defined at $x=0$ (the denominator becomes $0$), so it is not analytic there, and $x=0$ is a singular point. So we examine $xP(x),x^2Q(x)$:

$$ xP(x)=x\cdot\left(-\frac1x\right)=-1, \qquad x^2Q(x)=x^2\cdot0=0 $$

Both are analytic at $x=0$ (as constant functions). Therefore $x=0$ is a regular singular point. In this case $a_0=-1,b_0=0$ (and all the other $a_i,b_i$ are also $0$), so the indicial equation \eqref{eq:39-indicial} becomes

$$ \lambda(\lambda-1)+(-1)\lambda+0=\lambda^2-2\lambda=\lambda(\lambda-2)=0 \ \Longrightarrow\ \lambda=0,\,2 $$

Indeed, $y=1$ (a constant function, corresponding to $\lambda=0$) is a solution of this equation because $y''=0,\ y'=0$, and $y=x^2$ (corresponding to $\lambda=2$) also satisfies it because $y''=2,\ y'=2x$ give $2-\dfrac1x\cdot2x=2-2=0$; so both are certainly solutions. (Verified with SymPy that both satisfy the equation.)

Example 39.5: The indicial equation for $P(x)=\dfrac1x,\ Q(x)=-\dfrac2{x^2}$

For the equation $y''+P(x)y'+Q(x)y=0$ with $P(x)=\dfrac1x,\ Q(x)=-\dfrac2{x^2}$, verify that $x=0$ is a regular singular point, set up the indicial equation, and find the values of $\lambda$.

Solution. We have $xP(x)=x\cdot\dfrac1x=1$ and $x^2Q(x)=x^2\cdot\left(-\dfrac2{x^2}\right)=-2$, and both are analytic at $x=0$ as constant functions (independent of $x$). Hence $a_0=1,\ b_0=-2$ (all the other $a_i,b_i$ are $0$), and $x=0$ is a regular singular point. The indicial equation \eqref{eq:39-indicial-expand} is

$$ \lambda^2+(1-1)\lambda+(-2)=\lambda^2-2=0 \ \Longrightarrow\ \lambda=\pm\sqrt2 $$

We have $\lambda_1=\sqrt2,\ \lambda_2=-\sqrt2$, and the difference $\lambda_1-\lambda_2=2\sqrt2\approx2.83$ is not an integer. This is therefore case (i) of Theorem 39.1, and the two fundamental solutions are simply $y_1=x^{\sqrt2},\ y_2=x^{-\sqrt2}$ (when $a_0,b_0$ are constants and the other $a_i,b_i$ are $0$, the terms with $k\ge1$ may all be taken to be $0$, and the series is complete with its leading term alone). Indeed, if we substitute $y=x^{r}$ ($r$ a constant) directly into the equation, then since $y'=rx^{r-1},\ y''=r(r-1)x^{r-2}$,

$$ r(r-1)x^{r-2}+\frac1x\cdot rx^{r-1}-\frac2{x^2}\cdot x^{r}=\bigl\{r(r-1)+r-2\bigr\}x^{r-2}=(r^2-2)x^{r-2} $$

and we can confirm that this is exactly $0$ when $r=\pm\sqrt2$ (this is just the indicial equation $\lambda^2-2=0$ itself). (Verified with SymPy that both $r=\pm\sqrt2$ satisfy the equation.)

Example 39.6: Series solution of $y''-\dfrac{2x}{1-x^2}y'=0$ ($-1\lt x\lt1$)

Find the solution of the equation $\dfrac{d^2y}{dx^2}-\dfrac{2x}{1-x^2}\dfrac{dy}{dx}=0$ ($-1\lt x\lt1$) by the power series method.

Solution. Let $P(x)=-\dfrac{2x}{1-x^2},\ Q(x)=0$. In the range $-1\lt x\lt1$ the denominator $1-x^2$ never becomes $0$, so $P(x)$ (and, of course, $Q(x)=0$, which is a polynomial) is analytic at $x=0$, and $x=0$ is an ordinary point. Therefore the solution can be expressed in the form

$$ y=\sum_{k=0}^{\infty}a_k(x-0)^k $$

To find the coefficients, we differentiate $y$ once and twice.

$$ \frac{dy}{dx}=\sum_{k=1}^{\infty}a_k\cdot kx^{k-1},\qquad \frac{d^2y}{dx^2}=\sum_{k=2}^{\infty}a_k\cdot k(k-1)x^{k-2} $$

Substituting these into the original equation gives

$$ \sum_{k=2}^{\infty}a_k\,k(k-1)x^{k-2} - \frac{2x}{1-x^2}\sum_{k=1}^{\infty}a_k\,kx^{k-1} = 0 $$

The denominator $1-x^2$ on the right-hand side is a nuisance, so we multiply both sides by $1-x^2$ to clear it. At this point we split off by itself only the leading term ($k=1$) of the sum $\sum_{k=1}^{\infty}a_k kx^{k-1}$ in the second term—this is preparation for making it easy, later on, to align the exponents with the sum in the first term (which starts from $k=2$) and compare them.

$$ (1-x^2)\sum_{k=2}^{\infty}a_k\,k(k-1)x^{k-2} - 2x\left(a_1+\sum_{k=2}^{\infty}a_k\,kx^{k-1}\right) = 0 $$

We expand the left-hand side. Distributing $(1-x^2)$ over the sum in the first term (it splits into the part multiplied by $1$ and the part multiplied by $-x^2$), and distributing $-2x$ over each of $a_1$ and the sum inside the parentheses, the left-hand side becomes

$$ \sum_{k=2}^{\infty}a_k\,k(k-1)x^{k-2} - \sum_{k=2}^{\infty}a_k\,k(k-1)x^{k} - 2a_1x - 2\sum_{k=1}^{\infty}a_k\,kx^{k} = 0 $$

(In the last term, the part coming from the $a_1x^0$ that we split off earlier and the part coming from the remaining sum $\sum_{k=2}^\infty$ have been combined and written in the original form $\sum_{k=1}^\infty$.) Finally, we want to arrange all the terms in powers of $x^k$ and compare coefficients. The sum in the first term, $\sum_{k=2}^{\infty}a_k k(k-1)x^{k-2}$, has the power $x^{k-2}$, so we shift the index by $k\to k+2$ (this shift also makes the lower limit of the sum $k=0$, so we split off beforehand the $k=0$ term $2a_2$ and the $k=1$ term $6a_3x$), which gives

$$ 2a_2 + 6a_3x + \sum_{k=2}^{\infty}a_{k+2}(k+1)(k+2)x^{k} - \sum_{k=2}^{\infty}a_k\,k(k-1)x^{k} - 2a_1x - \sum_{k=1}^{\infty}2a_k\,kx^{k} = 0 $$

Now we collect the terms in $x^0,x^1$ separately (the coefficient of $x^1$ is obtained by combining $6a_3x$ with the $k=1$ term $-2a_1x$ of the last sum, giving $6a_3-2a_1=2(3a_3-a_1)$), and for the coefficient of $x^k$ ($k\ge2$) it suffices to compare the two sums directly, so we obtain

\begin{equation} 2a_2 + 2(3a_3-a_1)x + \sum_{k=2}^{\infty}\bigl\{(k+1)(k+2)a_{k+2} - k(k+1)a_k\bigr\}x^{k} = 0 \label{eq:39-ex6-identity} \end{equation}

Since Eq. \eqref{eq:39-ex6-identity} is an identity in $x$, the coefficient of each power must itself be $0$. That is,

$$ a_2=0,\qquad 3a_3-a_1=0,\qquad (k+1)(k+2)a_{k+2}-k(k+1)a_k=0\ \ (k=2,3,\ldots) $$

The last recurrence relation can be divided through by $k+1\neq0$ (automatic, since $k\ge2$), which gives $a_{k+2}=\dfrac{k}{k+2}a_k$. Starting from $a_2=0$, this recurrence makes all the even-degree coefficients $0$.

$$ a_2=a_4=a_6=\cdots=0 $$

On the other hand, the odd-degree coefficients are determined one after another starting from $a_1$.

$$ a_3=\frac13a_1,\qquad a_5=\frac35a_3=\frac15a_1,\qquad a_7=\frac57a_5=\frac17a_1 $$

(In general the pattern is $a_{2m+1}=\dfrac1{2m+1}a_1$—check that when we multiply the fractions $\dfrac{k}{k+2}$ one after another, all the intermediate factors cancel and disappear.) Therefore the series solution is

\begin{equation} y=a_0+a_1\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\cdots\right) \label{eq:39-ex6-series} \end{equation}

Here $a_0,a_1$ are two arbitrary constants that can be chosen freely according to the initial values, and they supply the freedom of two fundamental solutions.

Example 39.7: Series solution of $y''+\dfrac{2x}{1+x^2}y'=0$ (a problem similar to Example 39.6)

Find the solution of the equation $\dfrac{d^2y}{dx^2}+\dfrac{2x}{1+x^2}\dfrac{dy}{dx}=0$ by the power series method.

Solution. Since the denominator $1+x^2$ does not become $0$ for any real number $x$, $P(x)=\dfrac{2x}{1+x^2}$ is analytic for all real numbers, and $x=0$ is an ordinary point (only the sign differs from Example 39.6, and the idea is exactly the same). Let $y=\sum_{k=0}^{\infty}a_kx^k$ and substitute. Following exactly the same procedure as in Example 39.6 (multiply both sides by $1+x^2$ to clear the denominator, align the indices, and compare the coefficients of $x^k$), we obtain the identity

$$ 2a_2+2(3a_3+a_1)x+\sum_{k=2}^{\infty}\bigl\{(k+1)(k+2)a_{k+2}+k(k+1)a_k\bigr\}x^k=0 $$

(Compared with Eq. \eqref{eq:39-ex6-identity}, the identity of Example 39.6, several signs are reversed because the $-x^2$ term has become a $+x^2$ term.) Hence we obtain $a_2=0$, $3a_3=-a_1$, and the recurrence relation $a_{k+2}=-\dfrac{k}{k+2}a_k$ ($k\ge2$). Again all the even-degree coefficients are $0$, and the odd-degree coefficients are determined successively as follows:

$$ a_3=-\frac13a_1,\qquad a_5=-\frac35a_3=\frac15a_1,\qquad a_7=-\frac57a_5=-\frac17a_1 $$

Here the sign alternates as the coefficients are determined. Therefore the series solution is

$$ y=a_0+a_1\left(x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots\right) $$

The series in parentheses is exactly the Maclaurin expansion of the inverse tangent function ($\arctan x=\tan^{-1}x$), so $y=a_0+a_1\arctan x$ is the closed-form solution. Indeed, using $\dfrac{d}{dx}\arctan x=\dfrac1{1+x^2}$, a calculation exactly like that of Example 39.6 confirms that $y=a_0+a_1\arctan x$ satisfies the equation exactly (verified with SymPy). The radius of convergence of the expansion of $\arctan x$ is $1$, which again corresponds to the singular points of $P(x)=\dfrac{2x}{1+x^2}$ ($x=\pm i$; these are complex singular points, but they still make themselves felt in the real radius of convergence as a distance of $1$).

Example 39.8: Frobenius series solution of $y''+\dfrac3xy'-\dfrac3{x^2}y=0$ ($x\neq0$)

Find the solution of the equation $\dfrac{d^2y}{dx^2}+\dfrac3x\dfrac{dy}{dx}-\dfrac3{x^2}y=0$ ($x\neq0$).

Solution. Letting $P(x)=\dfrac3x,\ Q(x)=-\dfrac3{x^2}$, we have

$$ xP(x)=x\cdot\frac3x=3,\qquad x^2Q(x)=x^2\cdot\left(-\frac3{x^2}\right)=-3 $$

and both are constant functions (independent of $x$), hence analytic at $x=0$.

Therefore $x=0$ is a regular singular point, and the equation has a solution of the form $y=(x-0)^{\lambda}\displaystyle\sum_{k=0}^{\infty}a_k(x-0)^k$. Here $a_0=3,\ b_0=-3$. Differentiating $y$ once and then twice, we have

$$ \frac{dy}{dx}=\sum_{k=0}^{\infty}a_k(\lambda+k)x^{\lambda+k-1},\qquad \frac{d^2y}{dx^2}=\sum_{k=0}^{\infty}a_k(\lambda+k)(\lambda+k-1)x^{\lambda+k-2} $$

Substituting these into the original equation, we obtain

$$ \sum_{k=0}^{\infty}a_k(\lambda+k)(\lambda+k-1)x^{\lambda+k-2} + \frac3x\sum_{k=0}^{\infty}a_k(\lambda+k)x^{\lambda+k-1} - \frac3{x^2}\sum_{k=0}^{\infty}a_kx^{\lambda+k} = 0 $$

In the second and third terms on the right-hand side, if we bring the factors $\dfrac1x,\dfrac1{x^2}$, respectively, inside the sums, all the terms can be put into a form in which they have the same exponent $x^{\lambda+k-2}$ (note that $\dfrac3x\cdot x^{\lambda+k-1}=3x^{\lambda+k-2}$ and $-\dfrac3{x^2}\cdot x^{\lambda+k}=-3x^{\lambda+k-2}$). Factoring $x^{\lambda+k-2}$ out of all the terms in this way, we get

$$ \sum_{k=0}^{\infty}\Bigl\{a_k(\lambda+k)(\lambda+k-1) + 3a_k(\lambda+k) - 3a_k\Bigr\}x^{\lambda+k-2} = 0 $$

This is an identity in $x$, so for each $k$ the expression in the braces must be $0$.

\begin{equation} \bigl\{(\lambda+k)(\lambda+k-1)+3(\lambda+k)-3\bigr\}a_k=0 \qquad (k=0,1,2,\ldots) \label{eq:39-ex8-recursion} \end{equation}

Consider the case $k=0$. By the definition of a Frobenius series, $a_0\neq0$, so the expression in the braces must itself be $0$. This is exactly the indicial equation \eqref{eq:39-indicial-expand} with $a_0=3,b_0=-3$ substituted.

$$ \lambda(\lambda-1)+3\lambda-3=\lambda^2+2\lambda-3=0 \ \Longrightarrow\ \lambda=1,\,-3 $$

We have $\lambda_1=1,\ \lambda_2=-3$, and the difference $\lambda_1-\lambda_2=4$ is a positive integer—this is case (iii) of Theorem 39.1, which requires special care.

First we find $y_1$ from the larger root $\lambda=1$. Substituting $\lambda=1$ into Eq. \eqref{eq:39-ex8-recursion}, we get

$$ \bigl\{(1+k)(k)+3(1+k)-3\bigr\}a_k=(k^2+k+3k+3-3)a_k=(k^2+4k)a_k=k(k+4)a_k=0 $$

For $k=1,2,3,\ldots$ we have $k(k+4)\neq0$, so $a_k=0$ must hold for all of these $k$. Therefore only the leading term with $a_0$ remains in the series, and we obtain

$$ y_1=a_0x \quad\text{that is,}\quad y_1=x $$

as one of the fundamental solutions.

Next, a problem arises when we try to find $y_2$ from the other root $\lambda=-3$. As also noted in Theorem 39.1 of Section 39.4, when $\lambda_1-\lambda_2=4$ (a positive integer), there is a danger that the recurrence relation built from the smaller root will run into “resonance” partway through. Indeed, substituting $\lambda=-3$ into Eq. \eqref{eq:39-ex8-recursion}, the expression in the braces becomes $\{(k-3)(k-4)+3(k-3)-3\}=k(k-4)$ (that it factors was verified with SymPy), which is exactly $0$ at $k=4$ (= $\lambda_1-\lambda_2$). That is, in the recurrence relation coming from $\lambda=-3$, the coefficient $a_4$ is left undetermined at $k=4$—this phenomenon occurs because $\lambda_2+4=\lambda_1=1$ coincides with the other root. In this situation, if $a_4$ can be chosen freely, the resulting series ends up containing an extra constant multiple of $y_1=x$, and it is not necessarily a “truly new” solution that is linearly independent of $y_1$ (it is not guaranteed that the Wronskian is not identically $0$). So rather than pushing on in this way, we find $y_2$ by another method.

Here, the conceivable ways of expressing $y_2$ in terms of $y_1$ are either a constant multiple $y_2=Cy_1$ (this is merely a constant multiple of $y_1$ itself and carries no new information) or a form with a constant added, $y_2=C+y_1$ ($C$ is a constant). Let us actually try the latter. Letting $y_2=C+x$, we have $y_2'=1,\ y_2''=0$, so substituting this into the equation gives

$$ \frac3x\cdot1-\frac3{x^2}(C+x)=\frac3x-\frac{3C}{x^2}-\frac3x=-\frac{3C}{x^2} $$

For this to be $0$ for all $x$, we must have $C=0$. That is, even if we try the form $y_2=C+y_1$, all we obtain is $y_2=y_1$ (linear, or “$1$st-degree,” dependence), and no new solution independent of $y_1$ appears (verified with SymPy). Since both of the simple forms, a constant multiple and a constant added, fail, let us generalize the constant $C$ to a function $u(x)$ of $x$—

We use the method for the case in which one fundamental solution is known (Chapter 36; it is also called reduction of order). Since $y_1=x$ is known, we expect that another solution $y_2$ can be expressed, using some function $u(x)$, in the form $y_2=u(x)y_1=x\,u(x)$. The general solution can then be written, with $C_1$ the coefficient of $y_1$ and $C_2'$ the coefficient of $y_2$, as

$$ y=C_1y_1+C_2'y_2=y_1\bigl(C_1+C_2'u(x)\bigr) $$

Now let $y_2=x\,u(x)$ ($u(x)$ is an unknown function) and substitute this into the original equation. By the product rule, we have

$$ y_2'=u+xu',\qquad y_2''=2u'+xu'' $$

Substituting these into the equation $y''+\dfrac3xy'-\dfrac3{x^2}y=0$ gives

$$ (2u'+xu'')+\frac3x(u+xu')-\frac3{x^2}(xu)=xu''+2u'+\frac3xu+3u'-\frac3xu=xu''+5u'=0 $$

(Note that $\dfrac3xu$ and $-\dfrac3xu$ cancel exactly, and the coefficient of $u'$ combines into $2+3=5$.) Multiplying both sides by $x^2$ and simplifying, we get

$$ x^3u''+5x^2u'=0 $$

(This agrees with the result in the original notes.) Since $u$ itself does not appear and this is a first-order linear differential equation in $u'$, letting $w=u'$ gives $x^3w'+5x^2w=0$, that is, the separable equation $\dfrac{w'}{w}=-\dfrac5x$. Integrating both sides, we obtain

$$ \ln\abs w=-5\ln\abs x+C_1 \ \Longrightarrow\ w=\frac{C_2}{x^5} $$

Here $C_2=\pm e^{C_1}$; integrating $u'=w=\dfrac{C_2}{x^5}$ further, we have

$$ u=\int\frac{C_2}{x^5}\,dx=-\frac{C_2}{4x^4}+C_3=\frac{C_2'}{x^4}+C_3 \qquad (C_2'=-C_2/4) $$

What deserves attention is that no logarithmic term appeared at all in the course of this integration (if a $1/x$ form such as $w=C_2/x^1$ had appeared, integrating it would have produced $\ln x$). This is a concrete example of the “case in which the constant $C$ is $0$” mentioned in Theorem 39.1(iii) of Section 39.4. In this way we obtain

$$ y_2=x\,u(x)=x\left(C_3+\frac{C_2'}{x^4}\right)=C_3x+\frac{C_2'}{x^3} $$

On the right-hand side, the part $C_3x$ is merely a constant multiple of the already obtained $y_1=x$ and carries no new information. Therefore, a “truly new” second fundamental solution that is linearly independent of $y_1$ is

$$ y_2=\frac1{x^3} $$

(Indeed, substituting $y_2=1/x^3$ directly into the equation gives an identically $0$ result, as verified with SymPy. Moreover, the Wronskian $W(y_1,y_2)=y_1y_2'-y_2y_1'=x\cdot\left(-\dfrac3{x^4}\right)-\dfrac1{x^3}\cdot1=-\dfrac4{x^3}$ is not identically $0$, so $y_1,y_2$ are indeed linearly independent.)

From the above, the general solution is

\begin{equation} y=C_1y_1+C_2y_2=C_1x+\frac{C_2}{x^3} \label{eq:39-ex8-general} \end{equation}

(We also verified that SymPy's $\mathrm{dsolve}$ yields exactly the same general solution $y=C_1x+C_2/x^3$.)

Exercise 39.1: Radius of convergence of $\cos x$

Find the radius of convergence of the Maclaurin expansion of $\cos x$ (Eq. \eqref{eq:39-cos-series}) by d'Alembert's ratio test (Definition 39.1).

Hint: The coefficients of $\cos x$ appear only in the even-degree terms, and the coefficient of $x^{2k}$ is $a_{2k}=\dfrac{(-1)^k}{(2k)!}$. Compute $\left|\dfrac{a_{2k}}{a_{2k+2}}\right|$ and take the limit $k\to\infty$.

→ Solution

Exercise 39.2: Series solution of $y''-xy'-y=0$

For the equation $y''-xy'-y=0$, verify that $x=0$ is an ordinary point, and find the recurrence relation for the coefficients by the power series method. Furthermore, taking $a_0=1,a_1=0$, write down the terms of the series solution through the $x^6$ term.

Hint: In the same way as in Example 39.6, substitute $y=\sum a_kx^k$, align the exponents of $x^k$, and compare coefficients. Since there are no denominators to clear this time, it is simpler than Example 39.6.

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Exercise 39.3: Classifying a Bessel-type equation of half-integer order

For the equation $y''+\dfrac1xy'+\left(1-\dfrac1{4x^2}\right)y=0$, determine what kind of point $x=0$ is; if it is a regular singular point, set up the indicial equation and find the values of $\lambda$.

Hint: Take $P(x)=\dfrac1x,\ Q(x)=1-\dfrac1{4x^2}$ and examine what kind of series each of $xP(x),x^2Q(x)$ is at $x=0$ (since $x^2Q(x)=x^2-\dfrac14$ is a polynomial, it is analytic).

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Exercise 39.4: Complex roots of the indicial equation

For $P(x)=\dfrac1x,\ Q(x)=\dfrac1{x^2}$, set up the indicial equation and find the values of $\lambda$.

Hint: It suffices to substitute $a_0=1,b_0=1$ into the indicial equation \eqref{eq:39-indicial-expand}. Note that the resulting $\lambda$ need not be real.

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Exercise 39.5: Re-verifying the error in $\log(1+x)$

Regarding the sign error in the fourth term of the Maclaurin expansion of $\log(1+x)$ pointed out in the Caution box (box warn) in Section 39.2, actually compute the fourth derivative $f^{(4)}(0)$ of $f(x)=\log(1+x)$, and confirm by hand, for yourself, that the correct coefficient $a_4$ is $-\dfrac14$.

Hint: Differentiate successively, starting from $f'(x)=(1+x)^{-1}$, and follow the changes of sign. We have $a_4=f^{(4)}(0)/4!$ (Eq. \eqref{eq:39-maclaurin}).

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Exercise 39.6: What if $\lambda_1-\lambda_2$ were not an integer?

In the equation $y''+\dfrac3xy'-\dfrac3{x^2}y=0$ of Example 39.8, suppose that we had $Q(x)=-\dfrac2{x^2}$ instead of $Q(x)=-\dfrac3{x^2}$ (with $P(x)=\dfrac3x$ unchanged). What would the difference between the two roots of the indicial equation be? Also, explain how the procedure for finding the fundamental solution $y_2$ would differ from that of Example 39.8 in this case.

Hint: With $a_0=3,b_0=-2$, solve the indicial equation $\lambda^2+2\lambda-2=0$. Since the difference is no longer an integer, consider which case of Theorem 39.1 applies.

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