diffop - Maple Help

Linear Differential Operators

In Maple, one can work with differential operators by using a subset of the DEtools package. Differential operators in this case are polynomial objects of the form



with the coefficients ${a}_{i}\left(x\right)$ being rational functions over a field. Here $\mathrm{D}$ is an object that satisfies  properties such as $\mathrm{DF}\left(x\right)=1$ and $\mathrm{DF}\left(uv\right)=u\mathrm{DF}\left(v\right)+\mathrm{DF}\left(u\right)v$.  One can perform functions on these operators such as multiplying, forming symmetric products, taking one-sided greatest common divisors, factoring, and so on.

This functionality, due to Mark van Hoeij (University of Nijmegen), allows for significant future advances for applications such as finding closed form solutions of linear differential equations.  Some applications of this subpackage of DEtools are included below.

We use the with command to allow us to use the short form for the commands in the DEtools package.

 > $\mathrm{restart}$
 > $\mathrm{with}\left(\mathrm{DEtools}\right):$

Algebraic Operations with Differential Operators

Differential operators are a noncommutative domain with operations such as addition, multiplication, application, and so on. A differential operator $L$ in $C\left(x\right)\left[\mathrm{DF}\right]$ is an expression , where ${a}_{i}\left(x\right)$ are elements of C(x). An element $L$ in $C\left(x\right)\left[\mathrm{Dx}\right]$ corresponds to a linear homogeneous differential equation $L\left(y\left(x\right)\right)=0$.

Multiplication (see DEtools[mult]) in the ring $C\left(x\right)\left[\mathrm{Dx}\right]$ corresponds to composition of differential operators. So if $L=\mathrm{mult}\left(f,g\right)$, then $L\left(y\left(x\right)\right)=f\left(g\left(y\left(x\right)\right)\right)$. In particular, $\mathrm{mult}\left(\mathrm{DF},x\right)=x\cdot \mathrm{DF}+1$.

As examples of the types of algebraic operations you can do, consider the three differential operators:

 > $\mathrm{L1}:={x}^{2}{\mathrm{DF}}^{2}-x\mathrm{DF}+a-{x}^{2}$
 ${\mathrm{L1}}{≔}{{x}}^{{2}}{}{{\mathrm{DF}}}^{{2}}{-}{x}{}{\mathrm{DF}}{-}{{x}}^{{2}}{+}{a}$ (1.1)
 > $\mathrm{L2}:=x\mathrm{DF}-\left({x}^{2}-b\right)$
 ${\mathrm{L2}}{≔}{x}{}{\mathrm{DF}}{-}{{x}}^{{2}}{+}{b}$ (1.2)
 > $\mathrm{L3}:={\mathrm{DF}}^{2}-x$
 ${\mathrm{L3}}{≔}{{\mathrm{DF}}}^{{2}}{-}{x}$ (1.3)

You can multiply these operators and note that multiplication is noncommutative:

 > $\mathrm{L4}:=\mathrm{collect}\left(\mathrm{mult}\left(\mathrm{L1},\mathrm{L2},\left[\mathrm{DF},x\right]\right),\mathrm{DF}\right)$
 ${\mathrm{L4}}{≔}{{x}}^{{3}}{}{{\mathrm{DF}}}^{{3}}{+}\left({-}{{x}}^{{4}}{+}{b}{}{{x}}^{{2}}{+}{{x}}^{{2}}\right){}{{\mathrm{DF}}}^{{2}}{+}\left({-}{4}{}{{x}}^{{3}}{+}{a}{}{x}{-}{b}{}{x}{-}{x}\right){}{\mathrm{DF}}{+}{{x}}^{{4}}{-}{a}{}{{x}}^{{2}}{-}{b}{}{{x}}^{{2}}{+}{a}{}{b}$ (1.4)
 > $\mathrm{L5}:=\mathrm{collect}\left(\mathrm{mult}\left(\mathrm{L2},\mathrm{L1},\left[\mathrm{DF},x\right]\right),\mathrm{DF}\right)$
 ${\mathrm{L5}}{≔}{{x}}^{{3}}{}{{\mathrm{DF}}}^{{3}}{+}\left({-}{{x}}^{{4}}{+}{b}{}{{x}}^{{2}}{+}{{x}}^{{2}}\right){}{{\mathrm{DF}}}^{{2}}{+}\left({a}{}{x}{-}{b}{}{x}{-}{x}\right){}{\mathrm{DF}}{+}{{x}}^{{4}}{-}{a}{}{{x}}^{{2}}{-}{b}{}{{x}}^{{2}}{+}{a}{}{b}{-}{2}{}{{x}}^{{2}}$ (1.5)

The argument $\left[\mathrm{DF},x\right]$ tells the mult command that the multiplication is over the differential domain specified by $\mathrm{DF}$ and $x$, and so the variables a and b are constants. This can also be set by the environment variable $\mathrm{_Envdiffopdomain}≔\left[\mathrm{DF},x\right]$, if you expect to use the same domain throughout.

The concept of a one-sided lowest common multiple and greatest common divisor exists in such domains. For example,

 > $\mathrm{L6}:=\mathrm{LCLM}\left(\mathrm{L3},\mathrm{L2},\left[\mathrm{DF},x\right]\right)$
 ${\mathrm{L6}}{≔}{{\mathrm{DF}}}^{{3}}{+}\frac{\left({-}{{x}}^{{6}}{+}{3}{}{b}{}{{x}}^{{4}}{+}{{x}}^{{5}}{-}{3}{}{{b}}^{{2}}{}{{x}}^{{2}}{-}{b}{}{{x}}^{{3}}{-}{3}{}{{x}}^{{4}}{+}{{b}}^{{3}}{+}{{x}}^{{3}}{+}{3}{}{{b}}^{{2}}{+}{2}{}{b}\right){}{{\mathrm{DF}}}^{{2}}}{{x}{}\left({{x}}^{{4}}{-}{2}{}{b}{}{{x}}^{{2}}{-}{{x}}^{{3}}{+}{{b}}^{{2}}{+}{{x}}^{{2}}{+}{b}\right)}{-}{x}{}{\mathrm{DF}}{-}\frac{{-}{{x}}^{{6}}{+}{3}{}{b}{}{{x}}^{{4}}{+}{{x}}^{{5}}{-}{3}{}{{b}}^{{2}}{}{{x}}^{{2}}{-}{b}{}{{x}}^{{3}}{-}{2}{}{{x}}^{{4}}{+}{{b}}^{{3}}{-}{2}{}{b}{}{{x}}^{{2}}{+}{4}{}{{b}}^{{2}}{+}{{x}}^{{2}}{+}{3}{}{b}}{{{x}}^{{4}}{-}{2}{}{b}{}{{x}}^{{2}}{-}{{x}}^{{3}}{+}{{b}}^{{2}}{+}{{x}}^{{2}}{+}{b}}$ (1.6)

and

 > $\mathrm{L7}:=\mathrm{GCRD}\left(\mathrm{L4},\mathrm{L6},\left[\mathrm{DF},x\right]\right)$
 ${\mathrm{L7}}{≔}{\mathrm{DF}}{+}\frac{{-}{{x}}^{{2}}{+}{b}}{{x}}$ (1.7)

One can check this by using right or left division. In this case, you have

 > $\mathrm{rightdivision}\left(\mathrm{L6},\mathrm{L7},\left[\mathrm{DF},x\right]\right)$
 $\left[{{\mathrm{DF}}}^{{2}}{+}\frac{\left({-}{2}{}{{x}}^{{4}}{+}{{x}}^{{3}}{+}{2}{}{{b}}^{{2}}{+}{2}{}{b}\right){}{\mathrm{DF}}}{{x}{}\left({{x}}^{{4}}{-}{2}{}{b}{}{{x}}^{{2}}{-}{{x}}^{{3}}{+}{{b}}^{{2}}{+}{{x}}^{{2}}{+}{b}\right)}{-}\frac{{{x}}^{{5}}{-}{2}{}{b}{}{{x}}^{{3}}{-}{{x}}^{{4}}{+}{{b}}^{{2}}{}{x}{+}{2}{}{{x}}^{{3}}{+}{4}{}{b}{}{x}{-}{2}{}{{x}}^{{2}}{-}{6}{}{b}}{{{x}}^{{4}}{-}{2}{}{b}{}{{x}}^{{2}}{-}{{x}}^{{3}}{+}{{b}}^{{2}}{+}{{x}}^{{2}}{+}{b}}{,}{0}\right]$ (1.8)
 > $\mathrm{rightdivision}\left(\mathrm{L4},\mathrm{L7},\left[\mathrm{DF},x\right]\right)$
 $\left[{{x}}^{{3}}{}{{\mathrm{DF}}}^{{2}}{+}{{x}}^{{2}}{}{\mathrm{DF}}{-}{{x}}^{{3}}{+}{a}{}{x}{-}{x}{,}{0}\right]$ (1.9)

which in both cases gives a quotient with 0 remainder. You can check the computation by multiplying:

 > $\mathrm{collect}\left(\mathrm{L4}-\mathrm{mult}\left(\left[1\right],\mathrm{L7},\left[\mathrm{DF},x\right]\right),\mathrm{DF},\mathrm{normal}\right)$
 ${0}$ (1.10)

Factorization of Differential Operators

The section contains subsections on closed-form solutions, and on other factorization options.

Building Closed-form Solutions of Differential Equations

Consider the third-order differential equation

 > $\mathrm{ode}:=\left(-x-{x}^{3}\right)y\left(x\right)-2{x}^{2}\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)+\left({x}^{3}+x\right)\left(\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)\right)+{x}^{2}\left(\frac{{ⅆ}^{3}}{ⅆ{x}^{3}}y\left(x\right)\right)$
 ${\mathrm{ode}}{≔}\left({-}{{x}}^{{3}}{-}{x}\right){}{y}{}\left({x}\right){-}{2}{}{{x}}^{{2}}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}\left({{x}}^{{3}}{+}{x}\right){}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{{x}}^{{2}}{}\left(\frac{{{ⅆ}}^{{3}}}{{ⅆ}{{x}}^{{3}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)$ (2.1.1)

One can use factorization of differential operators to find closed form solutions for this ODE. One can convert the equation to a differential operator:

 > $\mathrm{dode}:=\mathrm{de2diffop}\left(\mathrm{ode},y\left(x\right),\left[\mathrm{DF},x\right]\right)$
 ${\mathrm{dode}}{≔}{{x}}^{{2}}{}{{\mathrm{DF}}}^{{3}}{+}\left({{x}}^{{3}}{+}{x}\right){}{{\mathrm{DF}}}^{{2}}{-}{2}{}{{x}}^{{2}}{}{\mathrm{DF}}{-}{{x}}^{{3}}{-}{x}$ (2.1.2)

and factor by:

 > $\mathrm{ode_fact}:=\mathrm{DFactor}\left(\mathrm{dode},\left[\mathrm{DF},x\right]\right)$
 ${\mathrm{ode_fact}}{≔}\left[{{x}}^{{2}}{}{\mathrm{DF}}{+}{x}{}\left({{x}}^{{2}}{+}{2}\right){,}{{\mathrm{DF}}}^{{2}}{-}\frac{{\mathrm{DF}}}{{x}}{-}\frac{{{x}}^{{2}}{-}{1}}{{{x}}^{{2}}}\right]$ (2.1.3)

This shows that one factorization (factorizations are not unique) has a right factor given by:

 > $\mathrm{right_fact}:={\mathrm{ode_fact}}_{2}$
 ${\mathrm{right_fact}}{≔}{{\mathrm{DF}}}^{{2}}{-}\frac{{\mathrm{DF}}}{{x}}{-}\frac{{{x}}^{{2}}{-}{1}}{{{x}}^{{2}}}$ (2.1.4)

Any solution of the corresponding differential equation is then also a solution to the original homogeneous equation. This results in

 > $\mathrm{ode2}:=\mathrm{diffop2de}\left(\mathrm{right_fact},\left[\mathrm{DF},x\right],y\left(x\right)\right)$
 ${\mathrm{ode2}}{≔}{-}\frac{\left({{x}}^{{2}}{-}{1}\right){}{y}{}\left({x}\right)}{{{x}}^{{2}}}{-}\frac{\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)}{{x}}{+}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)$ (2.1.5)
 > $\mathrm{dsolve}\left(\mathrm{ode2},y\left(x\right)\right)$
 ${y}{}\left({x}\right){=}\mathrm{c__1}{}{x}{}{\mathrm{BesselI}}{}\left({0}{,}{x}\right){+}\mathrm{c__2}{}{x}{}{\mathrm{BesselK}}{}\left({0}{,}{x}\right)$ (2.1.6)

The remaining solution can be determined by reduction of order using the above two solutions. This is actually the method used in dsolve to solve the original ODE:

 > $\mathrm{dsolve}\left(\mathrm{ode}\right)$
 ${y}{}\left({x}\right){=}\mathrm{c__1}{}{x}{}{\mathrm{BesselI}}{}\left({0}{,}{x}\right){+}\mathrm{c__2}{}{x}{}{\mathrm{BesselK}}{}\left({0}{,}{x}\right){+}\mathrm{c__3}{}{x}{}\left({\mathrm{BesselK}}{}\left({0}{,}{x}\right){}\left({\int }\frac{{\mathrm{BesselI}}{}\left({0}{,}{x}\right){}{{ⅇ}}^{{-}\frac{{{x}}^{{2}}}{{2}}}}{{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right){-}{\mathrm{BesselI}}{}\left({0}{,}{x}\right){}\left({\int }\frac{{\mathrm{BesselK}}{}\left({0}{,}{x}\right){}{{ⅇ}}^{{-}\frac{{{x}}^{{2}}}{{2}}}}{{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right)\right)$ (2.1.7)

More Factorization

The differential operator

 > $\mathrm{df}:={\mathrm{DF}}^{2}+\frac{21\left({x}^{2}-x+1\right)}{100{x}^{2}{\left(x-1\right)}^{2}}$
 ${\mathrm{df}}{≔}{{\mathrm{DF}}}^{{2}}{+}\frac{{21}{}\left({{x}}^{{2}}{-}{x}{+}{1}\right)}{{100}{}{{x}}^{{2}}{}{\left({x}{-}{1}\right)}^{{2}}}$ (2.2.1)

considered as an element of $Q\left(x\right)\left[\mathrm{DF}\right]$, is well known not to factor. We can check this by using the DFactor function:

 > $\mathrm{DFactor}\left(\mathrm{df},\left[\mathrm{DF},x\right]\right)$
 $\left[{{\mathrm{DF}}}^{{2}}{+}\frac{{21}{}\left({{x}}^{{2}}{-}{x}{+}{1}\right)}{{100}{}{{x}}^{{2}}{}{\left({x}{-}{1}\right)}^{{2}}}\right]$ (2.2.2)

In addition, its differential Galois field is known to be ${A}_{5}^{{\mathrm{SL}}_{2}}$. It follows from the work of Singer and Ulmer that the sixth symmetric power of order 7

 > $\mathrm{dff}:=\mathrm{symmetric_power}\left(\mathrm{df},6,\left[\mathrm{DF},x\right]\right)$
 ${\mathrm{dff}}{≔}{{\mathrm{DF}}}^{{7}}{+}\frac{{294}{}\left({{x}}^{{2}}{-}{x}{+}{1}\right){}{{\mathrm{DF}}}^{{5}}}{{25}{}{{x}}^{{2}}{}{\left({x}{-}{1}\right)}^{{2}}}{-}\frac{{147}{}\left({2}{}{{x}}^{{3}}{-}{3}{}{{x}}^{{2}}{+}{5}{}{x}{-}{2}\right){}{{\mathrm{DF}}}^{{4}}}{{5}{}{{x}}^{{3}}{}{\left({x}{-}{1}\right)}^{{3}}}{+}\frac{{441}{}\left({349}{}{{x}}^{{4}}{-}{698}{}{{x}}^{{3}}{+}{1547}{}{{x}}^{{2}}{-}{1198}{}{x}{+}{349}\right){}{{\mathrm{DF}}}^{{3}}}{{625}{}{{x}}^{{4}}{}{\left({x}{-}{1}\right)}^{{4}}}{-}\frac{{441}{}\left({1094}{}{{x}}^{{5}}{-}{2735}{}{{x}}^{{4}}{+}{7470}{}{{x}}^{{3}}{-}{8470}{}{{x}}^{{2}}{+}{4829}{}{x}{-}{1094}\right){}{{\mathrm{DF}}}^{{2}}}{{625}{}{{x}}^{{5}}{}{\left({x}{-}{1}\right)}^{{5}}}{+}\frac{{63}{}\left({1597268}{}{{x}}^{{6}}{-}{4791804}{}{{x}}^{{5}}{+}{15457733}{}{{x}}^{{4}}{-}{22929126}{}{{x}}^{{3}}{+}{19286733}{}{{x}}^{{2}}{-}{8620804}{}{x}{+}{1597268}\right){}{\mathrm{DF}}}{{62500}{}{{x}}^{{6}}{}{\left({x}{-}{1}\right)}^{{6}}}{-}\frac{{189}{}\left({277134}{}{{x}}^{{7}}{-}{969969}{}{{x}}^{{6}}{+}{3595389}{}{{x}}^{{5}}{-}{6563550}{}{{x}}^{{4}}{+}{7263809}{}{{x}}^{{3}}{-}{4817148}{}{{x}}^{{2}}{+}{1768603}{}{x}{-}{277134}\right)}{{31250}{}{{x}}^{{7}}{}{\left({x}{-}{1}\right)}^{{7}}}$ (2.2.3)

factors and has factors of order 3 and 4. However, even though it was known what the orders of the factors are, no previous algorithm could actually do the factorization. We now obtain (after some time)

 > $\mathrm{fact_df}:=\mathrm{DFactor}\left(\mathrm{dff},\left[\mathrm{DF},x\right]\right)$
 ${\mathrm{fact_df}}{≔}\left[{{\mathrm{DF}}}^{{4}}{+}\frac{{4}{}\left({2}{}{x}{-}{1}\right){}{{\mathrm{DF}}}^{{3}}}{{x}{}\left({x}{-}{1}\right)}{+}\frac{{3}{}\left({23}{}{{x}}^{{2}}{-}{23}{}{x}{+}{3}\right){}{{\mathrm{DF}}}^{{2}}}{{5}{}{{x}}^{{2}}{}\left({{x}}^{{2}}{-}{2}{}{x}{+}{1}\right)}{+}\frac{{3}{}\left({58}{}{{x}}^{{3}}{-}{87}{}{{x}}^{{2}}{+}{33}{}{x}{-}{2}\right){}{\mathrm{DF}}}{{50}{}{{x}}^{{3}}{}\left({x}{-}{1}\right){}\left({{x}}^{{2}}{-}{2}{}{x}{+}{1}\right)}{-}\frac{{96}{}\left({{x}}^{{4}}{-}{2}{}{{x}}^{{3}}{+}{3}{}{{x}}^{{2}}{-}{2}{}{x}{+}{1}\right)}{{625}{}{{x}}^{{4}}{}{\left({x}{-}{1}\right)}^{{2}}{}\left({{x}}^{{2}}{-}{2}{}{x}{+}{1}\right)}{,}{{\mathrm{DF}}}^{{3}}{-}\frac{{4}{}\left({2}{}{x}{-}{1}\right){}{{\mathrm{DF}}}^{{2}}}{{x}{}\left({x}{-}{1}\right)}{+}\frac{\left({749}{}{{x}}^{{2}}{-}{749}{}{x}{+}{249}\right){}{\mathrm{DF}}}{{25}{}{{x}}^{{2}}{}\left({{x}}^{{2}}{-}{2}{}{x}{+}{1}\right)}{-}\frac{{9}{}\left({266}{}{{x}}^{{3}}{-}{399}{}{{x}}^{{2}}{+}{265}{}{x}{-}{66}\right)}{{50}{}{{x}}^{{3}}{}\left({{x}}^{{3}}{-}{3}{}{{x}}^{{2}}{+}{3}{}{x}{-}{1}\right)}\right]$ (2.2.4)

One can check the factorization:

 > $L:=\mathrm{mult}\left(\mathrm{fact_df}\left[1\right],\mathrm{fact_df}\left[2\right],\left[\mathrm{DF},x\right]\right):$
 > $\mathrm{collect}\left(L-\mathrm{dff},\mathrm{DF},\mathrm{normal}\right)$
 ${0}$ (2.2.5)

Building of Differential Equations

One can take advantage of the differential operator tools in DEtools to construct solutions of differential equations from known equations. For example,

 > $\mathrm{df}:={x}^{2}{\mathrm{DF}}^{2}-x\mathrm{DF}+{x}^{2}-1$
 ${\mathrm{df}}{≔}{{x}}^{{2}}{}{{\mathrm{DF}}}^{{2}}{-}{x}{}{\mathrm{DF}}{+}{{x}}^{{2}}{-}{1}$ (3.1)

is a differential operator with Bessel function solutions. Indeed we have

 > $\mathrm{diffop2de}\left(\mathrm{df},y\left(x\right),\left[\mathrm{DF},x\right]\right)$
 $\left({{x}}^{{2}}{-}{1}\right){}{y}{}\left({x}\right){-}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){}{x}{+}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){}{{x}}^{{2}}$ (3.2)
 > $\mathrm{dsolve}\left(,y\left(x\right)\right)$
 ${y}{}\left({x}\right){=}\mathrm{c__1}{}{x}{}{\mathrm{BesselJ}}{}\left(\sqrt{{2}}{,}{x}\right){+}\mathrm{c__2}{}{x}{}{\mathrm{BesselY}}{}\left(\sqrt{{2}}{,}{x}\right)$ (3.3)

The third symmetric power of $\mathrm{df}$ then generates a differential equation that has the power combinations of the above as its solutions. For example,

 > $\mathrm{dff}:=\mathrm{symmetric_power}\left(\mathrm{df},3,\left[\mathrm{DF},x\right]\right)$
 ${\mathrm{dff}}{≔}{{\mathrm{DF}}}^{{4}}{-}\frac{{6}{}{{\mathrm{DF}}}^{{3}}}{{x}}{+}\frac{{5}{}\left({2}{}{{x}}^{{2}}{+}{1}\right){}{{\mathrm{DF}}}^{{2}}}{{{x}}^{{2}}}{-}\frac{{5}{}\left({6}{}{{x}}^{{2}}{-}{7}\right){}{\mathrm{DF}}}{{{x}}^{{3}}}{+}\frac{{3}{}\left({3}{}{{x}}^{{4}}{+}{2}{}{{x}}^{{2}}{-}{21}\right)}{{{x}}^{{4}}}$ (3.4)
 > $\mathrm{diffop2de}\left(\mathrm{dff},y\left(x\right),\left[\mathrm{DF},x\right]\right);$
 $\frac{{3}{}\left({3}{}{{x}}^{{4}}{+}{2}{}{{x}}^{{2}}{-}{21}\right){}{y}{}\left({x}\right)}{{{x}}^{{4}}}{-}\frac{{5}{}\left({6}{}{{x}}^{{2}}{-}{7}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{{{x}}^{{3}}}{+}\frac{{5}{}\left({2}{}{{x}}^{{2}}{+}{1}\right){}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{{{x}}^{{2}}}{-}\frac{{6}{}\left(\frac{{{ⅆ}}^{{3}}}{{ⅆ}{{x}}^{{3}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{{x}}{+}\frac{{{ⅆ}}^{{4}}}{{ⅆ}{{x}}^{{4}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)$ (3.5)
 > $\mathrm{dsolve}\left(,y\left(x\right)\right)$
 ${y}{}\left({x}\right){=}\mathrm{c__1}{}{{x}}^{{3}}{}{{\mathrm{BesselJ}}{}\left(\sqrt{{2}}{,}{x}\right)}^{{3}}{+}\mathrm{c__2}{}{{x}}^{{3}}{}{{\mathrm{BesselY}}{}\left(\sqrt{{2}}{,}{x}\right)}^{{3}}{+}\mathrm{c__3}{}{{x}}^{{3}}{}{\mathrm{BesselJ}}{}\left(\sqrt{{2}}{,}{x}\right){}{{\mathrm{BesselY}}{}\left(\sqrt{{2}}{,}{x}\right)}^{{2}}{+}\mathrm{c__4}{}{{x}}^{{3}}{}{{\mathrm{BesselJ}}{}\left(\sqrt{{2}}{,}{x}\right)}^{{2}}{}{\mathrm{BesselY}}{}\left(\sqrt{{2}}{,}{x}\right)$ (3.6)

We can construct products of solutions using the symmetric_product function. For example,

 > $\mathrm{df2}:=\mathrm{DF}-x:$
 > $\mathrm{dsolve}\left(\mathrm{diffop2de}\left(\mathrm{df2},y\left(x\right),\left[\mathrm{DF},x\right]\right),y\left(x\right)\right)$
 ${y}{}\left({x}\right){=}\mathrm{c__1}{}{{ⅇ}}^{\frac{{{x}}^{{2}}}{{2}}}$ (3.7)

produces an exponential solution. In this case

 > $\mathrm{df3}:=\mathrm{symmetric_product}\left(\mathrm{df},\mathrm{df2},\left[\mathrm{DF},x\right]\right)$
 ${\mathrm{df3}}{≔}{{x}}^{{2}}{}{{\mathrm{DF}}}^{{2}}{-}{2}{}{{x}}^{{3}}{}{\mathrm{DF}}{-}{x}{}{\mathrm{DF}}{+}{{x}}^{{4}}{+}{{x}}^{{2}}{-}{1}$ (3.8)
 > $\mathrm{dsolve}\left(\mathrm{diffop2de}\left(\mathrm{df3},y\left(x\right),\left[\mathrm{DF},x\right]\right),y\left(x\right)\right)$
 ${y}{}\left({x}\right){=}\mathrm{c__1}{}{{ⅇ}}^{\frac{{{x}}^{{2}}}{{2}}}{}{x}{}{\mathrm{BesselJ}}{}\left(\sqrt{{2}}{,}{x}\right){+}\mathrm{c__2}{}{{ⅇ}}^{\frac{{{x}}^{{2}}}{{2}}}{}{x}{}{\mathrm{BesselY}}{}\left(\sqrt{{2}}{,}{x}\right)$ (3.9)

gives a solution that is the product of solving df and df2.

To create an equation having an exponential solution, we can either explicitly construct operators from exponential equations, or we can use the exterior power operation. If an operator L has a right factor of order n, then the n-th exterior power of L has a right factor of order 1. For example

 > $L:=\mathrm{collect}\left(\mathrm{mult}\left(\mathrm{df}+x,\mathrm{df},\left[\mathrm{DF},x\right]\right),\mathrm{DF}\right)$
 ${L}{≔}{{x}}^{{4}}{}{{\mathrm{DF}}}^{{4}}{+}{2}{}{{x}}^{{3}}{}{{\mathrm{DF}}}^{{3}}{+}\left({2}{}{{x}}^{{4}}{+}{{x}}^{{3}}{-}{3}{}{{x}}^{{2}}\right){}{{\mathrm{DF}}}^{{2}}{+}\left({2}{}{{x}}^{{3}}{-}{{x}}^{{2}}{+}{3}{}{x}\right){}{\mathrm{DF}}{+}{{x}}^{{4}}{+}{{x}}^{{3}}{-}{2}{}{{x}}^{{2}}{-}{x}{+}{1}$ (3.10)
 > $M:=\mathrm{exterior_power}\left(L,2,\left[\mathrm{DF},x\right]\right)$
 ${M}{≔}{{\mathrm{DF}}}^{{6}}{+}\frac{{8}{}{{\mathrm{DF}}}^{{5}}}{{x}}{+}\frac{\left({4}{}{{x}}^{{2}}{+}{2}{}{x}{+}{7}\right){}{{\mathrm{DF}}}^{{4}}}{{{x}}^{{2}}}{+}\frac{\left({24}{}{{x}}^{{2}}{+}{7}{}{x}{-}{3}\right){}{{\mathrm{DF}}}^{{3}}}{{{x}}^{{3}}}{+}\frac{\left({25}{}{{x}}^{{2}}{+}{3}{}{x}{-}{7}\right){}{{\mathrm{DF}}}^{{2}}}{{{x}}^{{4}}}{+}\frac{\left({{x}}^{{2}}{+}{14}\right){}{\mathrm{DF}}}{{{x}}^{{5}}}{-}\frac{{{x}}^{{2}}{+}{14}}{{{x}}^{{6}}}$ (3.11)
 > $\mathrm{DFactor}\left(M,\left[\mathrm{DF},x\right]\right)$
 $\left[{\mathrm{DF}}{+}\frac{{4}}{{x}}{,}{{\mathrm{DF}}}^{{4}}{+}\frac{{5}{}{{\mathrm{DF}}}^{{3}}}{{x}}{+}\frac{{2}{}\left({2}{}{{x}}^{{2}}{+}{x}{-}{2}\right){}{{\mathrm{DF}}}^{{2}}}{{{x}}^{{2}}}{+}\frac{{3}{}\left({4}{}{x}{+}{1}\right){}{\mathrm{DF}}}{{{x}}^{{2}}}{+}\frac{{5}}{{{x}}^{{2}}}{,}{\mathrm{DF}}{-}\frac{{1}}{{x}}\right]$ (3.12)

shows this right-hand factor. This can also be seen with

 > $\mathrm{expsols}\left(\mathrm{diffop2de}\left(M,y\left(x\right),\left[\mathrm{DF},x\right]\right),y\left(x\right)\right)$
 $\left[{x}\right]$ (3.13)
 >