Solution. (a) We make the Ansatz

\[ u(x)=\sum _{k=0}^\infty a_kx^k,\qquad u'(x)=\sum _{k=0}^\infty a_kkx^{k-1},\qquad u''(x)=\sum _{k=0}^\infty a_kk(k-1)x^{k-2}. \]

Substituting this in the differential equation gives

\[ (1-x^2)\sum _{k=0}^\infty k(k-1)a_kx^{k-2}-2x\sum _{k=0}^\infty ka_kx^{k-1}-\lambda \sum _{k=0}^\infty a_kx^k=0. \]

We rewrite this as

\[ \sum _{k=0}^\infty k(k-1)a_kx^{k-2}-x^2\sum _{k=0}^\infty k(k-1)a_kx^{k-2}-2x\sum _{k=0}^\infty ka_kx^{k-1}-\lambda \sum _{k=0}^\infty a_kx^k=0, \]

and re-write further as

\[ \sum _{k=0}^\infty k(k-1)a_kx^{k-2}-\sum _{k=0}^\infty k(k-1)a_kx^k-\sum _{k=0}^\infty 2ka_kx^{k}-\lambda \sum _{k=0}^\infty a_kx^k=0. \]

To make the exponents line up, we shift the index in the first term: we define \(n:=k-2\) so that \(k=n+2\) and we obtain (noting that we can sum from \(n=0\) since the terms for \(n\in \{-2,-1\}\) equal zero)

\[ \sum _{n=0}^\infty (n+2)(n+1)a_{n+2}x^{n}. \]

Changing the name for the index back to \(k\), we obtain for the differential equation

\[ \sum _{k=0}^\infty (k+2)(k+1)a_{k+2}x^k-\sum _{k=0}^\infty k(k-1)a_kx^k-\sum _{k=0}^\infty 2ka_kx^k-\sum _{k=0}^\infty \lambda a_kx^k=0, \]

which gives

\[ \sum _{k=0}^\infty \left [ (k+2)(k+1)a_{k+2}-k(k-1)a_k-2ka_k-\lambda a_k \right ]x^k=0, \]

which gives (a power series is the zero function if and only if all coefficients are zero):

\[ (k+2)(k+1)a_{k+2}-k(k-1)a_k-2ka_k-\lambda a_k=0. \]

This gives the following relation for the coefficients:

\[ a_{k+2}=\frac {k(k-1)+2k+\lambda }{(k+2)(k+1)}a_k, \]

which we can rewrite as

\[ a_{k+2}=\frac {k(k+1)+\lambda }{(k+2)(k+1)}a_k. \]

(b) We see that (for a given \(\lambda \)) \(a_0\) determines all even coefficients and \(a_1\) determines all odd coefficients. In this way we obtain two linearly independent solutions (one by taking \(a_0=1\) and \(a_1=0\) which gives an even function and another by taking \(a_0=0\) and \(a_1=1\) which gives an odd function). □

Solution. We recall the recurrence relation obtained in Problem E.1:

\begin{equation} \label {eq:LegendreRecurrence}\tag {$\dagger $} a_{k+2}=\frac {k(k+1)+\lambda }{(k+2)(k+1)}a_k. \end{equation}

(a) When \(\lambda =-(n+1)n\) for some \(n\in \mN _0\), then from (\(\dagger \)) we obtain that \(a_{n+2}=0\), so that using (\(\dagger \)) again we have \(a_{n+4}=a_{n+6}=\ldots =0\). Hence the corresponding solution (if \(n\) is even the even solution and if \(n\) is odd the odd solution) is a polynomial of degree \(n\). When \(\lambda \) is not of the above form, then the sequence \((a_k)\) has infinitely many nonzero terms and therefore the power series does not reduce to a polynomial.

(b) Polynomials are continuously differentiable (on \(\mR \) and therefore in particular on \([-1,1]\)). It follows from Problem D.1 that the boundary condition obtained in Problem D.1 is satisfied by these polynomial solutions. □

Solution. (a) We make the Ansatz

\[ u(x)=\sum _{k=0}^\infty a_kx^k,\qquad u'(x)=\sum _{k=0}^\infty a_kkx^{k-1},\qquad u''(x)=\sum _{k=0}^\infty a_kk(k-1)x^{k-2}. \]

Substituting this in the differential equation gives

\[ (1-x^2)\sum _{k=0}^\infty k(k-1)a_kx^{k-2}-x\sum _{k=0}^\infty ka_kx^{k-1}-\lambda \sum _{k=0}^\infty a_kx^k=0. \]

We rewrite this as

\[ \sum _{k=0}^\infty k(k-1)a_kx^{k-2}-x^2\sum _{k=0}^\infty k(k-1)a_kx^{k-2}-x\sum _{k=0}^\infty ka_kx^{k-1}-\lambda \sum _{k=0}^\infty a_kx^k=0, \]

and re-write further as

\[ \sum _{k=0}^\infty k(k-1)a_kx^{k-2}-\sum _{k=0}^\infty k(k-1)a_kx^k-\sum _{k=0}^\infty ka_kx^{k}-\lambda \sum _{k=0}^\infty a_kx^k=0. \]

To make the exponents line up, we shift the index in the first term: we define \(n:=k-2\) so that \(k=n+2\) and we obtain (noting that we can sum from \(n=0\) since the terms for \(n\in \{-2,-1\}\) equal zero)

\[ \sum _{n=0}^\infty (n+2)(n+1)a_{n+2}x^{n}. \]

Changing the name for the index back to \(k\), we obtain for the differential equation

\[ \sum _{k=0}^\infty (k+2)(k+1)a_{k+2}x^k-\sum _{k=0}^\infty k(k-1)a_kx^k-\sum _{k=0}^\infty ka_kx^k-\sum _{k=0}^\infty \lambda a_kx^k=0, \]

which gives

\[ \sum _{k=0}^\infty \left [ (k+2)(k+1)a_{k+2}-k(k-1)a_k-ka_k-\lambda a_k \right ]x^k=0, \]

which gives (a power series is the zero function if and only if all coefficients are zero):

\[ (k+2)(k+1)a_{k+2}-k(k-1)a_k-ka_k-\lambda a_k=0. \]

This gives the following relation for the coefficients:

\[ a_{k+2}=\frac {k(k-1)+k+\lambda }{(k+2)(k+1)}a_k, \]

which we can rewrite as

\[ a_{k+2}=\frac {k^2+\lambda }{(k+2)(k+1)}a_k. \]

(b) We see that (for a given \(\lambda \)) \(a_0\) determines all even coefficients and \(a_1\) determines all odd coefficients. In this way we obtain two linearly independent solutions (one by taking \(a_0=1\) and \(a_1=0\) which gives an even function and another by taking \(a_0=0\) and \(a_1=1\) which gives an odd function). □

Solution. We recall the recurrence relation obtained in Problem E.3:

\begin{equation} \label {eq:ChebyshevRecurrence}\tag {$\dagger $} a_{k+2}=\frac {k^2+\lambda }{(k+2)(k+1)}a_k. \end{equation}

(a) When \(\lambda =-n^2\) for some \(n\in \mN _0\), then from (\(\dagger \)) we obtain that \(a_{n+2}=0\), so that using (\(\dagger \)) again we have \(a_{n+4}=a_{n+6}=\ldots =0\). Hence the corresponding solution (if \(n\) is even the even solution and if \(n\) is odd the odd solution) is a polynomial of degree \(n\). When \(\lambda \) is not of the above form, then the sequence \((a_k)\) has infinitely many nonzero terms and therefore the power series does not reduce to a polynomial.

(b) Polynomials are continuously differentiable (on \(\mR \) and therefore in particular on \([-1,1]\)). It follows from Problem D.4 that the boundary condition obtained in Problem D.4 is satisfied by these polynomial solutions. □

Solution. (a) We make the Ansatz

\[ u(x)=\sum _{k=0}^\infty a_kx^k,\qquad u'(x)=\sum _{k=0}^\infty a_kkx^{k-1},\qquad u''(x)=\sum _{k=0}^\infty a_kk(k-1)x^{k-2}. \]

Substituting this in the differential equation gives

\[ x\sum _{k=0}^\infty k(k-1)a_kx^{k-2}+(1-x)\sum _{k=0}^\infty ka_kx^{k-1}-\lambda \sum _{k=0}^\infty a_kx^k=0. \]

We rewrite this as

\[ \sum _{k=0}^\infty k(k-1)a_kx^{k-1}+\sum _{k=0}^\infty ka_kx^{k-1}-\sum _{k=0}^\infty ka_kx^{k}-\lambda \sum _{k=0}^\infty a_kx^k=0. \]

To make the exponents line up, we shift the index in the first and second terms: we define \(n:=k-1\) so that \(k=n+1\) and we obtain (noting that we can sum from \(n=0\) since the term for \(n=-1\) equal zero)

\[ \sum _{n=0}^\infty (n+1)na_{n+1}x^n+\sum _{n=0}^\infty (n+1)a_{n+1}x^n. \]

Changing the name for the index back to \(k\), we obtain for the differential equation

\[ \sum _{k=0}^\infty (k+1)ka_{k+1}x^k+\sum _{k=0}^\infty (k+1)a_{k+1}x^k-\sum _{k=0}^\infty ka_kx^k-\sum _{k=0}^\infty \lambda a_kx^k=0, \]

which gives

\[ \sum _{k=0}^\infty \left [ (k+1)ka_{k+1}+(k+1)a_{k+1}-ka_k-\lambda a_k \right ]x^k=0, \]

which gives (a power series is the zero function if and only if all coefficients are zero):

\[ (k+1)ka_{k+1}+(k+1)a_{k+1}-ka_k-\lambda a_k. \]

This gives the following relation for the coefficients:

\[ a_{k+1}=\frac {k+\lambda }{(k+1)k+(k+1)}a_k, \]

which we can rewrite as

\[ a_{k+1}=\frac {k+\lambda }{(k+1)^2}a_k. \]

(b) We see that (for a given \(\lambda \)) \(a_0\) determines all coefficients. In this way we obtain one linearly independent solutions (by taking \(a_0=1\)). We do not obtain two linearly independent solutions (for any \(\lambda \in \mC \)). □

Solution. We recall the recurrence relation obtained in Problem E.5:

\begin{equation} \label {eq:LaguerreRecurrence}\tag {$\dagger $} a_{k+1}=\frac {k+\lambda }{(k+1)^2}a_k. \end{equation}

(a) When \(\lambda =-n\) for some \(n\in \mN _0\), then from (\(\dagger \)) we obtain that \(a_{n+1}=0\), so that using (\(\dagger \)) again we have \(a_{n+2}=a_{n+3}=\ldots =0\). Hence the corresponding solution is a polynomial of degree \(n\). When \(\lambda \) is not of the above form, then the sequence \((a_k)\) has infinitely many nonzero terms and therefore the power series does not reduce to a polynomial.

(b) Polynomials are continuously differentiable (on \(\mR \) and therefore in particular on \([0,\infty )\phantom {]}\!\!\)). It follows from Problem D.6 that the boundary condition obtained in Problem D.6 is satisfied by these polynomial solutions. □

Solution. (i) We substitute the power series

\[ u(x)=\sum _{k=0}^\infty a_kx^k, \]

into \(xu''+(\alpha +1-x)u'=\eta u\). This gives

\[ x\sum _{k=0}^\infty k(k-1)a_kx^{k-2}+(\alpha +1-x)\sum _{k=0}^\infty ka_kx^{k-1} +\eta \sum _{k=0}^\infty a_kx^k=0. \]

Re-writing and shifting indices gives

\[ \sum _{k=0}^\infty \left [(k+1)(k+\alpha +1)a_{k+1}-\left (k-\eta \right )a_k\right ]x^k=0. \]

This gives the following relation for the coefficients:

\begin{equation} \label {eq:LaguerreassRecurrence}\tag {$\ast $} a_{k+1}=\frac {k-\eta }{(k+1)(k+\alpha +1)}a_k. \end{equation}

We note that (for a given \(\eta \)), \(a_0\) determines all the other coefficients uniquely and therefore we only obtain one linearly independent solution.

(ii) When \(\eta =n\) for some \(n\in \mN _0\), then from (\(\ast \)) we obtain that \(a_{n+1}=0\), so that using (\(\ast \)) again we have \(a_{n+2}=a_{n+3}=\ldots =0\). Hence the corresponding solution is a polynomial of degree \(n\).

(iii) We check that the coefficients

\[ a_k=\frac {(-1)^k}{k!(n-k)!\Gamma (\alpha +k+1)}, \]

satisfy (\(\ast \)) with \(\eta =n\). We have (using properties of factorials and the Gamma-function):

\begin{multline*} a_{k+1}=\frac {(-1)^{k+1}}{(k+1)!(n-k-1)!\Gamma (\alpha +k+2)} =\frac {-(-1)^k}{(k+1)k!\frac {(n-k)!}{n-k}(\alpha +k+1)\Gamma (\alpha +k+1)} \\ =\frac {-(-1)^k}{\frac {(k+1)(\alpha +k+1)}{n-k}~k!(n-k)!\Gamma (\alpha +k+1)} =-\frac {n-k}{(k+1)(\alpha +k+1)}~\frac {(-1)^k}{k!(n-k)!\Gamma (\alpha +k+1)} \\ =\frac {k-n}{(k+1)(\alpha +k+1)}~a_k, \end{multline*} as desired.

Multiplying all coefficients by \(\Gamma (\alpha +n+1)\) retains the relation (\(\ast \)). The leading coefficient is

\[ \Gamma (\alpha +n+1) \frac {(-1)^n}{n!(n-n)!\Gamma (\alpha +n+1)}=\frac {(-1)^n}{n!}, \]

as claimed. □

Solution. (a) We make the Ansatz

\[ u(x)=\sum _{k=0}^\infty a_kx^k,\qquad u'(x)=\sum _{k=0}^\infty a_kkx^{k-1},\qquad u''(x)=\sum _{k=0}^\infty a_kk(k-1)x^{k-2}. \]

Substituting this in the differential equation gives

\[ \sum _{k=0}^\infty k(k-1)a_kx^{k-2}-2x\sum _{k=0}^\infty ka_kx^{k-1}-\lambda \sum _{k=0}^\infty a_kx^k=0. \]

We rewrite this as

\[ \sum _{k=0}^\infty k(k-1)a_kx^{k-2}-\sum _{k=0}^\infty 2ka_kx^{k}-\lambda \sum _{k=0}^\infty a_kx^k=0. \]

To make the exponents line up, we shift the index in the first term: we define \(n:=k-2\) so that \(k=n+2\) and we obtain (noting that we can sum from \(n=0\) since the terms for \(n\in \{-2,-1\}\) equal zero)

\[ \sum _{n=0}^\infty (n+2)(n+1)a_{n+2}x^n. \]

Changing the name for the index back to \(k\), we obtain for the differential equation

\[ \sum _{k=0}^\infty (k+2)(k+1)a_{k+2}x^k-\sum _{k=0}^\infty 2ka_kx^{k}-\lambda \sum _{k=0}^\infty a_kx^k=0, \]

which gives

\[ \sum _{k=0}^\infty \left [ (k+2)(k+1)a_{k+2}-2ka_k-\lambda a_k \right ]x^k=0, \]

which gives (a power series is the zero function if and only if all coefficients are zero):

\[ (k+2)(k+1)a_{k+2}-2ka_k-\lambda a_k=0. \]

This gives the following relation for the coefficients:

\[ a_{k+2}=\frac {2k+\lambda }{(k+2)(k+1)}a_k. \]

(b) We see that (for a given \(\lambda \)) \(a_0\) determines all even coefficients and \(a_1\) determines all odd coefficients. In this way we obtain two linearly independent solutions (one by taking \(a_0=1\) and \(a_1=0\) which gives an even function and another by taking \(a_0=0\) and \(a_1=1\) which gives an odd function). □

Solution. We recall the recurrence relation obtained in Problem E.8:

\begin{equation} \label {eq:HermiteRecurrence}\tag {$\dagger $} a_{k+2}=\frac {2k+\lambda }{(k+2)(k+1)}a_k. \end{equation}

When \(\lambda =-2n\) for some \(n\in \mN _0\), then from (\(\dagger \)) we obtain that we obtain that \(a_{n+2}=0\), so that using (\(\dagger \)) again we have \(a_{n+4}=a_{n+6}=\ldots =0\). Hence the corresponding solution (if \(n\) is even the even solution and if \(n\) is odd the odd solution) is a polynomial of degree \(n\). When \(\lambda \) is not of the above form, then the sequence \((a_k)\) has infinitely many nonzero terms and therefore the power series does not reduce to a polynomial. □

Solution.

We make the Ansatz

\[ u(x)=x^\sigma \sum _{k=0}^\infty a_kx^k=\sum _{k=0}^\infty a_kx^{k+\sigma }. \]

We differentiate this term-by-term to obtain

\[ u'(x)=\sum _{k=0}^\infty (k+\sigma )a_kx^{k+\sigma -1},\qquad u''(x)=\sum _{k=0}^\infty (k+\sigma )(k+\sigma -1)a_kx^{k+\sigma -2}. \]

Substituting this into the given differential equation gives

\[ x^2\sum _{k=0}^\infty (k+\sigma )(k+\sigma -1)a_kx^{k+\sigma -2} +2x\sum _{k=0}^\infty (k+\sigma )a_kx^{k+\sigma -1} +\left (x^2-\sigma (\sigma +1)\right )\sum _{k=0}^\infty a_kx^{k+\sigma }=0, \]

which we re-write as

\[ \sum _{k=0}^\infty (k+\sigma )(k+\sigma -1)a_kx^{k+\sigma } +\sum _{k=0}^\infty 2(k+\sigma )a_kx^{k+\sigma } +\sum _{k=0}^\infty a_kx^{k+\sigma +2} -\sum _{k=0}^\infty \sigma (\sigma +1)a_kx^{k+\sigma }=0. \]

We shift the index in the third term by defining \(n:=k+2\) (so that \(k=n-2\)) to obtain for this third term

\[ \sum _{n=2}^\infty a_{n-2}x^{n+\sigma }. \]

We then obtain

\[ \sum _{k=0}^\infty k(k+2\sigma +1)a_kx^{k+\sigma } +\sum _{k=2}^\infty a_{k-2}x^{k+\sigma }=0. \]

We can now factor out \(x^\sigma \) and obtain

\[ \sum _{k=0}^\infty k(k+2\sigma +1)a_kx^{k} +\sum _{k=2}^\infty a_{k-2}x^{k}=0. \]

We can write this as

\[ \sum _{k=0}^\infty b_k x^k=0,\qquad b_k=\begin {cases} k(k+2\sigma +1)a_k&k\in \{0,1\}\\ k(k+2\sigma +1)a_k+a_{k-2}&k\geq 2. \end {cases} \]

We must then have that \(b_k=0\) for all \(k\). For \(k=0\) this gives \(0=0\), for \(k=1\) this gives \((2+2\sigma )a_1=0\), which gives \(a_1=0\). For \(k\geq 2\) we have the following relation between the coefficients

\[ k(k+2\sigma +1)a_k+a_{k-2}=0, \]

which gives

\[ a_k=\frac {-1}{k(k+2\sigma +1)}a_{k-2},\qquad k\geq 2. \]

Since \(a_1=0\) we obtain that \(a_{2k+1}=0\) for all \(k\) and we have that \(a_0\) uniquely determines \(a_{2k}\) for all \(k\). □

Solution. The recurrence relation when \(\sigma =0\) is

\[ a_k=\frac {-1}{k(k+1)}a_{k-2},\qquad k\geq 2,\qquad a_1=0,\qquad a_0=1. \]

The solution of this is

\[ a_{2n}=\frac {(-1)^n}{(2n+1)!},\qquad a_{2n+1}=0, \]

since we have with \(k=2n\) (so that \(k-2=2(n-1)\))

\begin{multline*} a_k=a_{2n}=\frac {(-1)^n}{(2n+1)!}a_0=\frac {-1}{(2n+1)2n}~\frac {(-1)^{n-1}}{(2n-1)!}a_0 =\frac {-1}{(2n+1)2n}~\frac {(-1)^{n-1}}{(2(n-1)+1)!}a_0 \\ =\frac {-1}{(2n+1)2n}~a_{2(n-1)} =\frac {-1}{(k+1)k}a_{k-2}. \end{multline*} We have

\[ \sin (x)=\sum _{k=0}^\infty \frac {(-1)^k}{(2k+1)!}x^{2k+1}, \]

so that

\[ \frac {\sin (x)}{x}=\sum _{k=0}^\infty \frac {(-1)^k}{(2k+1)!}x^{2k}, \]

which is indeed the series \(x^\sigma \sum _{k=0}^\infty a_kx^k\) with \(\sigma =0\) and \((a_k)\) as obtained above. □

Solution. We have

\[ u'(x)=\eta u_1'(x),\qquad u''(x)=\eta ^2 u_1''(x). \]

Hence

\[ x^2u''(x)+2xu'(x)+\left (\eta ^2x^2-\sigma (\sigma +1)\right )u(x) =x^2\eta ^2u_1''(\eta x)+2x\eta u_1'(\eta x)+\left (\eta ^2x^2-\sigma (\sigma +1)\right )u_1(\eta x). \]

With \(z:=\eta x\), the right-hand side is

\[ z^2u_1''(z)+2zu_1'(z)+\left (z^2-\sigma (\sigma +1)\right )u_1(z)=0, \]

where the last equality follows since \(u_1\) is a solution of the parametric form of the spherical Bessel equation with \(\eta =1\). □