The science of physics is built fundamentally upon differential equations. Many of the most useful differential equations appearing in physics are of second order, so that understanding the solutions that can arise and their interconnections has much to say about possible systems amenable to physical analysis. To this end consider a general second-order linear differential equation with polynomial coefficients,

${A}_{2}\left(z\right)\frac{{d}^{2}F}{d{z}^{2}}+{A}_{1}\left(z\right)\frac{dF}{dz}+{A}_{0}\left(z\right)F=0\phantom{\rule{5em}{0ex}}{A}_{i}=\sum _{k}{a}_{ik}{z}^{k}$

where the first index on each constant *a _{ij}* indicates which derivative it accompanies and the second the power of the independent variable. Since one can always divide through by any one of these constants, the solution has one less independent constant than those given.

The ideal general solution to this equation a function of all of the constant coefficients *a _{ij}* and the independent variable. Varying the coefficients would in principle allow one to see the change in the solution on a continuous analytic basis.

This presentation will seek general solutions to the equation for linear and quadratic coefficients using elementary transformations, with the intent of finding connections among the solutions. The techniques described here are reminiscent of solutions to differential equations in physics texts, which are typically broken into factors depending on how each of these affects the form of the differential equation. The presentation tries to make the nuts and bolts of the changes as clear as possible, without reference to any particular physical interpretation of the factors in the solution.

The simplest second-order equation is one with constant coefficients:

${a}_{20}\frac{{d}^{2}F}{d{z}^{2}}+{a}_{10}\frac{dF}{dz}+{a}_{00}F=0$

The well-known solutions to this equation are exponentials

${a}_{20}{\lambda}^{2}+{a}_{10}\lambda +{a}_{00}=0\phantom{\rule{4em}{0ex}}{\lambda}_{\pm}=\frac{-{a}_{10}\pm \sqrt{{a}_{10}^{2}-4{a}_{20}{a}_{00}}}{2{a}_{20}}$

If the discriminant under the square root is nonzero, the general solution is

$F\left(z\right)={c}_{1}exp\left[{\lambda}_{+}z\right]+{c}_{2}exp\left[{\lambda}_{-}z\right]$

and if the discriminant is zero the general solution is

$F\left(z\right)={c}_{1}exp\left[\lambda z\right]+{c}_{2}zexp\left[\lambda z\right]\phantom{\rule{3em}{0ex}}\lambda =-\frac{{a}_{10}}{2{a}_{20}}$

The next simplest second-order equation is the Euler equation

${a}_{22}{z}^{2}\frac{{d}^{2}F}{d{z}^{2}}+{a}_{11}z\frac{dF}{dz}+{a}_{00}F=0$

whose solutions are powers

${a}_{22}\lambda (\lambda -1)+{a}_{11}\lambda +{a}_{00}=0\phantom{\rule{3em}{0ex}}{\lambda}_{\pm}=\frac{{a}_{22}-{a}_{11}\pm \sqrt{({a}_{22}-{a}_{11}{)}^{2}-4{a}_{22}{a}_{00}}}{2{a}_{22}}$

If the discriminant under the square root is nonzero, the general solution is

$F\left(z\right)={c}_{1}{z}^{{\lambda}_{+}}+{c}_{2}{z}^{{\lambda}_{-}}$

and if the discriminant is zero the general solution is

$F\left(z\right)={c}_{1}{z}^{\lambda}+{c}_{2}{z}^{\lambda}lnz\phantom{\rule{5em}{0ex}}\lambda =\frac{{a}_{22}-{a}_{11}}{2{a}_{22}}$

These two simple solutions will serve as a point of reference for more complicated solutions.

There are several basic transformations that will be used to manipulate the various cases of the differential equation. For example, the coefficient of the first term can be moved onto the other terms by splitting off an exponential. Letting

${A}_{2}\frac{{d}^{2}f}{d{z}^{2}}+[{A}_{1}+2k{A}_{2}]\frac{df}{dz}+[{A}_{0}+k{A}_{1}+{k}^{2}{A}_{2}]f=0$

Similarly, the coefficient of the first term can be moved onto the other terms with additional factors of the independent variable by letting

${A}_{2}\frac{{d}^{2}f}{d{z}^{2}}+[{A}_{1}+4kz{A}_{2}]\frac{df}{dz}+[{A}_{0}+2kz{A}_{1}+(2k+4{k}^{2}{z}^{2}\left){A}_{2}\right]f=0$

The coefficient of the first term can be moved onto the other terms with additional inverse factors of the independent variable by letting

${A}_{2}\frac{{d}^{2}f}{d{z}^{2}}+[{A}_{1}+\frac{2k}{z}{A}_{2}]\frac{df}{dz}+[{A}_{0}+\frac{k}{z}{A}_{1}+\frac{k(k-1)}{{z}^{2}}{A}_{2}]f=0$

The relative powers of the independent variable on the first two terms of the equation can be adjusted by a change of variable

$\begin{array}{l}\frac{d}{dz}=k{z}^{k-1}\frac{d}{du}=k{u}^{(k-1)/k}\frac{d}{du}\\ \\ \frac{{d}^{2}}{d{z}^{2}}={k}^{2}{z}^{2(k-1)}\frac{{d}^{2}}{d{u}^{2}}+k(k-1){z}^{k-2}\frac{d}{du}\\ \phantom{\frac{{d}^{2}}{d{z}^{2}}}={k}^{2}{u}^{2(k-1)/k}\frac{{d}^{2}}{d{u}^{2}}+k(k-1){u}^{(k-2)/k}\frac{d}{du}\end{array}$

Useful cases of this transformation and the resultant differential equation include a square root variable

$\frac{{A}_{2}\left(u\right)}{4{u}^{2}}\frac{{d}^{2}F}{d{u}^{2}}+[\frac{{A}_{1}\left(u\right)}{2u}-\frac{{A}_{2}\left(u\right)}{4{u}^{3}}]\frac{dF}{du}+{A}_{0}\left(u\right)F=0$

a squared variable

$4u{A}_{2}\left(u\right)\frac{{d}^{2}F}{d{u}^{2}}+[2\sqrt{u}{A}_{1}\left(u\right)+2{A}_{2}\left(u\right)]\frac{dF}{du}+{A}_{0}\left(u\right)F=0$

and the fractional exponent

$\frac{9{u}^{2/3}}{4}{A}_{2}\left(u\right)\frac{{d}^{2}F}{d{u}^{2}}+[\frac{3{u}^{1/3}}{2}{A}_{1}\left(u\right)+\frac{3{A}_{2}\left(u\right)}{4{u}^{1/3}}]\frac{dF}{du}+{A}_{0}\left(u\right)F=0$

Any single constant in the linear equation can be altered by scaling the independent variable by a constant

${k}^{2}{A}_{2}\left(u\right)\frac{{d}^{2}F}{d{u}^{2}}+k{A}_{1}\left(u\right)\frac{dF}{du}+{A}_{0}\left(u\right)F=0$

and choosing the scale factor to achieve a desired value.

For purposes of comparison, solutions to the differential equation will be cast as much as possible in the form of confluent and Gauss hypergeometric functions. The differential equation for a confluent hypergeometric function is

$z\phantom{\rule{.2em}{0ex}}\frac{{d}^{2}F}{d{z}^{2}}+(c-z)\frac{dF}{dz}-aF=0$

Two linearly independent solutions to this equation are

${}_{1}F_{1}(a,c;z)\phantom{\rule{4em}{0ex}}{z}^{1-c}{}_{1}F_{1}(a-c+1,2-c;z)$

With the substitution $F\left(z\right)={e}^{z}f\left(z\right)$ the equation above becomes

$z\phantom{\rule{.2em}{0ex}}\frac{{d}^{2}f}{d{z}^{2}}+(c+z)\frac{df}{dz}+(c-a)f=0$

The solution of this last equation is the basis for an identity that allows one to change the sign of the independent variable in this function:

${}_{1}F_{1}(b,c;-z)={e}^{-z}{}_{1}F_{1}(c-b,c;z)$

The differential equation of the Gauss hypergeometric functions is

$z(1-z)\frac{{d}^{2}F}{d{z}^{2}}+[c-(a+b+1\left)z\right]\frac{dF}{dz}-abF=0$

Two linearly independent solutions to this equation are

${}_{2}F_{1}(a,b;c;z)\phantom{\rule{4em}{0ex}}{z}^{1-c}{}_{2}F_{1}(a-c+1,b-c+1;2-c;z)$

The presence of a constant coefficient on the linear part of the second term simplifies identification of solutions to some extent. The Gauss hypergeometric function is essentially different from the confluent hypergeometric function, so that solutions in terms of the former are distinct cases from solutions in terms of the latter.

First consider an equation that has a linear coefficient on the first term only:

$({a}_{20}+{a}_{21}z)\frac{{d}^{2}F}{d{z}^{2}}+{a}_{10}\frac{dF}{dz}+{a}_{00}F=0$

Treating the linear coefficient as a single variable, it could be moved onto the second term by splitting off an exponential, but this will also move it onto the third term. If instead a change of variable to a square root is applied, the resulting equation would have linear coefficients in the radical on both the first and third terms. Splitting off an exponential would then move this coefficient onto the second and third terms, on the last of which it could be removed by an appropriate choice of the accompanying constants. This means the solution to the equation is of the form

$F\left(u\right)={e}^{ku}f\left(u\right)\phantom{\rule{5em}{0ex}}u=m\sqrt{{a}_{20}+{a}_{21}z}$

Substituting this in the differential equation produces

$u\frac{{d}^{2}f}{d{u}^{2}}+[\frac{2{a}_{10}}{{a}_{21}}-1+2ku]\frac{df}{du}+\left[k\right(\frac{2{a}_{10}}{{a}_{21}}-1)+({k}^{2}+\frac{4{a}_{00}}{{m}^{2}{a}_{21}^{2}}\left)u\right]f=0$

The variable in the third term can now be removed with appropriate choices of constants. The multiplier of the independent variable will be kept real for comparison across cases, which means that the constant in the exponential must be imaginary. Choosing

$\begin{array}{c}F\left[u\right(z\left)\right]={c}_{1}{e}^{-iu/2}{}_{1}F_{1}[\frac{\lambda}{2}\phantom{\rule{.2em}{0ex}},\phantom{\rule{.2em}{0ex}}\lambda \phantom{\rule{.2em}{0ex}};iu]+{c}_{2}{e}^{-iu/2}{u}^{1-\lambda}{}_{1}F_{1}[1-\frac{\lambda}{2}\phantom{\rule{.2em}{0ex}},\phantom{\rule{.2em}{0ex}}2-\lambda \phantom{\rule{.2em}{0ex}};iu]\\ \\ u\left(z\right)=\frac{4\sqrt{{a}_{00}}}{{a}_{21}}\sqrt{{a}_{20}+{a}_{21}z}\phantom{\rule{4em}{0ex}}\lambda =\frac{2{a}_{10}}{{a}_{21}}-1\end{array}$

The presence of the constant *a*_{21} in the denominator indicates that the solution to this equation is essentially different from the equation with constant coefficients, since this constant cannot simply be set equal to zero. These solutions are equivalent to Bessel functions of order
$\pm \frac{\lambda -1}{2}$
multiplied by powers of the independent variable, but will be left in this more general form for ease of comparison.

Now consider an equation that has a linear coefficient on the second term only:

${a}_{20}\frac{{d}^{2}F}{d{z}^{2}}+({a}_{10}+{a}_{11}z)\frac{dF}{dz}+{a}_{00}F=0$

This equation can almost be transformed to the confluent hypergeometric equation by changing to a squared variable of the linear coefficient. The equation after this change is that for a confluent hypergeometric function of negative argument, and by the identity above this can be dealt with by including a negative exponential. For a solution of the form

$F\left(u\right)={e}^{-u}f\left(u\right)\phantom{\rule{5em}{0ex}}u=m({a}_{10}+{a}_{11}z{)}^{2}$

the equation becomes

$\begin{array}{l}u\frac{{d}^{2}f}{d{u}^{2}}+[\frac{1}{2}+(\frac{1}{2{a}_{20}{a}_{11}m}-2\left)u\right]\frac{df}{du}\\ \phantom{\rule{5em}{0ex}}-[\frac{1}{2}-\frac{{a}_{00}}{4{a}_{20}{a}_{11}^{2}m}+(\frac{1}{2{a}_{20}{a}_{11}m}-1\left)u\right]f=0\end{array}$

With the appropriate choice for *m* the general solution is

$\begin{array}{c}F\left[u\right(z\left)\right]={c}_{1}{e}^{-u}{}_{1}F_{1}[-\frac{\lambda}{2}\phantom{\rule{.2em}{0ex}},\phantom{\rule{.2em}{0ex}}\frac{1}{2}\phantom{\rule{.2em}{0ex}};u]+{c}_{2}{e}^{-u}{u}^{1/2}{}_{1}F_{1}[-\frac{\lambda}{2}+\frac{1}{2}\phantom{\rule{.2em}{0ex}},\phantom{\rule{.2em}{0ex}}\frac{3}{2}\phantom{\rule{.2em}{0ex}};u]\\ \\ u\left(z\right)=\frac{1}{2{a}_{20}{a}_{11}}({a}_{10}+{a}_{11}z{)}^{2}\phantom{\rule{4em}{0ex}}\lambda =\frac{{a}_{00}}{{a}_{11}}-1\end{array}$

These solutions become Hermite polynomials multiplied by the exponential when λ is an integer. In this case the constant *a*_{11} cannot simply be set equal to zero, indicating that its presence distinguishes this solution from other cases.

Next consider an equation that has a linear coefficient on the third term only:

${a}_{20}\frac{{d}^{2}F}{d{z}^{2}}+{a}_{10}\frac{dF}{dz}+({a}_{00}+{a}_{01}z)F=0$

Treating the linear coefficient as a single variable, changing to a variable with fractional exponent of three halves almost produces a simple equation with linear terms on the first and third terms. The coefficient of the second term would, however, remain complicated, but that is due to the presence of the original constant coefficient. This indicates a solution in two steps: first split off an exponential of the linear coefficient to remove the original constant on the second term, then attack the remainder of the equation. For a solution of the form

${a}_{20}\frac{{d}^{2}f}{d{z}^{2}}+({a}_{10}-2k{a}_{20})\frac{df}{dz}+({a}_{00}-k{a}_{10}+{k}^{2}{a}_{20}+{a}_{01}z)f=0$

and with the choice $k=\frac{{a}_{10}}{2{a}_{20}}$ becomes

${a}_{20}\frac{{d}^{2}f}{d{z}^{2}}+({a}_{00}-\frac{{a}_{10}^{2}}{4{a}_{20}}+{a}_{01}z)f=0$

This modified linear coefficient should now be used in the change of variable to a fractional exponent. After the change, the linear coefficients can be moved and adjusted with an exponential of the modified linear coefficient. For a solution of the form

$f\left(u\right)={e}^{lu}g\left(u\right)\phantom{\rule{5em}{0ex}}u=m({a}_{00}-\frac{{a}_{10}^{2}}{4{a}_{20}}+{a}_{01}z{)}^{\frac{3}{2}}$

the modified differential equation becomes

$u\frac{{d}^{2}g}{d{u}^{2}}+[\frac{1}{3}+2lu]\frac{dg}{du}+[\frac{l}{3}+({l}^{2}+\frac{4}{9{a}_{20}{a}_{01}^{2}{m}^{2}}\left)u\right]g=0$

Choosing

$\begin{array}{c}F\left[u\right(z\left)\right]={c}_{1}{e}^{-iu/2-\lambda z}{}_{1}F_{1}[\frac{1}{6}\phantom{\rule{.2em}{0ex}},\phantom{\rule{.2em}{0ex}}\frac{1}{3}\phantom{\rule{.2em}{0ex}};iu]+{c}_{2}{e}^{-iu/2-\lambda z}{u}^{2/3}{}_{1}F_{1}[\frac{5}{6}\phantom{\rule{.2em}{0ex}},\phantom{\rule{.2em}{0ex}}\frac{5}{3}\phantom{\rule{.2em}{0ex}};iu]\\ \\ u\left(z\right)=\frac{4}{3{a}_{01}\sqrt{{a}_{20}}}({a}_{00}-\frac{{a}_{10}^{2}}{4{a}_{20}}+{a}_{01}z{)}^{\frac{3}{2}}\phantom{\rule{4em}{0ex}}\lambda =\frac{{a}_{10}}{2{a}_{20}}\end{array}$

These solutions are proportional to Bessel functions of order one third multiplied by exponentials. Bessel functions of one-third order more frequently go by the name Airy functions. This result is not surprising, as the original differential equation is just the Airy differential equation with an additional first-derivative term.

Note that the exponentials in these solutions contain two functions of the independent variable. Also note that for this case both *a*_{01} and *a*_{20} appear in denominators, and cannot simply be set equal to zero.

Now on to equations with pairs of linear coefficients. First consider an equation that has linear coefficients on both the first and second terms:

$({a}_{20}+{a}_{21}z)\frac{{d}^{2}F}{d{z}^{2}}+({a}_{10}+{a}_{11}z)\frac{dF}{dz}+{a}_{00}F=0$

This equation is already recognizable as the confluent hypergeometric equation, but of negative argument that requires an exponential factor. For a solution of the form

$F\left(u\right)={e}^{-u}f\left(u\right)\phantom{\rule{5em}{0ex}}u=m({a}_{20}+{a}_{21}z)$

the equation becomes

$\begin{array}{l}u\frac{{d}^{2}F}{d{u}^{2}}+[\frac{{a}_{21}{a}_{10}-{a}_{20}{a}_{11}}{{a}_{21}^{2}}+(\frac{{a}_{11}}{m{a}_{21}^{2}}-2\left)u\right]\frac{dF}{du}\\ \phantom{\rule{5em}{0ex}}-[\frac{{a}_{21}{a}_{10}-{a}_{20}{a}_{11}}{{a}_{21}^{2}}-\frac{{a}_{00}}{m{a}_{21}^{2}}+(\frac{{a}_{11}}{m{a}_{21}^{2}}-1\left)u\right]F=0\end{array}$

With the appropriate choice for *m* the general solution is

$\begin{array}{c}F\left[u\right(z\left)\right]={c}_{1}{e}^{-u}{}_{1}F_{1}[\lambda -\frac{{a}_{00}}{{a}_{11}}\phantom{\rule{.2em}{0ex}},\phantom{\rule{.2em}{0ex}}\lambda \phantom{\rule{.2em}{0ex}};u]+{c}_{2}{e}^{-u}{u}^{1-\lambda}{}_{1}F_{1}[1-\frac{{a}_{00}}{{a}_{11}}\phantom{\rule{.2em}{0ex}},\phantom{\rule{.2em}{0ex}}2-\lambda \phantom{\rule{.2em}{0ex}};u]\\ \\ u\left(z\right)=\frac{{a}_{11}}{{a}_{21}^{2}}({a}_{20}+{a}_{21}z)\phantom{\rule{4em}{0ex}}\lambda =\frac{{a}_{21}{a}_{10}-{a}_{20}{a}_{11}}{{a}_{21}^{2}}\end{array}$

Note that both *a*_{21} and *a*_{11} appear in denominators, and so cannot simply be set equal to zero.

Now consider an equation that has linear coefficients on both the first and third terms:

$({a}_{20}+{a}_{21}z)\frac{{d}^{2}F}{d{z}^{2}}+{a}_{10}\frac{dF}{dz}+({a}_{00}+{a}_{01}z)F=0$

This equation is similar to the last, but the exponential must be allowed an undetermined constant. For a solution of the form

$F\left(u\right)={e}^{ku}f\left(u\right)\phantom{\rule{5em}{0ex}}u=m({a}_{20}+{a}_{21}z)$

the equation becomes

$u\frac{{d}^{2}F}{d{u}^{2}}+[\frac{{a}_{10}}{{a}_{21}}+2ku]\frac{dF}{du}+[\frac{{a}_{21}{a}_{00}-{a}_{20}{a}_{01}}{m{a}_{21}^{3}}+k\frac{{a}_{10}}{{a}_{21}}+({k}^{2}+\frac{{a}_{01}}{{m}^{2}{a}_{21}^{3}}\left)u\right]F=0$

Choosing

$\begin{array}{c}F\left[u\right(z\left)\right]={c}_{1}{e}^{-iu/2}{}_{1}F_{1}[\frac{\lambda}{2}+i\alpha \phantom{\rule{.2em}{0ex}},\phantom{\rule{.2em}{0ex}}\lambda \phantom{\rule{.2em}{0ex}};iu]+{c}_{2}{e}^{-iu/2}{u}^{1-\lambda}{}_{1}F_{1}[1-\frac{\lambda}{2}+i\alpha \phantom{\rule{.2em}{0ex}},\phantom{\rule{.2em}{0ex}}2-\lambda \phantom{\rule{.2em}{0ex}};iu]\\ \\ u\left(z\right)=2\sqrt{\frac{{a}_{01}}{{a}_{21}^{3}}}({a}_{20}+{a}_{21}z)\phantom{\rule{3em}{0ex}}\lambda =\frac{{a}_{10}}{{a}_{21}}\phantom{\rule{3em}{0ex}}\alpha =\frac{{a}_{21}{a}_{00}-{a}_{20}{a}_{01}}{2\sqrt{{a}_{01}{a}_{21}^{3}}}\end{array}$

Note that in this case both *a*_{21} and *a*_{01} appear in denominators, and so cannot simply be set equal to zero.

The solutions for these last two cases look very similar, and as well they should since they are essentially the same equation with solutions differing by an exponential factor. Write the two equations together,

$\begin{array}{l}({a}_{20}+{a}_{21}z)\frac{{d}^{2}F}{d{z}^{2}}+({a}_{10}+{a}_{11}z)\frac{dF}{dz}+{a}_{00}F=0\\ ({b}_{20}+{b}_{21}z)\frac{{d}^{2}F}{d{z}^{2}}+{b}_{10}\frac{dF}{dz}+({b}_{00}+{b}_{01}z)F=0\end{array}$

and split off an exponential in each:

$\begin{array}{l}({a}_{20}+{a}_{21}z)\frac{{d}^{2}f}{d{z}^{2}}+[{a}_{10}+2{k}_{a}{a}_{20}+({a}_{11}+2{k}_{a}{a}_{21}\left)z\right]\frac{df}{dz}\\ \phantom{\rule{3.5em}{0ex}}+[{a}_{00}+{k}_{a}{a}_{10}+{k}_{a}^{2}{a}_{20}+({k}_{a}{a}_{11}+{k}_{a}^{2}{a}_{21}\left)z\right]f=0\\ ({b}_{20}+{b}_{21}z)\frac{{d}^{2}f}{d{z}^{2}}+[{b}_{10}+2{k}_{b}{b}_{20}+2{k}_{b}{b}_{21}z]\frac{df}{dz}\\ \phantom{\rule{4em}{0ex}}+[{b}_{00}+{k}_{b}{b}_{10}+{k}_{b}^{2}{b}_{20}+({b}_{01}+{k}_{b}^{2}{b}_{21}\left)z\right]f=0\end{array}$

Each equation can be turned into the other by choosing

${k}_{a}=-\frac{{a}_{11}}{2{a}_{21}}\phantom{\rule{5em}{0ex}}{k}_{b}=-i\phantom{\rule{.3em}{0ex}}\sqrt{\frac{{b}_{01}}{{b}_{21}}}$

These values correspond to exponentials in *z* multiplying the appropriate general solutions, and the remaining differences in multiplicative constants follow from these choices. This also means that the equation with linear coefficients on all terms

$({a}_{20}+{a}_{21}z)\frac{{d}^{2}F}{d{z}^{2}}+({a}_{10}+{a}_{11}z)\frac{dF}{dz}+({a}_{00}+{a}_{01}z)F=0$

has a similar solution, and can be transformed into either of the previous two cases. For a solution of the form

$F\left(u\right)={e}^{ku}f\left(u\right)\phantom{\rule{5em}{0ex}}u=m({a}_{20}+{a}_{21}z)$

the equation becomes

$\begin{array}{l}u\frac{{d}^{2}F}{d{u}^{2}}+[\frac{{a}_{21}{a}_{10}-{a}_{20}{a}_{11}}{{a}_{21}^{2}}+(\frac{{a}_{11}}{m{a}_{21}^{2}}+2k\left)u\right]\frac{dF}{du}\\ \phantom{\rule{1.5em}{0ex}}+[\frac{{a}_{21}{a}_{00}-{a}_{20}{a}_{01}}{m{a}_{21}^{3}}+k\frac{{a}_{21}{a}_{10}-{a}_{20}{a}_{11}}{{a}_{21}^{2}}+({k}^{2}+k\frac{{a}_{11}}{m{a}_{21}^{2}}+\frac{{a}_{01}}{{m}^{2}{a}_{21}^{3}}\left)u\right]F=0\end{array}$

Choosing constants to bring the second term as close as possible to confluent hypergeometric form while also removing the variable in the third term means

$\frac{{a}_{11}}{m{a}_{21}^{2}}+2k=-1\phantom{\rule{4em}{0ex}}{k}^{2}+k\frac{{a}_{11}}{m{a}_{21}^{2}}+\frac{{a}_{01}}{{m}^{2}{a}_{21}^{3}}=0$

Choosing the positive sign for *m* in the solution to these two equations, the general solution of the differential equation is

$\begin{array}{c}F\left[u\right(z\left)\right]={c}_{1}{e}^{ku}{}_{1}F_{1}[\frac{\lambda}{2}+\alpha \phantom{\rule{.2em}{0ex}},\phantom{\rule{.2em}{0ex}}\lambda \phantom{\rule{.2em}{0ex}};iu]+{c}_{2}{e}^{ku}{u}^{1-\lambda}{}_{1}F_{1}[1-\frac{\lambda}{2}+\alpha \phantom{\rule{.2em}{0ex}},\phantom{\rule{.2em}{0ex}}2-\lambda \phantom{\rule{.2em}{0ex}};iu]\\ \\ u\left(z\right)=\frac{\sqrt{{a}_{11}^{2}-4{a}_{21}{a}_{01}}}{{a}_{21}^{2}}({a}_{20}+{a}_{21}z)\phantom{\rule{3em}{0ex}}k=-\frac{{a}_{11}}{2\sqrt{{a}_{11}^{2}-4{a}_{21}{a}_{01}}}-\frac{1}{2}\\ \lambda =\frac{{a}_{21}{a}_{10}-{a}_{20}{a}_{11}}{{a}_{21}^{2}}\phantom{\rule{3em}{0ex}}\alpha =\frac{{a}_{21}{a}_{11}\lambda -2({a}_{21}{a}_{00}-{a}_{20}{a}_{01})}{2{a}_{21}\sqrt{{a}_{11}^{2}-4{a}_{21}{a}_{01}}}\end{array}$

In this case either *a*_{11} or *a*_{01} can be set equal to zero to recover previous cases, but both cannot be set equal to zero simultaneously, and neither can *a*_{21} be set equal to zero.

The last case to consider is an equation that has linear coefficients on both the second and third terms:

${a}_{20}\frac{{d}^{2}F}{d{z}^{2}}+({a}_{10}+{a}_{11}z)\frac{dF}{dz}+({a}_{00}+{a}_{01}z)F=0$

This can be converted to a previous case and is then solved. For a solution of the form

$\begin{array}{l}{a}_{20}\frac{{d}^{2}f}{d{z}^{2}}+({a}_{10}-2k{a}_{20}+{a}_{11}z)\frac{df}{dz}\\ \phantom{\rule{3em}{0ex}}+[{a}_{00}-k{a}_{10}+{k}^{2}{a}_{20}+({a}_{01}-k{a}_{11}\left)z\right]f=0\end{array}$

and with the choice $k=\frac{{a}_{01}}{{a}_{11}}$ becomes

${a}_{20}\frac{{d}^{2}f}{d{z}^{2}}+[{a}_{10}-\frac{2{a}_{20}{a}_{01}}{{a}_{11}}+{a}_{11}z]\frac{df}{dz}+[{a}_{00}-\frac{{a}_{10}{a}_{01}}{{a}_{11}}+\frac{{a}_{20}{a}_{01}^{2}}{{a}_{11}^{2}}]f=0$

The equation is now a form solved above used a squared variable, but with different constants. The general solution is thus

$\begin{array}{c}F\left[u\right(z\left)\right]={c}_{1}{e}^{-u-kz}{}_{1}F_{1}[-\frac{\lambda}{2}\phantom{\rule{.2em}{0ex}},\phantom{\rule{.2em}{0ex}}\frac{1}{2}\phantom{\rule{.2em}{0ex}};u]+{c}_{2}{e}^{-u-kz}{u}^{1/2}{}_{1}F_{1}[-\frac{\lambda}{2}+\frac{1}{2}\phantom{\rule{.2em}{0ex}},\phantom{\rule{.2em}{0ex}}\frac{3}{2}\phantom{\rule{.2em}{0ex}};u]\\ \\ u\left(z\right)=\frac{1}{2{a}_{20}{a}_{11}}({a}_{10}-\frac{2{a}_{20}{a}_{01}}{{a}_{11}}+{a}_{11}z{)}^{2}\phantom{\rule{4em}{0ex}}k=\frac{{a}_{01}}{{a}_{11}}\\ \lambda =\frac{{a}_{00}}{{a}_{11}}-\frac{{a}_{10}{a}_{01}}{{a}_{11}^{2}}+\frac{{a}_{20}{a}_{01}^{2}}{{a}_{11}^{3}}-1\end{array}$

In this case the constant *a*_{01} can be set to zero to recover a previous case, but the constant *a*_{11} cannot be set equal to zero.

Consider cases of the differential equation with coefficients quadratic in the independent variable. For a more generalized form of the Gauss hypergeometric equation,

$({a}_{20}+{a}_{21}z+{a}_{22}{z}^{2})\frac{{d}^{2}F}{d{z}^{2}}+({a}_{10}+{a}_{11}z)\frac{dF}{dz}+{a}_{00}F=0$

first identify the roots of the quadratic coefficient on the first term:

${a}_{22}{r}^{2}+{a}_{21}r+{a}_{20}=0\phantom{\rule{3em}{0ex}}{r}_{\pm}=\frac{-{a}_{21}\pm \sqrt{{a}_{21}^{2}-4{a}_{22}{a}_{20}}}{2{a}_{22}}$

For a nonzero discriminant under the radical, the equation can be written

${a}_{22}(z-{r}_{+})(z-{r}_{-})\frac{{d}^{2}F}{d{z}^{2}}+({a}_{10}+{a}_{11}z)\frac{dF}{dz}+{a}_{00}F=0$

Now map the singularity at *r*_{+} to zero and the singularity at *r*_{−} to one with a linear fractional, or Möbius, transformation. A transformation that takes any two points *z*_{1} and *z*_{2} into any two points *w*_{1} and *w*_{2} is constructed by writing

$(z-{z}_{1})(w-{w}_{2})=(z-{z}_{2})(w-{w}_{1})$

since substituting the values for *z* produces the correct values for *w*. The transformation to be employed is

$w=\frac{z-{r}_{+}}{{r}_{-}-{r}_{+}}\phantom{\rule{4em}{0ex}}z=({r}_{-}-{r}_{+})w+{r}_{+}$

under which the differential equation becomes

$w(1-w)\frac{{d}^{2}F}{d{w}^{2}}+[\frac{{a}_{10}+{a}_{11}{r}_{+}}{{a}_{22}({r}_{+}-{r}_{-})}-\frac{{a}_{11}}{{a}_{22}}w]\frac{dF}{dw}-\frac{{a}_{00}}{{a}_{22}}F=0$

where the entire equation has been multiplied by negative one for conformity with the Gauss hypergeometric differential equation. The general solution is

$\begin{array}{c}F\left[w\right(z\left)\right]={c}_{1}{}_{2}F_{1}(a,b;c;w)+{c}_{2}{w}^{1-c}{}_{2}F_{1}(a-c+1,b-c+1;2-c;w)\\ \\ w\left(z\right)=\frac{z-{r}_{+}}{{r}_{-}-{r}_{+}}\phantom{\rule{5em}{0ex}}{r}_{\pm}=\frac{-{a}_{21}\pm \sqrt{{a}_{21}^{2}-4{a}_{22}{a}_{20}}}{2{a}_{22}}\\ a=\frac{{a}_{11}-{a}_{22}+\sqrt{({a}_{11}-{a}_{22}{)}^{2}-4{a}_{22}{a}_{00}}}{2{a}_{22}}\\ b=\frac{{a}_{11}-{a}_{22}-\sqrt{({a}_{11}-{a}_{22}{)}^{2}-4{a}_{22}{a}_{00}}}{2{a}_{22}}\\ c=\frac{{a}_{10}+{a}_{11}{r}_{+}}{{a}_{22}({r}_{+}-{r}_{-})}\end{array}$

Note that either of the two constants *a*_{21} or *a*_{11} can be set equal to zero without changing the nature of the solution, so that this form holds for any of the differential equations

$\begin{array}{l}({a}_{20}+{a}_{22}{z}^{2})\frac{{d}^{2}F}{d{z}^{2}}+({a}_{10}+{a}_{11}z)\frac{dF}{dz}+{a}_{00}F=0\\ ({a}_{20}+{a}_{21}z+{a}_{22}{z}^{2})\frac{{d}^{2}F}{d{z}^{2}}+{a}_{10}\frac{dF}{dz}+{a}_{00}F=0\\ ({a}_{20}+{a}_{21}z+{a}_{22}{z}^{2})\frac{{d}^{2}F}{d{z}^{2}}+{a}_{11}z\frac{dF}{dz}+{a}_{00}F=0\end{array}$

The constant *a*_{22} cannot be set equal to zero, and this is definitive for this form. Since the parameter *c* of the hypergeometric function occurs in denominators, only the second half of the general solution remains valid when setting both *a*_{21} and *a*_{11} equal to zero; the other linearly independent solution for this subcase can be written as a rather involved integral but will be omitted here. The constant *a*_{20} also cannot be set equal to zero, so that this solution is distinct from that of the Euler equation.

If the discriminant above is zero, so that the roots of the quadratic coefficient coincide, the equation is of the form

${a}_{22}(z+\frac{{a}_{21}}{2{a}_{22}}{)}^{2}\frac{{d}^{2}F}{d{z}^{2}}+({a}_{10}+{a}_{11}z)\frac{dF}{dz}+{a}_{00}F=0$

The solution to this equation is straightforward but not obvious. Transforming to an inverse variable leads to a quadratic coefficient on the first term and linear and quadratic coefficients on the second term. Splitting off a power of the inverse variable will then give constant and linear coefficients on the third term without changing the structure of the other coefficients. Removing the constant part of the third-term coefficient will leave an extra power of the independent variable that can be factored out of the entire equation, leaving the confluent hypergeometric equation.

With the inverse variable $w=(z+\frac{{a}_{21}}{2{a}_{22}}{)}^{-1}$ the equation becomes

${w}^{2}\frac{{d}^{2}F}{d{w}^{2}}+\left[\right(2-\frac{{a}_{11}}{{a}_{22}})w-\frac{{a}_{10}}{{a}_{22}}{w}^{2}]\frac{dF}{dw}+\frac{{a}_{00}}{{a}_{22}}F=0$

and with a solution of the form $F\left(w\right)={w}^{\lambda}f\left(w\right)$ one has

$\begin{array}{l}{w}^{2}\frac{{d}^{2}F}{d{w}^{2}}+\left[\right(2(\lambda +1)-\frac{{a}_{11}}{{a}_{22}})w-\frac{{a}_{10}}{{a}_{22}}{w}^{2}]\frac{dF}{dw}\\ \phantom{\rule{5em}{0ex}}+[\lambda (\lambda +1-\frac{{a}_{11}}{{a}_{22}})+\frac{{a}_{00}}{{a}_{22}}-k\frac{{a}_{10}}{{a}_{22}}w]F=0\end{array}$

An appropriate choice of λ will now remove the constant part of the coefficient on the third term. This leaves behind a confluent hypergeometric equation in terms of a scaled variable. The general solution can be written compactly as

$\begin{array}{c}F\left[w\right(z\left)\right]={c}_{1}{w}^{{k}_{+}}{}_{1}F_{1}({\lambda}_{+},{c}_{+};w)+{c}_{2}{w}^{{k}_{-}}{}_{1}F_{1}({\lambda}_{-},{c}_{-};w)\\ \\ w\left(z\right)=\frac{{a}_{10}}{{a}_{22}}(z+\frac{{a}_{21}}{2{a}_{22}}{)}^{-1}\\ {\lambda}_{\pm}=\frac{{a}_{11}-{a}_{22}\pm \sqrt{({a}_{11}-{a}_{22}{)}^{2}-4{a}_{22}{a}_{00}}}{2{a}_{22}}\\ {c}_{\pm}=1\pm \frac{\sqrt{({a}_{11}-{a}_{22}{)}^{2}-4{a}_{22}{a}_{00}}}{{a}_{22}}\end{array}$

Before considering the case of a quadratic coefficient on the second term, first consider a quadratic coefficient on the third term with a linear on the second term:

${a}_{20}\frac{{d}^{2}F}{d{z}^{2}}+({a}_{10}+{a}_{11}z)\frac{dF}{dz}+({a}_{00}+{a}_{01}z+{a}_{02}{z}^{2})F=0$

The coefficient of the first term can be brought onto the other terms by splitting off exponential, and in order to remove the quadratic part of the coefficient on the third term it should be an quadratic exponential. For a solution of the form $F\left(z\right)={e}^{-l{z}^{2}}f\left(z\right)$ the equation becomes

$\begin{array}{l}{a}_{20}\frac{{d}^{2}F}{d{z}^{2}}+[{a}_{10}+({a}_{11}-4l{a}_{20}\left)z\right]\frac{dF}{dz}\\ \phantom{\rule{2em}{0ex}}+[{a}_{00}-2l{a}_{20}+({a}_{01}-2l{a}_{10})z+({a}_{02}-2l{a}_{11}+4{l}^{2}{a}_{20}\left){z}^{2}\right]F=0\end{array}$

The appropriate choice of *l* removes the quadratic part of the third-term coefficient. The equation is thereby reduced to a case of the confluent hypergeometric equation solved above, but with three of the constants shifted. Taking the positive root in determining *l*, the general solution is

$\begin{array}{c}F\left[u\right(z\left)\right]={c}_{1}{e}^{-u-kz-l{z}^{2}}{}_{1}F_{1}[-\frac{\lambda}{2}\phantom{\rule{.2em}{0ex}},\phantom{\rule{.2em}{0ex}}\frac{1}{2}\phantom{\rule{.2em}{0ex}};u]+{c}_{2}{e}^{-u-kz-l{z}^{2}}{u}^{1/2}{}_{1}F_{1}[-\frac{\lambda}{2}+\frac{1}{2}\phantom{\rule{.2em}{0ex}},\phantom{\rule{.2em}{0ex}}\frac{3}{2}\phantom{\rule{.2em}{0ex}};u]\\ \\ u\left(z\right)=\frac{1}{2{a}_{20}\sqrt{{a}_{11}^{2}-4{a}_{20}{a}_{02}}}({a}_{10}-\frac{2{a}_{20}({a}_{01}-2l{a}_{10})}{\sqrt{{a}_{11}^{2}-4{a}_{20}{a}_{02}}}+\sqrt{{a}_{11}^{2}-4{a}_{20}{a}_{02}}\phantom{\rule{.3em}{0ex}}z{)}^{2}\\ k=\frac{{a}_{01}-2l{a}_{10}}{\sqrt{{a}_{11}^{2}-4{a}_{20}{a}_{02}}}\phantom{\rule{4em}{0ex}}l=\frac{{a}_{11}+\sqrt{{a}_{11}^{2}-4{a}_{20}{a}_{02}}}{4{a}_{20}}\\ \lambda =\frac{{a}_{00}-2l{a}_{20}}{\sqrt{{a}_{11}^{2}-4{a}_{20}{a}_{02}}}-\frac{{a}_{10}({a}_{01}-2l{a}_{10})}{{a}_{11}^{2}-4{a}_{20}{a}_{02}}+\frac{{a}_{20}({a}_{01}-2l{a}_{10}{)}^{2}}{({a}_{11}^{2}-4{a}_{20}{a}_{02}{)}^{3/2}}-1\end{array}$

The constants *a*_{10}, *a*_{11} and *a*_{01} can be set to zero individually or simultaneously without changing the structure of the solution, so that it describes equations of the type

$\begin{array}{l}{a}_{20}\frac{{d}^{2}F}{d{z}^{2}}+{a}_{10}\frac{dF}{dz}+({a}_{00}+{a}_{01}z+{a}_{02}{z}^{2})F=0\\ {a}_{20}\frac{{d}^{2}F}{d{z}^{2}}+{a}_{11}z\frac{dF}{dz}+({a}_{00}+{a}_{01}z+{a}_{02}{z}^{2})F=0\\ {a}_{20}\frac{{d}^{2}F}{d{z}^{2}}+({a}_{10}+{a}_{11}z)\frac{dF}{dz}+({a}_{00}+{a}_{02}{z}^{2})F=0\end{array}$

but the constants *a*_{20} and *a*_{02} cannot be set equal to zero. There is also a singularity for

The last distinct case with a single quadratic coefficient is when the coefficient is on the middle term of the equation:

${a}_{20}\frac{{d}^{2}F}{d{z}^{2}}+({a}_{10}+{a}_{11}z+{a}_{12}{z}^{2})\frac{dF}{dz}+{a}_{00}F=0$

This equation is a form of the triconfluent Heun equation; other forms of Heun equations will appear for combinations of quadratic and linear coefficients. Since Heun functions are more general than hypergeometric functions they cannot be written in terms of them, while hypergeometric functions can always be written in terms of Heun functions. This simple looking equation represents a hefty increase in complexity, and forms a natural ending point of seeking solutions through elementary transformations.

It is instructive at this point to summarize which cases have the same type of solution, apart from exponential multipliers. Using a capital “C” for a constant coefficient, “L” for a linear coefficient and “Q” for a quadratic coefficient, the cases considered fall into the following groupings:

{ C , C , C }

{ L , C , C }

{ C , L , C } , { C , L , L } , { C , C , Q } and { C , L , Q }

{ C , C , L }

{ L , L , C } , { L , C , L } and { L , L , L }

{ Q , C , C } and { Q , L , C }

The reason solutions falls into these groups is not at all obvious from what has been presented so far. It is also not clear why a variety of powers of the independent variable appear as the arguments of solutions. Both situations become understandable with a discussion of the singularities of the equations, which will be the subject of another presentation.

*Uploaded 2013.07.08 — Updated 2013.08.01*
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