Added May 26, 2019.
Problem Chapter 6.6.3.1, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y,z)\)
Mathematica ✓
ClearAll["Global`*"]; pde = a*D[w[x, y,z], x] + b*D[w[x, y,z], y] +c*Tan[gamma*z]*D[w[x,y,z],z]==0; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y,z], {x, y,z}], 60*10]];
Maple ✓
restart; pde := a*diff(w(x,y,z),x)+ b*diff(w(x,y,z),y)+c*tan(gamma*z)*diff(w(x,y,z),z)= 0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y,z))),output='realtime'));
Hand solution
Solve
Using Lagrange-charpit
From first two pair of equation we obtain \(\frac {b}{a}x-y=C_{1}\) and from \(\frac {dx}{a}=\frac {dz}{c\tan \left ( \gamma z\right ) }\) we obtain \(\frac {c}{a}x-\frac {1}{\gamma }\ln \left ( \sin \left ( \gamma z\right ) \right ) =C_{2}\). Since \(dw=0\) and \(w=C_{3}\). Where \(C_{1},C_{2},C_{3}\) are constants. But \(C_{3}=F\left ( C_{1},C_{2}\right ) \) where \(F\) is arbitrary function. Hence
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Added May 26, 2019.
Problem Chapter 6.6.3.2, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y,z)\)
Mathematica ✓
ClearAll["Global`*"]; pde = a*D[w[x, y,z], x] + b*Tan[beta*y]*D[w[x, y,z], y] +c*Tan[lambda*x]*D[w[x,y,z],z]==0; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y,z], {x, y,z}], 60*10]];
Maple ✓
restart; pde := a*diff(w(x,y,z),x)+ b*tan(beta*y)*diff(w(x,y,z),y)+c*tan(lambda*x)*diff(w(x,y,z),z)= 0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y,z))),output='realtime'));
Hand solution
Solve
Using Lagrange-charpit
From first two pair of equations, integrating gives \(\frac {b}{a}x-\frac {1}{\beta }\ln \left ( \sin \left ( \beta y\right ) \right ) =C_{1}\) and from \(\frac {dx}{a}=\frac {dz}{c\tan \left ( \lambda x\right ) }\) we obtain \(\frac {c}{a}\tan \left ( \lambda x\right ) dx=dz\). Integrating gives \(-\frac {c}{a\lambda }\ln \left ( \cos \left ( \lambda x\right ) \right ) -z=C_{2}\). Since \(dw=0\) then \(w=C_{3}\). Where \(C_{1},C_{2},C_{3}\) are constants. But \(C_{3}=F\left ( C_{1},C_{2}\right ) \) where \(F\) is arbitrary function. Hence
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Added May 26, 2019.
Problem Chapter 6.6.3.3, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y,z)\)
Mathematica ✓
ClearAll["Global`*"]; pde = a*D[w[x, y,z], x] + b*Tan[beta*y]*D[w[x, y,z], y] +c*Tan[gamma*z]*D[w[x,y,z],z]==0; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y,z], {x, y,z}], 60*10]];
Maple ✓
restart; pde := a*diff(w(x,y,z),x)+ b*tan(beta*y)*diff(w(x,y,z),y)+c*tan(gamma*z)*diff(w(x,y,z),z)= 0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y,z))),output='realtime'));
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Added May 26, 2019.
Problem Chapter 6.6.3.4, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y,z)\)
Mathematica ✓
ClearAll["Global`*"]; pde = a*D[w[x, y,z], x] + b*Tan[beta*y]*D[w[x, y,z], y] +c*Tan[gamma*z]*D[w[x,y,z],z]==0; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y,z], {x, y,z}], 60*10]];
Maple ✓
restart; pde := a*diff(w(x,y,z),x)+ b*tan(beta*y)*diff(w(x,y,z),y)+c*tan(gamma*z)*diff(w(x,y,z),z)= 0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y,z))),output='realtime'));
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Added May 26, 2019.
Problem Chapter 6.6.3.5, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y,z)\)
Mathematica ✗
ClearAll["Global`*"]; pde = mu*nu*Tan[lambda*x]*D[w[x, y,z], x] + lambda*nu*Tan[mu*y]*D[w[x, y,z], y] +lambda*mu*Tan[nu*z]*D[w[x,y,z],z]==0; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y,z], {x, y,z}], 60*10]];
Failed
Maple ✓
restart; pde := mu*nu*tan(lambda*x)*diff(w(x,y,z),x)+ lambda*nu*tan(mu*y)*diff(w(x,y,z),y)+lambda*mu*tan(nu*z)*diff(w(x,y,z),z)= 0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y,z))),output='realtime'));
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