Ordinary and Partial Differential Equations
Maple is the world leader in finding exact solutions to ordinary and partial differential equations. Maple 2020 extends that lead even further with new algorithms and techniques for solving more ODEs and PDEs, including general solutions, and solutions with initial conditions and/or boundary conditions. For Maple 2020, there are significant improvements both in dsolve and in pdsolve for the exact solution of ODEs and PDEs, with and without initial or boundary conditions. For ODEs, a new algorithm for computing hypergeometric solutions to 2ndorder linear ODEs is capable of solving new classes of problems that were previously out of reach. For PDEs, improvements include a significantly extended ability to solve problems with boundary and initial conditions, also using integral Mellin and Hankel transforms, and two new methods for finding general PDE solutions, by mapping the PDE onto an ODE with auxiliary functions and by using decomposition into first integrals. As well, when the PDE depends on parameters, new symmetry techniques allow you to compute parameterized symmetries and to reduce the number of these parameters, something frequently convenient in different contexts.
Hypergeometric solutions for second-order linear ODEs
Exact solutions to PDEs with Boundary / Initial conditions
Mellin and Hankel transform solutions for PDE with Boundary Conditions
A general solution to a PDE calculated via rewriting the PDE as an ODE with arbitrary auxiliary functions
Solving a PDE by making use of first integrals
PDE parameterized symmetries
Parameter-continuous symmetry transformations
Previous Maple releases already have algorithms for computing hypergeometric solutions for linear ODEs. The new algorithms implemented in Maple 2020, however, are more general. Suppose an ODE admits solutions of the form
yx = r__0x⋅Yx+r__1x⋅Y 'x
where r__0x and r__1x are rational functions of x, and Yxis the solution of some other linear ODE. Then the new algorithm can compute that other linear ODE and solve it, provided that it admits solutions of the form
Yx=ⅇ∫rxⅆxF12a,b;c;fx
where rx and fx are rational functions of x, F12 is the hypergeometric function and a, b, and c are arbitrary constants. The new algorithms use modular reduction, Hensel lifting, rational function reconstruction, and rational number reconstruction.
This algorithm is more general than previous algorithms in that it has no restrictions neither on the degree of the pullback function fx nor the number of singularities of the input equation. The implementation follows the presentation by Imamoglu, E. and van Hoeij, M. "Computing Hypergeometric Solutions of Second Order Linear Differential Equations using Quotients of Formal Solutions and Integral Bases", Journal of Symbolic Computation, 83, (2017): 254-271.
Examples
A problem that admits solutions of the form Yx=ⅇ∫rxⅆxF12a,b;c;fx
restart;
ode__1 ≔ ⅆ2ⅆx2yx+424x2−1ⅆⅆxyxx16x2−1 +2192x3−56x2−6x+1yxx24x−116x2−1= 0:
dsolveode__1
yx=_C1F12−12,12;1;4x+124x−12x2+_C2F1212,32;2;−16x4x−12x4x−12
An example where the solutions are of the form yx = r__0x⋅Yx+r__1x⋅Y 'x
ode__2 ≔ ⅆ2ⅆx2yx+2112x3−168x2+73x−9ⅆⅆxyxx−1+2x16x2−16x+3+4576x5−1024x4+488x3−15x2−27x+3yx−1+2x2x21+4x16x2−16x+3=0:
dsolveode__2
yx=_C1−4x+1F1212,12;1;16x22x−1x2+−16x2+1F1232,32;2;16x22x3−x2+_C2−4x+1F1212,12;1;−16x2+12x−1x2+16x2−1F1232,32;2;−16x2+12x3−x2
An example with a parameter a in the coefficients
ode__3≔ax+1ⅆ2ⅆx2yx+288ax4−18+−432a+224x3+190a−336x2+−24a+146xⅆⅆxyx32x4−48x3+22x2−3x+4096ax6+12+−7232a+2304x5+3568a−4096x4+−160a+1952x3+−218a−60x2+30a−108xyx2x−12x24x+116x2−16x+3=0:
dsolveode__3
yx=_C14x−34F1232,32;2;16x2x−F1232,32;2;16x2+F1212,12;1;16x216x2−14x−12x−1ax+1x232x3−48x2+22x−3+_C24x−3−4F1232,32;2;−16x2+1x+F1212,12;1;−16x2+1+F1232,32;2;−16x2+116x2−14x−12x−1ax+1x232x3−48x2+22x−3
Verify this solution
odetest,ode__3
0
Some ODEs with 4 regular singular points that can be solved in terms of HeunG functions can also be solved using the new algorithm. This is possible whenever through a gauge transformation one of the singularities can be removed. The following ODE is of that kind, and the default solution returned by dsolve is in terms of HeunG
ode__4 ≔ ⅆ2ⅆx2yx+x4−44x3+1206x2−44x+1yx4x2−34x+12x2=0:
dsolveode__4
yx=_C1x−x2+34x−134HG242−17+122,3362−254−17+122317+1222,32,32,32,1,x−17+122−17+122+_C2x−x+17+122−x2+34x−114HG242−17+122,62−6−17+122317+1222,1,1,12,1,x−17+122−17+122
To get a solution in terms of hypergeometric functions you can indicate the method to be used, in this case the new method hypergeometricsols
dsolveode__4,hypergeometricsols
yx=_C1ⅇ−∫−x3+6x2−34x+1x+15x2+51x−1x2−34x+1xx+1ⅆx2F1213,23;1;x3+30x2−24x+1+x2−7x+1x2−34x+12x+13+_C2ⅇ−∫−x3+6x2−34x+1x+15x2+51x−1x2−34x+1xx+1ⅆx2F1213,23;1;−x2+7x−1x2−34x+1+x3−24x2+30x+12x+13
The integrals can actually be computed
value
yx=_C1ⅇarctanh36−36x12x+12−36x2+lnx2−34x+14−lnx+12+lnx2F1213,23;1;x3+30x2−24x+1+x2−7x+1x2−34x+12x+13+_C2ⅇarctanh36−36x12x+12−36x2+lnx2−34x+14−lnx+12+lnx2F1213,23;1;−x2+7x−1x2−34x+1+x3−24x2+30x+12x+13
odetest,ode__4
Maple 2019 included a significant leap in the computation of exact solutions for PDE with Boundary / Initial conditions. For Maple 2020, another jump ahead in the solving capabilities happened. The examples below belong to different classes of problems out of reach of the Maple 2019 developments.
An example where the solution involves products of Bessel and Hankel functions
pde__1≔∂∂tur,t=∂2∂r2ur,t+∂∂rur,tr:
iv__1≔ur,0=1−r,u1,t=0:
pdsolvepde__1,iv__1
ur,t=casesplit/ansSum−BesselJ0,λnrπBesselJ1,λnStruveH0,λn−BesselJ0,λnStruveH1,λnsinhλn2t−coshλn2tλn2BesselJ0,λn2+BesselJ1,λn2,n=1..∞,Andλn=BesselJZeros0,n,0<λn
A similar problem but in three variables; the solution is a double infinite sum
pde__2≔∂∂tur,z,t=∂2∂r2ur,z,t+∂∂rur,z,tr+∂2∂z2ur,z,t:
iv__2≔ur,0,t=0,ur,1,t=0,u1,z,t=0,ur,z,0=fr,z:
pdsolvepde__2,iv__2
ur,z,t=casesplit/ansSumSum4BesselJ0,λn1rsinnπzexp−tπ2n2+λn12IntBesselJ0,λn1rrIntsinnπzfr,z,z=0..1,AllSolutions,r=0..1,AllSolutionshypergeom12,1,2,−λn12,n=1..∞,n1=1..∞,Andλn1=BesselJZeros0,n1,0≤λn1
A large number of new infinite series solutions are now computable for different classes of problems (the number of boundary or initial conditions, whether they are periodic or involve derivatives of the unknown, etc.)
pde__3≔∂2∂t2ux,t=4∂2∂x2ux,t:
iv__3≔ux,0=0,D2ux,0=sinx2,u−π,t=0,uπ,t=0:
pdsolvepde__3,iv__3
ux,t=315∑n=5∞16sinnx+π2sinnt−1n−1πn2n2−16π−160cos3x2sin3t+672cosx2sint315π
pde__4≔∂2∂x2ux,y+∂2∂y2ux,y=0:
iv__4≔u0,y=0,ux,1=A,ux,0=0:
pdsolvepde__4,iv__4,HINT=boundedseriesx=∞assuming0<x,0<y
ux,y=Ay+∑n=1∞2−1nAⅇ−nπxsinnπynπ
pde__5≔∂∂tut,x=∂2∂x2ut,x:
iv__5≔u0,x=1,ut,−1=0,ut,1=0:
pdsolvepde__5,iv__5
ut,x=∑n=1∞−2ⅇ−π2n2t4sinnπx+12−1n−1nπ
An example with periodic conditions in both the unknown and its first derivative, where different methods combine resulting in a solution involving an arbitrary function of x
pde__6≔∂∂tux,t=∂2∂x2ux,t:
iv__6≔u−π,t=uπ,t,D1u−π,t=D1uπ,t:
pdsolvepde__6,iv__6
ux,t=2∑n=1∞ⅇ−n2t∫−ππ_F1xsinnxⅆxsinnx+∫−ππ_F1xcosnxⅆxcosnxππ+∫−ππ_F1xⅆx2π
The detection of particular values of the summation index for which the solution branches with a different expression is now better:
pde__7≔∂2∂t2ux,t+2∂∂tux,t=∂2∂x2ux,t:
iv__7≔D2ux,0=0,ux,0=fx,u0,t=0,uπ,t=0:
pdsolvepde__7,iv__7,ux,tassuming0<t
ux,t=∑n=1∞2ⅇ−tt+1∫0πsinxfxⅆxsinxπn=1−1+−n2+1ⅇ−−n2+1+1t+ⅇ−1+−n2+1t−n2+1+1∫0πsinnxfxⅆxsinnx−n2+1πotherwise
A problem with two boundary and two initial conditions, all for the unknown ur,t, where the solution method consists of splitting the problem into two PDE & BC problems and the solution is the sum of those two solutions
pde__8≔∂2∂r2ur,t+∂∂rur,tr+∂2∂t2ur,tr2=0:
iv__8≔ur,π6=π6,ur,π2=0,u1,t=0,u2,t=t:
pdsolvepde__8,iv__8
ur,t=∑n1=1∞−−r3n1+r−3n18n1−1n1−13sinn1−6t+π264n1−1n1+∑n=1∞−−1n−1sinnπlnrln2ⅇnπ7π−6t6ln2−ⅇnπ6t+π6ln23nⅇnπ2ln2−ⅇnπ23ln2
Similar example, but involving conditions for the two first derivatives of the unknown ur,t
pde__9≔∂2∂r2ur,t+∂∂rur,tr+∂2∂t2ur,tr2=0:
iv__9≔ur,π6=π6,D2ur,π2=0,u1,t=0,D1u2,t=t:
pdsolvepde__9,iv__9
ur,t=∑n1=0∞16cos12t−2πn14+3t2+π4−r32+3n1+r−32−3n1π+4−1n123n1+12+232+3n1πn1926n1+6n13+96n12+48n1+864n1+812+n13π+∑n=0∞2sin1+2nπlnr2ln2ⅇ1+2nπ−6t+5π12ln2+ⅇ−−6t+π1+2nπ12ln231+2nⅇ1+2nπ23ln2+1
A problem with initial conditions involving arbitrary functions evaluated at algebraic expressions, as fx,−a2−b2, and boundary conditions that are periodic as ua,t=ub,t
pde__10≔∂∂tux,t=k∂2∂x2ux,t:
iv__10≔ux,0=fx,−a2−b2,ua,t=ub,t,∂∂aua,t=∂∂bub,t:
pdsolvepde__10,iv__10,ux,tassuminga<b
ux,t=−b+a∑n=1∞−2ⅇ−4kπ2n2t−b+a2∫abfx,−a2−b2sin2nπ−x+a−b+aⅆxsin2nπ−x+a−b+a+∫abfx,−a2−b2cos2nπ−x+a−b+aⅆxcos2nπ−x+a−b+a−b+a−∫abfx,−a2−b2ⅆx−b+a
Similar problem involving an arbitrary function in the initial conditions
pde__11≔∂∂tυξ,t=k∂2∂ξ2υξ,t:
iv__11≔D1υa,t=0,υl+a,t=0,υξ,0=fξ,a:
pdsolvepde__11,iv__11,υξ,tassuming0<a,0<l
υξ,t=∑n=0∞2cos1+2nπ−ξ+a2lⅇ−kπ21+2n2t4l2∫al+afξ,acos1+2nπ−ξ+a2lⅆξl
Some simple, however tricky problems with nonlinear boundary conditions
pde__12≔−∂2∂r2ur,t:
iv__12≔D1u0,t+D1u0,tu0,t=0:
pdsolvepde__12,iv__12
ur,t=_C2,ur,t=_C1r−1
Problems with more than one evaluation point for the same variable, periodic, or involving piecewise initial conditions
pde__13≔∂2∂t2υξ,y,t=∂2∂ξ2υξ,y,t+∂2∂y2υξ,y,t:
iv__13≔υπ3,y,t=0,υ4π3,y,t=0,υξ,0,t=0,υξ,π,t=0,D3υξ,y,0=0,υξ,y,0=ξ−π34π3−ξyπ−y:
pdsolvepde__13,iv__13
υξ,y,t=∑n1=1∞∑n=1∞16−−1n1+n+−1n1+−1n−1sinnπ−3ξ3sinn1ycosn2+n12tπ2n3n13
pde__14≔∂2∂t2ux,t+4π∂∂tux,t3=16∂2∂x2ux,t:
iv__14≔u0,t=0,D1u3,t=0,D2ux,0=0,ux,0=x100≤xandx≤11101<xandx≤3:
pdsolvepde__14,iv__14
ux,t=∑n=0∞4sinπx6ⅇ−2πt3πt+325π2n=033sinπn3+cosπn3sin1+2nπx62nn+1+Iⅇ2I3π−2nn+1+It−ⅇ2I3π2nn+1+It−2nn+1+I10n+1nπ21+2n2otherwise
In previous Maple releases, the Fourier and Laplace transforms were used to compute exact solutions to PDE problems with boundary conditions. Now, Mellin and Hankel transforms are also used for that same purpose.
Examples:
pde__1≔x2∂2∂x2ux,y+x∂∂xux,y+∂2∂y2ux,y=0: iv__1≔ux,0=0,ux,1=10≤xandx<101<x:
sol ≔ pdsolvepde__1,iv__1
ux,y=invmellinsinsysinss,s,x
As usual, you can let pdsolve choose the solving method, or indicate the method yourself:
pde__2≔∂2∂r2ur,t+∂∂rur,tr+∂2∂t2ur,t=−Q__0qr: iv__2≔ur,0=0:
pdsolvepde__2,iv__2
ur,t=Q__0−hankelexp−sthankelqr,r,s,0s2,s,r,0+hankelhankelqr,r,s,0s2,s,r,0
It is sometimes preferable to see these solutions in terms of more familiar integrals. For that purpose, use
convert,Int, only=hankel
ur,t=Q__0−∫0∞ⅇ−st∫0∞qrrJ0rsⅆrJ0rssⅆs+∫0∞∫0∞qrrJ0rsⅆrJ0rssⅆs
An example where the hankel transform is computed:
pde__3≔c2∂2∂r2ur,t+∂∂rur,tr=∂2∂t2ur,t:iv__3≔ur,0=Aaa2+r2−12,D2ur,0=0:
pdsolvepde__3, iv__3,method=Hankel assuming r > 0, t > 0, a > 0
ur,t=Aa−c2t2+2Iact+a2+r2+−c2t2−2Iact+a2+r22−c2t2−2Iact+a2+r2−c2t2+2Iact+a2+r2
Under certain conditions, it is possible to rewrite a PDE as an ODE for one unknown and some auxiliary arbitrary functions. When that is possible, the solving process involves as many steps as there are independent variables in the PDE. Each step consists of solving only a single intermediate ODE (and no PDEs). In this way, a general solution is obtained for the original PDE.
Below are four examples, all of them nonlinear. For the 4th one the solving method is shown step by step.
Example 1:
pde__1≔∂2∂t∂xwt,x−a∂∂xwt,x2wt,x−∂∂twt,xwt,x+b+cwt,x∂∂xwt,x−c∂∂twt,x2awt,x−kwt,x−bc2a−c24awt,x:
pdsolvepde__1;
wt,x=∫−cⅇ2k∫_F1x−4ka+b2+ⅇ−−4ka+b2t4_F1xak−_F1xb2+_F1x−4ka+b2b+ⅇ−−4ka+b2tb+−4ka+b2ⅇ−−4ka+b2tⅆx2aⅆx+_F2tⅇ∫−2k_F1x−4ka+b2+ⅇ−−4ka+b2t4_F1xak−_F1xb2+_F1x−4ka+b2b+ⅇ−−4ka+b2tb+−4ka+b2ⅇ−−4ka+b2tⅆx
Example 2:
pde__2≔wt,x3∂4∂t2∂x2wt,x−2wt,x2∂∂twt,x∂3∂t∂x2wt,x+∂∂xwt,x∂3∂t2∂xwt,x−2wt,x∂2∂t∂xwt,x−2∂∂twt,x∂∂xwt,x2+2wt,x∂2∂x2wt,x+∂∂xwt,x2∂∂twt,x2−wt,x∂2∂t2wt,xwt,x∂2∂x2wt,x−2∂∂xwt,x2:
pdsolvepde__2;
wt,x=_F4tⅇ_F1xt−_F2x−_F3tx
Example 3:
pde__3≔4 ∂5∂t3∂x2yx,t∂∂tyx,t+6∂4∂t2∂x2yx,t∂2∂t2yx,t+4∂3∂t∂x2yx,t∂3∂t3yx,t+∂2∂x2yx,t∂4∂t4yx,t+6∂3∂t2∂xyx,t2+8∂2∂t∂xyx,t∂4∂t3∂xyx,t+2∂∂xyx,t∂5∂t4∂xyx,t+yx,t∂6∂t4∂x2yx,t:
pdsolvepde__3;
yx,t=RootOf_F1xt3−6_F2xt2−6_F3xt+6_F5tx+3_Z2−6_F4x−3_F6t
Example 4
pde__4≔∂3∂t∂x∂zEt,x,zEt,x,z−∂2∂x∂zEt,x,z∂∂tEt,x,zEt,x,z2−∂2∂t∂xEt,x,z∂∂zEt,x,zEt,x,z2−∂∂xEt,x,z∂2∂t∂zEt,x,zEt,x,z2+2∂∂xEt,x,z∂∂zEt,x,z∂∂tEt,x,zEt,x,z3:
Step-by-step solving process for pde__4
Step 1. pde__4 can be written as an ODE with respect to t by taking the derivatives of Et,x,zwith respect to x and z as auxiliary arbitrary functions of t according to
auxiliary_functions ≔A__xt=∂∂xEt,x,z,A__zt=∂∂zEt,x,z,A__xzt=∂2∂x∂zEt,x,z,E__0t=Et,x,z:
The resulting ODE is
ODE__t≔ⅆⅆtA__xztE__0t−A__xztⅆⅆtE__0tE__0t2−ⅆⅆtA__xtA__ztE__0t2+2A__xtA__ztⅆⅆtE__0tE__0t3−A__xtⅆⅆtA__ztE__0t2:
Despite the presence of the auxiliary arbitrary functions A__xt, A__zt and A__xzt, the problem is in reach of dsolve's algorithms, so solve this ODE__t for E__0t
dsolveODE__t,E__0t
E__0t=−A__xzt−A__xzt2+4_C1A__xtA__zt2_C1,E__0t=−A__xzt+A__xzt2+4_C1A__xtA__zt2_C1
The solution above reduced the order of the original problem pde__4 by one. Depending on the problem, more than one reduction of order may occur.
Step 2. We need to express the integration constant with respect to t in the context of a problem in t, x, and z, that is, _C1 is an arbitrary function of x and z. For illustration purposes take for instance the first solution
_C1=solve1,_C1;
_C1=−A__xztE__0t−A__xtA__ztE__0t2
subs_C1 = _F1x,t,
_F1x,t=−A__xztE__0t−A__xtA__ztE__0t2
Restoring the original functions, we get
subsauxiliary_functions,
_F1x,t=−∂2∂x∂zEt,x,zEt,x,z−∂∂xEt,x,z∂∂zEt,x,zEt,x,z2
This pde in t, x, and z can now be written in terms of two auxiliary functions just of x as
auxiliary_functions ≔ A__zx = ∂∂zEt,x,z, E__0x=Et,x,z:
Resulting in
ODE__x≔_F1x,z=−ⅆⅆxA__zxE__0x−ⅆⅆx E__0xA__zxE__0x2
ODE__x≔_F1x,z=−ⅆⅆxA__zxE__0x−ⅆⅆxE__0xA__zxE__0x2
Solve this ODE
dsolveODE__x,E__0x
E__0x=A__zx∫−_F1x,zⅆx+_C1
Step 3. Proceed as in the previous step, isolating _C1 , rewriting it as an arbitrary function of z and t and restoring the original functions
_C1=solve,_C1;
_C1=−E__0x∫−_F1x,zⅆx−A__zxE__0x
subs_C1 = _F2t,z,
_F2t,z=−E__0x∫−_F1x,zⅆx−A__zxE__0x
Although not necessary, a simplification is possible in that the integral of an arbitrary function is also an arbitrary function, so we can remove the integration by redefining _F1 → D1_F1
eval,Int=int,_F1 = D1_F1
_F2t,z=−−E__0x_F1x,z−A__zxE__0x
Restore the original variables
_F2t,z=−−Et,x,z_F1x,z−∂∂zEt,x,zEt,x,z
This resulting pde is already an ODE in disguise, from where the general solution to the original problem pde__4 is
pdsolve
Et,x,z=_F3t,xⅇ∫_F2t,z−_F1x,zⅆz
This solution can be simplified further by redefining _F2 → D2_F2 and _F1 → D2_F1
eval,Int=int, _F2 = D2_F2, _F1 = D2_F1
Et,x,z=_F3t,xⅇ_F2t,z−_F1x,z
Verify this solution with the original pde__4
pdetest,pde__4
Doing all these steps in one go
pdsolvepde__4
Any PDE can be solved iterating the two steps below related to its first integrals, when they can be computed.
Step 1: Write the differential equation for a first integral - a function Φ of the PDE's jet space - by equating to 0 its total derivative with respect to (any) one of the independent variables of the original problem. The resulting PDE is of first order and can be tackled using the characteristic strip method.
Step 2: When the PDE for Φ can be solved, the solution is an arbitrary function of the differential invariants of the PDE for Φ. Equate one such differential invariant to an arbitrary function of the remaining independent variables of the problem, resulting in a pde of reduced order with respect to the original PDE problem, a first integral. At this point one can either repeat the process to calculate more first integrals in order to attempt an algebraic elimination (multiple reduction of order) or move again to Step 1 with the reduced order PDE.
withPDEtools:
pde__1≔∂∂xux,y−∂∂yux,y
pde__1≔∂∂xux,y−∂2∂x∂yux,y
FirstIntegralSolverpde__1,ux,y;
ux,y=∫_F1yⅇ−yⅆy+_F2xⅇy
pdetest,
The solving process step-by-step
Construct an expression for the first integral ΦQ, where Q=x,y,u,ux,uy, the jet space of pde__1
Q≔x,y,u,ux,uy
declareΦQ
Phix,y,u,ux,uywill now be displayed asΦ
Take the total derivative of ΦQ with respect to x
D_DxPhiQ,x,ux,y=0
diffΦx,y,u,ux,uy,x+diffΦx,y,u,ux,uy,uux+diffΦx,y,u,ux,uy,uxux,x+diffΦx,y,u,ux,uy,uyux,y=0
Take this PDE modulo pde__1, and that is all the formulation of the problem
isolateToJetpde__1,ux,y, ux,y
ux,y=ux
PDE__Φ ≔subs,
diffΦx,y,u,ux,uy,x+diffΦx,y,u,ux,uy,uux+diffΦx,y,u,ux,uy,uxux,x+diffΦx,y,u,ux,uy,uyux=0
Solve PDE__Φ, and in doing so, the solution cannot depend on ux,x (it can only depend on Q = x,y,u,ux,uy). You can achieve that using the ivars option of pdsolve
pdsolvePDE__Φ,ivars=ux,x
Φx,y,u,ux,uy=_F1y,−u+uy
Since Φ is a first integral, the arguments of the arbitrary function entering this solution are also first integrals. Take the second one, which involves 1st order derivatives, and equate it to an arbitrary function all the independent variables but x (the variable used when taking the total derivative of Φ). In this case, there is only one other variable, y.
op1,2,2,=_F1y
−u+uy=_F1y
Rewrite this result in function notation
FromJet,ux,y
−ux,y+diffux,y,y=_F1y
This result is already an ODE in disguise; solve it to arrive at the general solution of pde__1
A more involved example:
pde__2≔A__1∂2∂x__4∂x__1wx__1,x__2,x__3,x__4+A__2∂2∂x__4∂x__2wx__1,x__2,x__3,x__4+A__3∂2∂x__4∂x__3wx__1,x__2,x__3,x__4+A__6+A__5∂∂x__4wx__1,x__2,x__3,x__4+A__7A__1∂∂x__1wx__1,x__2,x__3,x__4+A__2∂∂x__2wx__1,x__2,x__3,x__4+A__3∂∂x__3wx__1,x__2,x__3,x__4+A__5wx__1,x__2,x__3,x__4+A__4+A__8A__1∂∂x__1wx__1,x__2,x__3,x__4+A__2∂∂x__2wx__1,x__2,x__3,x__4+A__3∂∂x__3wx__1,x__2,x__3,x__4+A__5wx__1,x__2,x__3,x__4+A__42:
PDEtools:-FirstIntegralSolverpde__2,wx__1,x__2,x__3,x__4;
wx__1,x__2,x__3,x__4=−12IntatexpA__5_aA__12A__3A__4A__8+A__3A__7+tan12A__324A__6A__8−A__72_F1_a,A__1x__2+A__2_a−A__2x__1A__1,A__1x__3+A__3_a−A__3x__1A__1+x__4A__3A__324A__6A__8−A__72A__1A__3A__8,_a=x__1−2_F2A__1x__2−A__2x__1A__1,A__1x__3−A__3x__1A__1,x__4exp−A__5x__1A__1
