For 1st order ODEs, the simplest problem beyond the reach of complete solving algorithms is known as Abel equations. These are equations of the form

where $y\equiv y\left(x\right)$ is the unknown and the $f_i\equiv f_i\left(x\right)$ and $g_j\equiv g_j\left(x\right)$ are arbitrary functions of $x$. The biggest subclass of Abel equations known to be solvable was discovered by our research team and is the AIR 4-parameter class. New in Maple 16, two additional 1-parameter classes of Abel equations, beyond the AIR class, are now also solvable.

Examples

PDEtools[declare](y(x), prime=x);

$y\left(x\right)\mathrm{will\; now\; be\; displayed\; as}y$

$\mathrm{derivatives\; with\; respect\; to}x\mathrm{of\; functions\; of\; one\; variable\; will\; now\; be\; displayed\; with\; \text{'}}$

This equation, of type Abel 2nd kind, depending on one parameter $\mathrm{\alpha }$, is now solved in terms of hypergeometric functions

ode[1] := diff(y(x),x) = -y(x)*(-3+x)*((-9+2*alpha)*y(x)-9)/(x*(54-36*x+6*x^2-3*alpha+2*alpha*x)*y(x)+9*alpha+81-54*x+9*x^2);

dsolve(ode[1]);

$\mathrm{c\_\_1}+\frac{27\left({\left(xy+3\right)}^2\left(\frac{243}{8}+\left(\mathrm{\alpha }+\frac{9}{4}\right)\mathrm{\alpha }{y}^4\left(\mathrm{\alpha }+9\right)\mathrm{LerchPhi}\left(-\frac{2y\mathrm{\alpha }}{9}+y+1\,1\,\frac{8\mathrm{\alpha }-9}{-9+2\mathrm{\alpha }}\right)+\frac{3\left(-\frac{81}{4}-{\mathrm{\alpha }}^2-\frac{45}{4}\mathrm{\alpha }\right)y^3}{2}+\frac{27\left(\mathrm{\alpha }+9\right)y^2}{8}-\frac{243y}{8}\right){\left(\frac{2y\mathrm{\alpha }}{9}-y-1\right)}^{\frac{8\mathrm{\alpha }-9}{-9+2\mathrm{\alpha }}}-\frac{81\left(\frac{3}{2}+\left(x-\frac{3}{2}\right)y\right)\left(-\frac{9}{2}+\left(\mathrm{\alpha }-\frac{9}{2}\right)y\right){\left(\frac{2y\mathrm{\alpha }}{9}-y-1\right)}^{\frac{6\mathrm{\alpha }}{-9+2\mathrm{\alpha }}}}{2}\right)}{2{\left(\mathrm{\alpha }-\frac{9}{2}\right)}^4{\left(xy+3\right)}^2{y}^4}=0$

The related class of Abel equations that is now entirely solvable consists of the set of equations that can be obtained from equation (2) by changing variables

where $G\left(x\right)$ and the four $P_i\left(x\right),Q_j\left(x\right)$ are arbitrary rational functions of $x$; this is the most general transformation that preserves the form of Abel equations and thus generates Abel ODE classes.

The following ODE family depending on an arbitrary function $G\left(x\right)$ is representative of the next difficult problem beyond Abel equations, that is, a problem involving 4th powers of $y$ in the right-hand side, now under focus

ode[2] := diff(y(x),x) = 1/2*(G(x)^2*y(x)^4-diff(G(x),x)*y(x)^2+x)/G(x)/y(x);

${\mathrm{ode}}_2≔\mathrm{y\text{'}}=\frac{{G\left(x\right)}^2{y}^4-\mathrm{G\text{'}}{y}^2+x}{2G\left(x\right)y}$

We solve it here in implicit form to avoid square roots obscuring the solution

sol[2] := dsolve(ode[2], y(x), implicit);

${\mathrm{sol}}_2≔\mathrm{c\_\_1}+\frac{\mathrm{AiryBi}\left(-x\right)G\left(x\right)y^2-\mathrm{AiryBi}\left(1\,-x\right)}{\mathrm{AiryAi}\left(-x\right)G\left(x\right)y^2-\mathrm{AiryAi}\left(1\,-x\right)}=0$

odetest(sol[2], ode[2], y(x));

$0$

A generalization of the problem above, solvable in Maple 16, involving an arbitrary function $G\left(y\right)$

ode[3] := diff(y(x),x) = 2*G(y(x))*x/(2-D(G)(y(x))*x^2+G(y(x))^2*x^4-2*y(x)*G(y(x))*x^2);

${\mathrm{ode}}_3≔\mathrm{y\text{'}}=\frac{2G\left(y\right)x}{2-\mathrm{D}\left(G\right)\left(y\right)x^2+{G\left(y\right)}^2{x}^4-2yG\left(y\right)x^2}$

sol[3] := dsolve(ode[3]);

${\mathrm{sol}}_3≔\mathrm{c\_\_1}+\frac{-G\left(y\right)x^2+2y}{-2{\ⅇ}^{y^2}+\mathrm{erfi}\left(y\right)\left(-G\left(y\right)x^2+2y\right)\sqrt{\mathrm{\pi }}}=0$

odetest(sol[3], ode[3]);

$0$

Using new algorithms developed by our research team, the dsolve command in Maple 16 can additionally solve two new nonlinear ODE families for each of the 2nd, 3rd and 4th order problems, with the ODE families involving arbitrary functions of the independent ($x$) or dependent ($y$) variables.

Examples

A 4th order ODE family

ode[4] := diff(y(x), x$4) = 5*y(x)*diff(y(x), x$3) + (10*diff(y(x), x) - 10*y(x)^2-1/x)*diff(y(x), x$2) - 15*y(x)*diff(y(x), x)^2+(10*y(x)^3 + 3/x*y(x))*diff(y(x),x) - y(x)^5 - 1/x*y(x)^3;

${\mathrm{ode}}_4≔\mathrm{y\text{'}\text{'}\text{'}\text{'}}=5y\mathrm{y\text{'}\text{'}\text{'}}+\left(10\mathrm{y\text{'}}-10y^2-\frac{1}{x}\right)\mathrm{y\text{'}\text{'}}-15y{\mathrm{y\text{'}}}^2+\left(10y^3+\frac{3y}{x}\right)\mathrm{y\text{'}}-y^5-\frac{y^3}{x}$

This ODE has no point symmetries; the determining PDE for the symmetry infinitesimals only admits both of them equal to zero:

PDEtools:-DeterminingPDE(ode[4]);

$\left\{{\mathrm{\_\η}}_y\left(x\,y\right)=0\,{\mathrm{\_\ξ}}_x\left(x\,y\right)=0\right\}$

Using new algorithms, this problem is nevertheless solvable in explicit form in terms of Bessel functions

dsolve(ode[4]);

$y=\frac{-\mathrm{BesselJ}\left(3\,2\sqrt{x}\right)x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\mathrm{c\_\_4}-\mathrm{BesselY}\left(3\,2\sqrt{x}\right)x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-2\mathrm{c\_\_3}x-\mathrm{c\_\_2}}{\mathrm{c\_\_1}+\mathrm{c\_\_2}x+\mathrm{c\_\_3}{x}^2+\mathrm{c\_\_4}{x}^2\mathrm{BesselJ}\left(4\,2\sqrt{x}\right)+x^2\mathrm{BesselY}\left(4\,2\sqrt{x}\right)}$

Another problem not admitting point symmetries, of 3rd order

ode[5] := diff(y(x), x$3) = (4*y(x)-1/x)*diff(y(x), x$2) + 3*diff(y(x), x)^2 + (-6*y(x)^2+3/x*y(x)-x)*diff(y(x), x) + y(x)^4 - 1/x*y(x)^3 + y(x)^2*x - y(x);

${\mathrm{ode}}_5≔\mathrm{y\text{'}\text{'}\text{'}}=\left(4y-\frac{1}{x}\right)\mathrm{y\text{'}\text{'}}+3{\mathrm{y\text{'}}}^2+\left(-6y^2+\frac{3y}{x}-x\right)\mathrm{y\text{'}}+y^4-\frac{y^3}{x}+y^{2}x-y$

dsolve(ode[5]);

$y=\frac{-240\mathrm{c\_\_2}x\mathrm{hypergeom}\left(\left[\frac{1}{3}\,\frac{1}{3}\right]\,\left[\frac{2}{3}\,\frac{2}{3}\,\frac{4}{3}\right]\,-\frac{x^3}{9}\right)+15\mathrm{c\_\_2}{x}^4\mathrm{hypergeom}\left(\left[\frac{4}{3}\,\frac{4}{3}\right]\,\left[\frac{5}{3}\,\frac{5}{3}\,\frac{7}{3}\right]\,-\frac{x^3}{9}\right)-480\mathrm{c\_\_3}{x}^2\mathrm{hypergeom}\left(\left[\frac{2}{3}\,\frac{2}{3}\right]\,\left[1\,\frac{4}{3}\,\frac{5}{3}\right]\,-\frac{x^3}{9}\right)+16\mathrm{c\_\_3}{x}^5\mathrm{hypergeom}\left(\left[\frac{5}{3}\,\frac{5}{3}\right]\,\left[2\,\frac{7}{3}\,\frac{8}{3}\right]\,-\frac{x^3}{9}\right)-720\mathrm{MeijerG}\left(\left[\left[1\right]\,\left[\right]\right]\,\left[\left[\frac{2}{3}\,\frac{2}{3}\,\frac{1}{3}\right]\,\left[\right]\right]\,\frac{x^3}{9}\right)}{240x\left(\mathrm{c\_\_1}+\mathrm{c\_\_2}x\mathrm{hypergeom}\left(\left[\frac{1}{3}\,\frac{1}{3}\right]\,\left[\frac{2}{3}\,\frac{2}{3}\,\frac{4}{3}\right]\,-\frac{x^3}{9}\right)+\mathrm{c\_\_3}{x}^2\mathrm{hypergeom}\left(\left[\frac{2}{3}\,\frac{2}{3}\right]\,\left[1\,\frac{4}{3}\,\frac{5}{3}\right]\,-\frac{x^3}{9}\right)+\mathrm{MeijerG}\left(\left[\left[1\,1\right]\,\left[\right]\right]\,\left[\left[\frac{2}{3}\,\frac{2}{3}\,\frac{1}{3}\right]\,\left[0\right]\right]\,\frac{x^3}{9}\right)\right)}$

A 2nd order nonlinear ODE problem for which point symmetries exist but are of no use for integration purposes (they involve an unsolved 4th order linear ODE).

ode[6] := diff(y(x), x, x) = x*(diff(y(x), x))^3*y(x)+x^2*(x-1)*(diff(y(x), x))^3+(-3*x+1)*(diff(y(x), x))^2;

${\mathrm{ode}}_6≔\mathrm{y\text{'}\text{'}}=x{\mathrm{y\text{'}}}^{3}y+x^2\left(x-1\right){\mathrm{y\text{'}}}^3+\left(-3x+1\right){\mathrm{y\text{'}}}^2$

sol[6] := dsolve(ode[6]);

${\mathrm{sol}}_6≔\frac{\left(\mathrm{AiryAi}\left(\frac{1}{4}-y\right)\mathrm{c\_\_2}+\mathrm{AiryBi}\left(\frac{1}{4}-y\right)\right){\ⅇ}^{-\frac{y}{2}}}{{\int }_{\phantom{\mathrm{`\; `}}}^y{\ⅇ}^{-\frac{\mathrm{\_a}}{2}}\left(\mathrm{AiryAi}\left(\frac{1}{4}-\mathrm{\_a}\right)\mathrm{c\_\_2}+\mathrm{AiryBi}\left(\frac{1}{4}-\mathrm{\_a}\right)\right)\ⅆ\mathrm{\_a}+\mathrm{c\_\_1}}+x=0$

odetest(sol[6], ode[6]);

$0$

A 3rd order ODE problem illustrating improvements in existing algorithms: in previous releases this equation was only solved in terms of uncomputed integrals of unresolved RootOf expressions; now the solution is computed explicitly as a rational function

ode[7] := diff(y(x), x$3) + 4*y(x)*diff(y(x), x$2) + 3*diff(y(x), x)^2 + 6*y(x)^2*diff(y(x), x) + y(x)^4 = 0;

${\mathrm{ode}}_7≔\mathrm{y\text{'}\text{'}\text{'}}+4y\mathrm{y\text{'}\text{'}}+3{\mathrm{y\text{'}}}^2+6\mathrm{y\text{'}}{y}^2+y^4=0$

sol[7] := dsolve(ode[7]);

odetest(sol[7], ode[7]);

$0$

dsolve can now solve ODEs that involve anticommutative variables set using the Physics package using the approach explained in PerformOnAnticommutativeSystem.

with(Physics, Setup);

$\left[\mathrm{Setup}\right]$

Set first $\mathrm{\theta }$ and $Q$ as suffixes for variables of type/anticommutative (see Physics[Setup])

Setup(anticommutativepre = {theta, Q});

$\mathrm{*\; Partial\; match\; of\; \text{'}}\mathrm{anticommutativepre}\mathrm{\text{'}\; against\; keyword\; \text{'}}\mathrm{anticommutativeprefix}\text{'}$

$\mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$

$\left[\mathrm{anticommutativeprefix}=\left\{Q\,\mathrm{\theta }\right\}\right]$

Consider this ordinary differential equation for the anticommutative function $Q$ of a commutative variable $x$

diff(Q(x), x, x) - Q(x)*diff(Q(x), x) = 0;

$\mathrm{Q\text{'}\text{'}}-Q\left(x\right)\mathrm{Q\text{'}}=0$

Its solution using dsolve involves an anticommutative constant $\mathrm{\_lambda1}$, analogous to the commutative constants $\mathrm{c\_\_1}$

dsolve((23));

$Q\left(x\right)=\tan \left(\frac{\sqrt{\mathrm{c\_\_1}\mathrm{\_\λ1}}\left(\mathrm{c\_\_2}+x\right)\sqrt{2}}{2}\right)\sqrt{\mathrm{c\_\_1}\mathrm{\_\λ1}}\sqrt{2}$

Both dsolve and pdsolve can now solve PDEs that involve anticommutative variables set using the Physics package using the approach explained in PerformOnAnticommutativeSystem.

with(PDEtools), with(Physics):

Set first $\mathrm{\theta }$ and $Q$ as suffixes for variables of type/anticommutative (see Physics[Setup])

Setup(anticommutativepre = {theta, Q});

$\mathrm{*\; Partial\; match\; of\; \text{'}}\mathrm{anticommutativepre}\mathrm{\text{'}\; against\; keyword\; \text{'}}\mathrm{anticommutativeprefix}\text{'}$

$\mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$

$\left[\mathrm{anticommutativeprefix}=\left\{Q\,\mathrm{\theta }\right\}\right]$

Consider this partial differential equation for the anticommutative function $Q$ of commutative and anticommutative variables $x,\mathrm{\theta }$

diff(Q(x, y,theta), x, theta) = 0;

$Q_{x,\mathrm{\theta }}=0$

Its solution using pdsolve

pdsolve((26));

$Q\left(x\,y\,\mathrm{\theta }\right)=\mathrm{f\_\_2}\left(x\,y\right)\mathrm{\_\λ2}+\mathrm{f\_\_4}\left(y\right)\mathrm{\theta }$

Note the introduction of an anticommutative constant $\mathrm{\_lambda2}$, analogous to the commutative constants $\mathrm{\_Cn}$ where n is an integer. The arbitrary functions $\mathrm{\_Fn}$ introduced are all commutative as usual and the Grassmannian parity (on right-hand-side if compared with the one on the left-hand-side) is preserved

Physics:-GrassmannParity((27));

$1=1$

A PDE system example with one unknown anticommutative function $Q$ of four variables, two commutative and two anticommutative; to avoid redundant typing in the input that follows and redundant display of information on the screen, use PDEtools:-declare, and PDEtools:-diff_table, that also handles anticommutative variables by automatically using Physics:-diff when Physics is loaded

PDEtools:-declare(Q(x, y, theta[1], theta[2]));

$Q\left(x\,y\,{\mathrm{\theta }}_1\,{\mathrm{\theta }}_2\right)\mathrm{will\; now\; be\; displayed\; as}Q$

q := PDEtools:-diff_table(Q(x, y, theta[1], theta[2])):

Now we can enter derivatives directly as the function's name indexed by the differentiation variables and see the display the same way; two PDEs

pde[1] := q[x, y, theta[1]] + q[x, y, theta[2]] - q[y, theta[1], theta[2]] = 0;

${\mathrm{pde}}_1≔Q_{x,y,{\mathrm{\theta }}_1}+Q_{x,y,{\mathrm{\theta }}_2}-Q_{y,{\mathrm{\theta }}_1,{\mathrm{\theta }}_2}=0$

pde[2] := q[theta[1]] = 0;

${\mathrm{pde}}_2≔Q_{{\mathrm{\theta }}_1}=0$

The solution to this system:

pdsolve([pde[1], pde[2]]);

$Q=\mathrm{f\_\_4}\left(x\,y\right)\mathrm{\_\λ3}+\left(\mathrm{f\_\_9}\left(x\right)+\mathrm{f\_\_8}\left(y\right)\right){\mathrm{\theta }}_2$

The dsubs command also works with anticommutative variables, though natively without using the approach explained in PerformOnAnticommutativeSystem. By inspection, it is clear that the derivatives in pde[2] can be substituted in pde[1] reducing the problem to a simpler one:

dsubs(pde[2], pde[1]);

$Q_{x,y,{\mathrm{\theta }}_2}=0$

pdsolve((33));

$Q=\mathrm{f\_\_4}\left(x\,y\right)\mathrm{\_\λ3}+\mathrm{f\_\_12}\left(x\,y\right){\mathrm{\theta }}_1+\left(\mathrm{f\_\_9}\left(x\right)+\mathrm{f\_\_8}\left(y\right)\right){\mathrm{\theta }}_2+\left(\mathrm{f\_\_11}\left(x\right)+\mathrm{f\_\_10}\left(y\right)\right)\mathrm{\_\λ4}{\mathrm{\theta }}_1{\mathrm{\theta }}_2$

Substituting this result for $Q$ back into pde[2], then multiplying by ${\mathrm{\theta }}_1$ and subtracting from the above also leads to the PDE system solution.

Using differential elimination techniques and the approach explained in PerformOnAnticommutativeSystem, ReducedForm in this example arrives at the same result as dsubs

PDEtools:-ReducedForm(pde[1], pde[2]);

$\left[Q_{x,y,{\mathrm{\theta }}_2}\right]\mathbf{where}\left[\right]$

Another command updated to handle anticommutative variables via PerformOnAnticommutativeSystem is casesplit; with this linear system in $Q$ and its derivatives it returns

casesplit([pde[1], pde[2]]);

$\left[Q_{x,y,{\mathrm{\theta }}_2}=0\,Q_{{\mathrm{\theta }}_1}=0\right]\mathbf{where}\left[\right]$

With the core functionality in place, most the symmetry commands that work on top can now also handle anticommutative variables, most of them natively, without going through PerformOnAnticommutativeSystem.

Set for instance the generic form of the infinitesimals for a PDE system like this one formed by pde[1] and pde[2]. For this purpose, we need anticommutative infinitesimals for the dependent variable $Q$ and two of the independent variables, ${\mathrm{\theta }}_1$ and ${\mathrm{\theta }}_2$; we use here the capital greek letters $\mathrm{\Xi }$ and $\mathrm{{\rm H}}$ for the anticommutative infinitesimal symmetry generators and the corresponding lower case greek letters for commutative ones

Setup(anticommutativepre = {Xi, Eta}, additionally);

$\mathrm{*\; Partial\; match\; of\; \text{'}}\mathrm{anticommutativepre}\mathrm{\text{'}\; against\; keyword\; \text{'}}\mathrm{anticommutativeprefix}\text{'}$

$\mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$

$\left[\mathrm{anticommutativeprefix}=\left\{\mathrm{{\rm H}}\,Q\,\mathrm{\Xi }\,\mathrm{\_\λ}\,\mathrm{\theta }\right\}\right]$

S := [xi[1], xi[2], Xi[1], Xi[2], Eta](x, y, theta[1], theta[2]);

$S≔\left[{\mathrm{\xi }}_1\left(x\,y\,{\mathrm{\theta }}_1\,{\mathrm{\theta }}_2\right)\,{\mathrm{\xi }}_2\left(x\,y\,{\mathrm{\theta }}_1\,{\mathrm{\theta }}_2\right)\,{\mathrm{\Xi }}_1\left(x\,y\,{\mathrm{\theta }}_1\,{\mathrm{\theta }}_2\right)\,{\mathrm{\Xi }}_2\left(x\,y\,{\mathrm{\theta }}_1\,{\mathrm{\theta }}_2\right)\,\mathrm{{\rm H}}\left(x\,y\,{\mathrm{\theta }}_1\,{\mathrm{\theta }}_2\right)\right]$

PDEtools:-declare(S);

$\mathrm{{\rm H}}\left(x\,y\,{\mathrm{\theta }}_1\,{\mathrm{\theta }}_2\right)\mathrm{will\; now\; be\; displayed\; as}\mathrm{{\rm H}}$

$\mathrm{\Xi }\left(x\,y\,{\mathrm{\theta }}_1\,{\mathrm{\theta }}_2\right)\mathrm{will\; now\; be\; displayed\; as}\mathrm{\Xi }$

$\mathrm{\xi }\left(x\,y\,{\mathrm{\theta }}_1\,{\mathrm{\theta }}_2\right)\mathrm{will\; now\; be\; displayed\; as}\mathrm{\xi }$

The corresponding InfinitesimalGenerator

InfinitesimalGenerator(S, Q(x,y,theta[1], theta[2]));

$f\mapsto {\mathrm{\xi }}_1\left(x\,y\,{\mathrm{\theta }}_1\,{\mathrm{\theta }}_2\right)\left(\frac{\ⅆf}{\ⅆx}\right)+{\mathrm{\xi }}_2\left(x\,y\,{\mathrm{\theta }}_1\,{\mathrm{\theta }}_2\right)\left(\frac{\ⅆf}{\ⅆy}\right)+{\mathrm{\Xi }}_1\left(x\,y\,{\mathrm{\theta }}_1\,{\mathrm{\theta }}_2\right)\left(\frac{\ⅆf}{\ⅆ{\mathrm{\theta }}_1}\right)+{\mathrm{\Xi }}_2\left(x\,y\,{\mathrm{\theta }}_1\,{\mathrm{\theta }}_2\right)\left(\frac{\ⅆf}{\ⅆ{\mathrm{\theta }}_2}\right)+\mathrm{{\rm H}}\left(x\,y\,{\mathrm{\theta }}_1\,{\mathrm{\theta }}_2\right)\left(\frac{\ⅆf}{\ⅆQ}\right)$

The table-function that returns the prolongation of the infinitesimal for $Q$ is computed with Eta_k, assign it here to the lower case $\mathrm{\eta }$ to use more familiar notation (recall $q\left[\right]=Q\left(x\,y\,{\mathrm{\theta }}_1\,{\mathrm{\theta }}_2\right)$)

eta := Eta_k(S, q[]);

$\mathrm{\eta }≔\mathrm{\eta }$

The first prolongations of $\mathrm{\eta }$ with respect to $x$ and  ${\mathrm{\theta }}_1$

eta[Q, [x]];

${\mathrm{{\rm H}}}_x-{\left({\mathrm{\xi }}_1\right)}_x{Q}_x-{\left({\mathrm{\xi }}_2\right)}_x{Q}_y-Q_{{\mathrm{\theta }}_1}{\left({\mathrm{\Xi }}_1\right)}_x-Q_{{\mathrm{\theta }}_2}{\left({\mathrm{\Xi }}_2\right)}_x$

eta[Q, [theta[1]]];

${\mathrm{{\rm H}}}_{{\mathrm{\theta }}_1}+Q_x{\left({\mathrm{\xi }}_1\right)}_{{\mathrm{\theta }}_1}+Q_y{\left({\mathrm{\xi }}_2\right)}_{{\mathrm{\theta }}_1}-Q_{{\mathrm{\theta }}_1}{\left({\mathrm{\Xi }}_1\right)}_{{\mathrm{\theta }}_1}-Q_{{\mathrm{\theta }}_2}{\left({\mathrm{\Xi }}_2\right)}_{{\mathrm{\theta }}_1}$

The second mixed prolongations of $\mathrm{\eta }$ with respect to $x,y$ and  $x,{\mathrm{\theta }}_1$

eta[Q, [x, y]];

${\mathrm{{\rm H}}}_{x,y}-{\left({\mathrm{\xi }}_1\right)}_{x,y}{Q}_x-{\left({\mathrm{\xi }}_2\right)}_{x,y}{Q}_y-Q_{{\mathrm{\theta }}_1}{\left({\mathrm{\Xi }}_1\right)}_{x,y}-Q_{{\mathrm{\theta }}_2}{\left({\mathrm{\Xi }}_2\right)}_{x,y}-{\left({\mathrm{\xi }}_1\right)}_y{Q}_{x,x}-{\left({\mathrm{\xi }}_2\right)}_y{Q}_{x,y}-Q_{x,{\mathrm{\theta }}_1}{\left({\mathrm{\Xi }}_1\right)}_y-Q_{x,{\mathrm{\theta }}_2}{\left({\mathrm{\Xi }}_2\right)}_y-{\left({\mathrm{\xi }}_1\right)}_x{Q}_{x,y}-{\left({\mathrm{\xi }}_2\right)}_x{Q}_{y,y}-Q_{y,{\mathrm{\theta }}_1}{\left({\mathrm{\Xi }}_1\right)}_x-Q_{y,{\mathrm{\theta }}_2}{\left({\mathrm{\Xi }}_2\right)}_x$

eta[Q, [x, theta[1]]];

${\mathrm{{\rm H}}}_{x,{\mathrm{\theta }}_1}+Q_x{\left({\mathrm{\xi }}_1\right)}_{x,{\mathrm{\theta }}_1}+Q_y{\left({\mathrm{\xi }}_2\right)}_{x,{\mathrm{\theta }}_1}-Q_{{\mathrm{\theta }}_1}{\left({\mathrm{\Xi }}_1\right)}_{x,{\mathrm{\theta }}_1}-Q_{{\mathrm{\theta }}_2}{\left({\mathrm{\Xi }}_2\right)}_{x,{\mathrm{\theta }}_1}+Q_{x,x}{\left({\mathrm{\xi }}_1\right)}_{{\mathrm{\theta }}_1}+Q_{x,y}{\left({\mathrm{\xi }}_2\right)}_{{\mathrm{\theta }}_1}-Q_{x,{\mathrm{\theta }}_1}{\left({\mathrm{\Xi }}_1\right)}_{{\mathrm{\theta }}_1}-Q_{x,{\mathrm{\theta }}_2}{\left({\mathrm{\Xi }}_2\right)}_{{\mathrm{\theta }}_1}-Q_{x,{\mathrm{\theta }}_1}{\left({\mathrm{\xi }}_1\right)}_x-Q_{y,{\mathrm{\theta }}_1}{\left({\mathrm{\xi }}_2\right)}_x-Q_{{\mathrm{\theta }}_1,{\mathrm{\theta }}_2}{\left({\mathrm{\Xi }}_2\right)}_x$

The DeterminingPDE for this system

DeterminingPDE([pde[1], pde[2]], S);

$\left\{-{\left({\mathrm{\xi }}_2\right)}_{{\mathrm{\theta }}_1}=0\,-{\left({\mathrm{\xi }}_2\right)}_{{\mathrm{\theta }}_2}=0\,-{\left({\mathrm{\xi }}_1\right)}_{{\mathrm{\theta }}_1,{\mathrm{\theta }}_2}{\mathrm{\theta }}_1-{\left({\mathrm{\xi }}_1\right)}_{{\mathrm{\theta }}_2}=0\,-{\left({\mathrm{\xi }}_1\right)}_{{\mathrm{\theta }}_1}+{\left({\mathrm{\xi }}_1\right)}_{{\mathrm{\theta }}_1,{\mathrm{\theta }}_2}{\mathrm{\theta }}_2=0\,-{\mathrm{\theta }}_1{\left({\mathrm{\xi }}_1\right)}_{y,{\mathrm{\theta }}_1}-{\mathrm{\theta }}_2{\left({\mathrm{\xi }}_1\right)}_{y,{\mathrm{\theta }}_2}+{\left({\mathrm{\xi }}_1\right)}_y=0\,-{\mathrm{\theta }}_1{\left({\mathrm{\xi }}_2\right)}_{x,{\mathrm{\theta }}_1}-{\mathrm{\theta }}_2{\left({\mathrm{\xi }}_2\right)}_{x,{\mathrm{\theta }}_2}+{\left({\mathrm{\xi }}_2\right)}_x=0\,-{\mathrm{\theta }}_1{\left({\mathrm{\xi }}_1\right)}_{x,x,{\mathrm{\theta }}_1}-{\mathrm{\theta }}_2{\left({\mathrm{\xi }}_1\right)}_{x,x,{\mathrm{\theta }}_2}+{\left({\mathrm{\xi }}_1\right)}_{x,x}=0\,{\left({\mathrm{\Xi }}_1\right)}_x-{\left({\mathrm{\Xi }}_1\right)}_{x,{\mathrm{\theta }}_2}{\mathrm{\theta }}_2-{\left({\mathrm{\Xi }}_1\right)}_{x,{\mathrm{\theta }}_1}{\mathrm{\theta }}_1=0\,{\left({\mathrm{\Xi }}_2\right)}_x-{\left({\mathrm{\Xi }}_2\right)}_{x,{\mathrm{\theta }}_2}{\mathrm{\theta }}_2-{\left({\mathrm{\Xi }}_2\right)}_{x,{\mathrm{\theta }}_1}{\mathrm{\theta }}_1=0\,{\mathrm{{\rm H}}}_{{\mathrm{\theta }}_1}=0\,{\mathrm{{\rm H}}}_{{\mathrm{\theta }}_1,{\mathrm{\theta }}_2}=0\,{\left({\mathrm{\Xi }}_1\right)}_{{\mathrm{\theta }}_1,{\mathrm{\theta }}_2}=0\,{\left({\mathrm{\Xi }}_2\right)}_{{\mathrm{\theta }}_1,{\mathrm{\theta }}_2}=0\,{\left({\mathrm{\xi }}_1\right)}_{{\mathrm{\theta }}_1,{\mathrm{\theta }}_2}=0\,{\left({\mathrm{\xi }}_2\right)}_{{\mathrm{\theta }}_1,{\mathrm{\theta }}_2}=0\,{\left({\mathrm{\Xi }}_2\right)}_{x,{\mathrm{\theta }}_2}=0\,{\left({\mathrm{\Xi }}_2\right)}_{y,{\mathrm{\theta }}_2}=0\,{\mathrm{{\rm H}}}_{x,y,{\mathrm{\theta }}_2}=0\,{\left({\mathrm{\Xi }}_1\right)}_{{\mathrm{\theta }}_1}=-{\mathrm{\theta }}_1{\left({\mathrm{\xi }}_1\right)}_{x,{\mathrm{\theta }}_1}-{\mathrm{\theta }}_2{\left({\mathrm{\xi }}_1\right)}_{x,{\mathrm{\theta }}_2}+{\left({\mathrm{\xi }}_1\right)}_x\,{\left({\mathrm{\Xi }}_1\right)}_{{\mathrm{\theta }}_2}={\left({\mathrm{\Xi }}_2\right)}_{{\mathrm{\theta }}_2}+{\mathrm{\theta }}_1{\left({\mathrm{\xi }}_1\right)}_{x,{\mathrm{\theta }}_1}+{\mathrm{\theta }}_2{\left({\mathrm{\xi }}_1\right)}_{x,{\mathrm{\theta }}_2}-{\left({\mathrm{\xi }}_1\right)}_x\,{\left({\mathrm{\Xi }}_2\right)}_{{\mathrm{\theta }}_1}=0\right\}$

To compute now the exact form of the symmetry infinitesimals you can either solve this PDE system for the commutative and anticommutative functions using pdsolve, or directly pass the system to Infinitesimals that will perform all the steps automatically

pdsolve((46));

$\left\{\mathrm{{\rm H}}=\mathrm{f\_\_8}\left(x\,y\right)\mathrm{\_\λ5}+\left(\mathrm{f\_\_32}\left(x\right)+\mathrm{f\_\_31}\left(y\right)\right){\mathrm{\theta }}_2\,{\mathrm{\Xi }}_1=\mathrm{f\_\_28}\left(y\right)\mathrm{\_\λ7}+\mathrm{c\_\_2}{\mathrm{\theta }}_1+\left(\mathrm{c\_\_1}-\mathrm{c\_\_2}\right){\mathrm{\theta }}_2\,{\mathrm{\Xi }}_2=\mathrm{f\_\_29}\left(y\right)\mathrm{\_\λ9}+\mathrm{c\_\_1}{\mathrm{\theta }}_2\,{\mathrm{\xi }}_1=\mathrm{c\_\_2}x+\mathrm{c\_\_3}\,{\mathrm{\xi }}_2=\mathrm{f\_\_30}\left(y\right)\right\}$

Substituting into S, you get the symmetry infinitesimal lists

subs((47), S);

$\left[\mathrm{c\_\_2}x+\mathrm{c\_\_3}\,\mathrm{f\_\_30}\left(y\right)\,\mathrm{f\_\_28}\left(y\right)\mathrm{\_\λ7}+\mathrm{c\_\_2}{\mathrm{\theta }}_1+\left(\mathrm{c\_\_1}-\mathrm{c\_\_2}\right){\mathrm{\theta }}_2\,\mathrm{f\_\_29}\left(y\right)\mathrm{\_\λ9}+\mathrm{c\_\_1}{\mathrm{\theta }}_2\,\mathrm{f\_\_8}\left(x\,y\right)\mathrm{\_\λ5}+\left(\mathrm{f\_\_32}\left(x\right)+\mathrm{f\_\_31}\left(y\right)\right){\mathrm{\theta }}_2\right]$

Infinitesimals also works with anticommutative variables natively; the related result comes specialized

Infinitesimals([pde[1], pde[2]], q[], S);

$\left[1\,\mathrm{f\_\_34}\left(y\right)\,\mathrm{f\_\_33}\left(y\right)\mathrm{\_\λ7}\,\mathrm{f\_\_2}\left(y\right)\mathrm{\_\λ9}\,\mathrm{f\_\_35}\left(x\,y\right)\mathrm{\_\λ5}+\left(\mathrm{f\_\_12}\left(x\right)+\mathrm{f\_\_4}\left(y\right)\right){\mathrm{\theta }}_2\right],\left[x\,\mathrm{f\_\_9}\left(y\right)\,\mathrm{f\_\_36}\left(y\right)\mathrm{\_\λ7}+{\mathrm{\theta }}_1-{\mathrm{\theta }}_2\,\mathrm{f\_\_8}\left(y\right)\mathrm{\_\λ9}\,\mathrm{f\_\_37}\left(x\,y\right)\mathrm{\_\λ5}+\left(\mathrm{f\_\_11}\left(x\right)+\mathrm{f\_\_10}\left(y\right)\right){\mathrm{\theta }}_2\right],\left[0\,\mathrm{f\_\_40}\left(y\right)\,\mathrm{f\_\_38}\left(y\right)\mathrm{\_\λ7}+{\mathrm{\theta }}_2\,\mathrm{f\_\_39}\left(y\right)\mathrm{\_\λ9}+{\mathrm{\theta }}_2\,\mathrm{f\_\_43}\left(x\,y\right)\mathrm{\_\λ5}+\left(\mathrm{f\_\_42}\left(x\right)+\mathrm{f\_\_41}\left(y\right)\right){\mathrm{\theta }}_2\right]$

To verify this result you can use SymmetryTest - it also handles anticommutative variables natively.

map(SymmetryTest, [(49)], [pde[1], pde[2]]);

$\left[\left\{0\right\}\,\left\{0\right\}\,\left\{0\right\}\right]$

To see these three list of symmetry infinitesimals with a label in the left-hand-side, you can use map

map[3](zip, `=`, S, [(49)]);

$\left[\left[{\mathrm{\xi }}_1=1\,{\mathrm{\xi }}_2=\mathrm{f\_\_34}\left(y\right)\,{\mathrm{\Xi }}_1=\mathrm{f\_\_33}\left(y\right)\mathrm{\_\λ7}\,{\mathrm{\Xi }}_2=\mathrm{f\_\_2}\left(y\right)\mathrm{\_\λ9}\,\mathrm{{\rm H}}=\mathrm{f\_\_35}\left(x\,y\right)\mathrm{\_\λ5}+\left(\mathrm{f\_\_12}\left(x\right)+\mathrm{f\_\_4}\left(y\right)\right){\mathrm{\theta }}_2\right]\,\left[{\mathrm{\xi }}_1=x\,{\mathrm{\xi }}_2=\mathrm{f\_\_9}\left(y\right)\,{\mathrm{\Xi }}_1=\mathrm{f\_\_36}\left(y\right)\mathrm{\_\λ7}+{\mathrm{\theta }}_1-{\mathrm{\theta }}_2\,{\mathrm{\Xi }}_2=\mathrm{f\_\_8}\left(y\right)\mathrm{\_\λ9}\,\mathrm{{\rm H}}=\mathrm{f\_\_37}\left(x\,y\right)\mathrm{\_\λ5}+\left(\mathrm{f\_\_11}\left(x\right)+\mathrm{f\_\_10}\left(y\right)\right){\mathrm{\theta }}_2\right]\,\left[{\mathrm{\xi }}_1=0\,{\mathrm{\xi }}_2=\mathrm{f\_\_40}\left(y\right)\,{\mathrm{\Xi }}_1=\mathrm{f\_\_38}\left(y\right)\mathrm{\_\λ7}+{\mathrm{\theta }}_2\,{\mathrm{\Xi }}_2=\mathrm{f\_\_39}\left(y\right)\mathrm{\_\λ9}+{\mathrm{\theta }}_2\,\mathrm{{\rm H}}=\mathrm{f\_\_43}\left(x\,y\right)\mathrm{\_\λ5}+\left(\mathrm{f\_\_42}\left(x\right)+\mathrm{f\_\_41}\left(y\right)\right){\mathrm{\theta }}_2\right]\right]$