Added March 9, 2019.
Problem Chapter 4.6.5.1, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y)\)
\[ w_x + a w_y = (b \sin (\lambda x)+k \cos (\mu y)) w \]
Mathematica ✓
ClearAll["Global`*"]; pde = D[w[x, y], x] + a*D[w[x, y], y] == (b*Sin[lambda*x] + k*Cos[mu*y])*w[x, y]; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
\[\left \{\left \{w(x,y)\to c_1(y-a x) e^{\frac {k \sin (\mu y)}{a \mu }-\frac {b \cos (\lambda x)}{\lambda }}\right \}\right \}\]
Maple ✓
restart; pde := diff(w(x,y),x)+ a*diff(w(x,y),y) = (b*sin(lambda*x)+k*cos(mu*y))*w(x,y); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y))),output='realtime'));
\[w \left (x , y\right ) = \mathit {\_F1} \left (-a x +y \right ) {\mathrm e}^{\frac {-a b \mu \cos \left (\lambda x \right )+k \lambda \sin \left (\mu y \right )}{a \lambda \mu }}\]
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Added March 9, 2019.
Problem Chapter 4.6.5.2, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y)\)
\[ w_x + a \sin (\mu y) w_y = (b \sin (\lambda x)+k \tan (\mu y)) w \]
Mathematica ✓
ClearAll["Global`*"]; pde = D[w[x, y], x] + a*D[w[x, y], y] == (b*Sin[lambda*x] + k*Tan[mu*y])*w[x, y]; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
\[\left \{\left \{w(x,y)\to e^{-\frac {b \cos (\lambda x)}{\lambda }} c_1(y-a x) \cos ^{-\frac {k}{a \mu }}(\mu y)\right \}\right \}\]
Maple ✓
restart; pde := diff(w(x,y),x)+ a*diff(w(x,y),y) = (b*sin(lambda*x)+k*tan(mu*y))*w(x,y); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y))),output='realtime'));
\[w \left (x , y\right ) = \left (\cos ^{-\frac {k}{a \mu }}\left (\mu y \right )\right ) \mathit {\_F1} \left (-a x +y \right ) {\mathrm e}^{-\frac {b \cos \left (\lambda x \right )}{\lambda }}\]
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Added March 9, 2019.
Problem Chapter 4.6.5.3, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y)\)
\[ w_x + a \sin (\mu y) w_y = b \tan (\lambda x) w \]
Mathematica ✓
ClearAll["Global`*"]; pde = D[w[x, y], x] + a*Sin[mu*y]*D[w[x, y], y] == b*Tan[lambda*x]*w[x, y]; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
\[\left \{\left \{w(x,y)\to \cos ^{-\frac {b}{\lambda }}(\lambda x) c_1\left (\frac {\log \left (\tan \left (\frac {\mu y}{2}\right )\right )}{\mu }-a x\right )\right \}\right \}\]
Maple ✓
restart; pde := diff(w(x,y),x)+ a*sin(mu*y)*diff(w(x,y),y) = b*tan(lambda*x)*w(x,y); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y))),output='realtime'));
\[w \left (x , y\right ) = \left (\tan ^{2}\left (\lambda x \right )+1\right )^{\frac {b}{2 \lambda }} \mathit {\_F1} \left (\frac {\ln \left (\RootOf \left (\mu y -\arctan \left (\frac {2 \mathit {\_Z} \,{\mathrm e}^{a \mu x}}{\mathit {\_Z}^{2} {\mathrm e}^{2 a \mu x}+1}, -\frac {\mathit {\_Z}^{2} {\mathrm e}^{2 a \mu x}-1}{\mathit {\_Z}^{2} {\mathrm e}^{2 a \mu x}+1}\right )\right )\right )}{a \mu }\right )\]
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Added March 9, 2019.
Problem Chapter 4.6.5.4, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y)\)
\[ w_x + a \tan (\mu y) w_y = b \sin (\lambda x) w \]
Mathematica ✓
ClearAll["Global`*"]; pde = D[w[x, y], x] + a*Tan[mu*y]*D[w[x, y], y] == b*Sin[lambda*x]*w[x, y]; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
\[\left \{\left \{w(x,y)\to e^{-\frac {b \cos (\lambda x)}{\lambda }} c_1\left (\frac {\log (\sin (\mu y))}{\mu }-a x\right )\right \}\right \}\]
Maple ✓
restart; pde := diff(w(x,y),x)+ a*tan(mu*y)*diff(w(x,y),y) = b*sin(lambda*x)*w(x,y); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y))),output='realtime'));
\[w \left (x , y\right ) = \mathit {\_F1} \left (\frac {\ln \left (\mathrm {csgn}\left (\frac {1}{\cos \left (\mu y \right )}\right ) {\mathrm e}^{-a \mu x} \sin \left (\mu y \right )\right )}{a \mu }\right ) {\mathrm e}^{-\frac {b \cos \left (\lambda x \right )}{\lambda }}\]
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Added March 9, 2019.
Problem Chapter 4.6.5.5, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y)\)
\[ \sin (\lambda x) w_x + a w_y = b \cos (\mu y) w \]
Mathematica ✓
ClearAll["Global`*"]; pde = Sin[lambda*x]*D[w[x, y], x] + a*D[w[x, y], y] == b*Cos[mu*y]*w[x, y]; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
\[\left \{\left \{w(x,y)\to e^{\frac {b \sin (\mu y)}{a \mu }} c_1\left (\frac {-a \log \left (\sin \left (\frac {\lambda x}{2}\right )\right )+a \log \left (\cos \left (\frac {\lambda x}{2}\right )\right )+\lambda y}{\lambda }\right )\right \}\right \}\]
Maple ✓
restart; pde := sin(lambda*x)*diff(w(x,y),x)+ a*diff(w(x,y),y) = b*cos(mu*y)*w(x,y); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y))),output='realtime'));
\[w \left (x , y\right ) = \mathit {\_F1} \left (\frac {-a \ln \left (-\cot \left (\lambda x \right )+\csc \left (\lambda x \right )\right )+\lambda y}{\lambda }\right ) {\mathrm e}^{\frac {b \sin \left (\mu y \right )}{a \mu }}\]
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Added March 9, 2019.
Problem Chapter 4.6.5.6, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y)\)
\[ \cot (\lambda x) w_x + a w_y = b \tan (\mu y) w \]
Mathematica ✓
ClearAll["Global`*"]; pde = Cot[lambda*x]*D[w[x, y], x] + a*D[w[x, y], y] == b*Tan[mu*y]*w[x, y]; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
\[\left \{\left \{w(x,y)\to \cos ^{-\frac {b}{a \mu }}(\mu y) c_1\left (\frac {a \log (\cos (\lambda x))}{\lambda }+y\right )\right \}\right \}\]
Maple ✓
restart; pde := cot(lambda*x)*diff(w(x,y),x)+ a*diff(w(x,y),y) = b*tan(mu*y)*w(x,y); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y))),output='realtime'));
\[w \left (x , y\right ) = \left (\cos ^{-\frac {b}{a \mu }}\left (\mu y \right )\right ) \mathit {\_F1} \left (\frac {-a \ln \left (\cot ^{2}\left (\lambda x \right )+1\right )+2 a \ln \left (\cot \left (\lambda x \right )\right )+2 \lambda y}{2 \lambda }\right )\]
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