6.3.28 8.4

6.3.28.1 [1014] Problem 1

problem number 1014

Added Feb. 17, 2019.

Problem Chapter 3.8.4.1 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

Solve for \(w(x,y)\) \[ w_x + a w_y = f(x,y) \]

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[w[x, y], x] + a*D[w[x, y], y] == f[x, y]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 

\[\left \{\left \{w(x,y)\to \int _1^xf(K[1],-a x+y+a K[1])dK[1]+c_1(y-a x)\right \}\right \}\]

Maple ✓

restart; 
pde :=  diff(w(x,y),x) +a*diff(w(x,y),y) =  f(x,y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
 

\[w \left ( x,y \right ) =\int ^{x}\!f \left ( {\it \_a}, \left ( {\it \_a}-x \right ) a+y \right ) {d{\it \_a}}+{\it \_F1} \left ( -ax+y \right ) \]

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6.3.28.2 [1015] Problem 2

problem number 1015

Added Feb. 17, 2019.

Problem Chapter 3.8.4.2 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

Solve for \(w(x,y)\) \[ a x w_x + b y w_y = f(x,y) \]

Mathematica ✓

ClearAll["Global`*"]; 
pde =  a*x*D[w[x, y], x] + b*y*D[w[x, y], y] == f[x, y]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 

\[\left \{\left \{w(x,y)\to \int _1^x\frac {f\left (K[1],x^{-\frac {b}{a}} y K[1]^{\frac {b}{a}}\right )}{a K[1]}dK[1]+c_1\left (y x^{-\frac {b}{a}}\right )\right \}\right \}\]

Maple ✓

restart; 
pde :=  a*x*diff(w(x,y),x) +b*y*diff(w(x,y),y) =  f(x,y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
 

\[w \left ( x,y \right ) =\int ^{x}\!{\frac {1}{{\it \_a}\,a}f \left ( {\it \_a},y{x}^{-{\frac {b}{a}}}{{\it \_a}}^{{\frac {b}{a}}} \right ) }{d{\it \_a}}+{\it \_F1} \left ( y{x}^{-{\frac {b}{a}}} \right ) \]

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6.3.28.3 [1016] Problem 3

problem number 1016

Added Feb. 17, 2019.

Problem Chapter 3.8.4.3 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

Solve for \(w(x,y)\) \[ f(x) w_x + g(x) y w_y = h(x,y) \]

Mathematica ✓

ClearAll["Global`*"]; 
pde =  f[x]*D[w[x, y], x] + g[x]*y*D[w[x, y], y] == h[x, y]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 

\[\left \{\left \{w(x,y)\to c_1\left (y \exp \left (-\int _1^x\frac {g(K[1])}{f(K[1])}dK[1]\right )\right )+\int _1^x\frac {h\left (K[2],\exp \left (\int _1^{K[2]}\frac {g(K[1])}{f(K[1])}dK[1]-\int _1^x\frac {g(K[1])}{f(K[1])}dK[1]\right ) y\right )}{f(K[2])}dK[2]\right \}\right \}\]

Maple ✓

restart; 
pde :=  f(x)*diff(w(x,y),x) +g(x)*y*diff(w(x,y),y) =  h(x,y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
 

\[w \left ( x,y \right ) =\int ^{x}\!{\frac {1}{f \left ( {\it \_b} \right ) }h \left ( {\it \_b},y{{\rm e}^{-\int \!{\frac {g \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x+\int \!{\frac {g \left ( {\it \_b} \right ) }{f \left ( {\it \_b} \right ) }}\,{\rm d}{\it \_b}}} \right ) }{d{\it \_b}}+{\it \_F1} \left ( y{{\rm e}^{-\int \!{\frac {g \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}} \right ) \]

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6.3.28.4 [1017] Problem 4

problem number 1017

Added Feb. 17, 2019.

Problem Chapter 3.8.4.4 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

Solve for \(w(x,y)\) \[ f(x) w_x + (g_1(x) y+ g_0(x)) w_y = h(x,y) \]

Mathematica ✓

ClearAll["Global`*"]; 
pde =  f[x]*D[w[x, y], x] + (g1[x]*y + g0[x])*D[w[x, y], y] == h[x, y]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 

\[\left \{\left \{w(x,y)\to c_1\left (y \exp \left (-\int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right )-\int _1^x\frac {\exp \left (-\int _1^{K[2]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \text {g0}(K[2])}{f(K[2])}dK[2]\right )+\int _1^x\frac {h\left (K[3],\exp \left (\int _1^{K[3]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \left (\exp \left (-\int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) y-\int _1^x\frac {\exp \left (-\int _1^{K[2]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \text {g0}(K[2])}{f(K[2])}dK[2]+\int _1^{K[3]}\frac {\exp \left (-\int _1^{K[2]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \text {g0}(K[2])}{f(K[2])}dK[2]\right )\right )}{f(K[3])}dK[3]\right \}\right \}\]

Maple ✓

restart; 
pde :=  f(x)*diff(w(x,y),x) +(g1(x)*y+g0(x))*diff(w(x,y),y) =  h(x,y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
 

\[w \left ( x,y \right ) =\int ^{x}\!{\frac {1}{f \left ( {\it \_f} \right ) }h \left ( {\it \_f}, \left ( \int \!{\frac {{\it g0} \left ( {\it \_f} \right ) }{f \left ( {\it \_f} \right ) }{{\rm e}^{-\int \!{\frac {{\it g1} \left ( {\it \_f} \right ) }{f \left ( {\it \_f} \right ) }}\,{\rm d}{\it \_f}}}}\,{\rm d}{\it \_f}-\int \!{\frac {{\it g0} \left ( x \right ) }{f \left ( x \right ) }{{\rm e}^{-\int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}}\,{\rm d}x+y{{\rm e}^{-\int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}} \right ) {{\rm e}^{\int \!{\frac {{\it g1} \left ( {\it \_f} \right ) }{f \left ( {\it \_f} \right ) }}\,{\rm d}{\it \_f}}} \right ) }{d{\it \_f}}+{\it \_F1} \left ( -\int \!{\frac {{\it g0} \left ( x \right ) }{f \left ( x \right ) }{{\rm e}^{-\int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}}\,{\rm d}x+y{{\rm e}^{-\int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}} \right ) \]

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6.3.28.5 [1018] Problem 5

problem number 1018

Added Feb. 17, 2019.

Problem Chapter 3.8.4.5 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

Solve for \(w(x,y)\) \[ f(x) w_x + (g_1(x) y+ g_0(x) y^k) w_y = h(x,y) \]

Mathematica ✓

ClearAll["Global`*"]; 
pde =  f[x]*D[w[x, y], x] + (g1[x]*y + g0[x]*y^k)*D[w[x, y], y] == h[x, y]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 

\[\left \{\left \{w(x,y)\to c_1\left ((k-1) \int _1^x\frac {\exp \left ((k-1) \int _1^{K[2]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \text {g0}(K[2])}{f(K[2])}dK[2]+y^{1-k} \exp \left ((k-1) \int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right )\right )+\int _1^x\frac {h\left (K[3],\left (\exp \left (-\int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]-(k-1) \int _1^{K[3]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) y^{-k} \left (\exp \left (\int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) (k-1) \int _1^x\frac {\exp \left ((k-1) \int _1^{K[2]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \text {g0}(K[2])}{f(K[2])}dK[2] y^k-\exp \left (\int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) (k-1) \int _1^{K[3]}\frac {\exp \left ((k-1) \int _1^{K[2]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \text {g0}(K[2])}{f(K[2])}dK[2] y^k+\exp \left (k \int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) y\right )\right ){}^{\frac {1}{1-k}}\right )}{f(K[3])}dK[3]\right \}\right \}\]

Maple ✓

restart; 
pde :=  f(x)*diff(w(x,y),x) +(g1(x)*y+g0(x)*y^k)*diff(w(x,y),y) =  h(x,y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
 

\[w \left ( x,y \right ) =\int ^{x}\!{\frac {1}{f \left ( {\it \_f} \right ) }h \left ( {\it \_f}, \left ( \left ( 1-k \right ) \int \!{\frac {{\it g0} \left ( {\it \_f} \right ) }{f \left ( {\it \_f} \right ) }{{\rm e}^{ \left ( k-1 \right ) \int \!{\frac {{\it g1} \left ( {\it \_f} \right ) }{f \left ( {\it \_f} \right ) }}\,{\rm d}{\it \_f}}}}\,{\rm d}{\it \_f}+ \left ( k-1 \right ) \int \!{\frac {{\it g0} \left ( x \right ) }{f \left ( x \right ) }{{\rm e}^{ \left ( k-1 \right ) \int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}}\,{\rm d}x+{y}^{1-k}{{\rm e}^{ \left ( k-1 \right ) \int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}} \right ) ^{- \left ( k-1 \right ) ^{-1}}{{\rm e}^{\int \!{\frac {{\it g1} \left ( {\it \_f} \right ) }{f \left ( {\it \_f} \right ) }}\,{\rm d}{\it \_f}}} \right ) }{d{\it \_f}}+{\it \_F1} \left ( \left ( k-1 \right ) \int \!{\frac {{\it g0} \left ( x \right ) }{f \left ( x \right ) }{{\rm e}^{ \left ( k-1 \right ) \int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}}\,{\rm d}x+{y}^{1-k}{{\rm e}^{ \left ( k-1 \right ) \int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}} \right ) \]

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6.3.28.6 [1019] Problem 6

problem number 1019

Added Feb. 17, 2019.

Problem Chapter 3.8.4.6 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

Solve for \(w(x,y)\)

\[ f(x) w_x + (g_1(x)+ g_0(x) e^{\lambda y}) w_y = h(x,y) \]

Mathematica ✗

ClearAll["Global`*"]; 
pde =  f[x]*D[w[x, y], x] + (g1[x] + g0[x]*Exp[lambda*y])*D[w[x, y], y] == h[x, y]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 

Failed

Maple ✓

restart; 
pde :=  f(x)*diff(w(x,y),x) +(g1(x)+g0(x)*exp(lambda*y))*diff(w(x,y),y) =  h(x,y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
 

\[w \left ( x,y \right ) =\int ^{x}\!{\frac {1}{f \left ( {\it \_f} \right ) }h \left ( {\it \_f},{\frac {1}{\lambda } \left ( \ln \left ( \left ( \lambda \,\int \!{\frac {{\it g0} \left ( x \right ) }{f \left ( x \right ) }{{\rm e}^{\lambda \,\int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}}\,{\rm d}x-\int \!{\frac {{\it g0} \left ( {\it \_f} \right ) }{f \left ( {\it \_f} \right ) }{{\rm e}^{\lambda \,\int \!{\frac {{\it g1} \left ( {\it \_f} \right ) }{f \left ( {\it \_f} \right ) }}\,{\rm d}{\it \_f}}}}\,{\rm d}{\it \_f}\lambda +{{\rm e}^{-\lambda \, \left ( y-\int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x \right ) }} \right ) ^{-1} \right ) +\lambda \,\int \!{\frac {{\it g1} \left ( {\it \_f} \right ) }{f \left ( {\it \_f} \right ) }}\,{\rm d}{\it \_f} \right ) } \right ) }{d{\it \_f}}+{\it \_F1} \left ( {\frac {1}{\lambda } \left ( -\lambda \,\int \!{\frac {{\it g0} \left ( x \right ) }{f \left ( x \right ) }{{\rm e}^{\lambda \,\int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}}\,{\rm d}x-{{\rm e}^{-\lambda \, \left ( y-\int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x \right ) }} \right ) } \right ) \]

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6.3.28.7 [1020] Problem 7

problem number 1020

Added Feb. 17, 2019.

Problem Chapter 3.8.4.7 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

Solve for \(w(x,y)\)

\[ f_1(x) g_1(y) w_x + f_2(x) g_2(y) w_y = h(x,y) \]

Mathematica ✗

ClearAll["Global`*"]; 
pde =  f1[x]*g1[y]*D[w[x, y], x] + f2[x]*g2[y]*D[w[x, y], y] == h[x, y]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 

Failed

Maple ✓

restart; 
pde :=  f1(x)*g1(y)*diff(w(x,y),x) +f2(x)*g2(y)*diff(w(x,y),y) =  h(x,y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
 

\[w \left ( x,y \right ) =\int ^{x}\!{\frac {1}{{\it f1} \left ( {\it \_f} \right ) }h \left ( {\it \_f},\RootOf \left ( \int \!{\frac {{\it f2} \left ( {\it \_f} \right ) }{{\it f1} \left ( {\it \_f} \right ) }}\,{\rm d}{\it \_f}-\int ^{{\it \_Z}}\!{\frac {{\it g1} \left ( {\it \_a} \right ) }{{\it g2} \left ( {\it \_a} \right ) }}{d{\it \_a}}-\int \!{\frac {{\it f2} \left ( x \right ) }{{\it f1} \left ( x \right ) }}\,{\rm d}x+\int \!{\frac {{\it g1} \left ( y \right ) }{{\it g2} \left ( y \right ) }}\,{\rm d}y \right ) \right ) \left ( {\it g1} \left ( \RootOf \left ( \int \!{\frac {{\it f2} \left ( {\it \_f} \right ) }{{\it f1} \left ( {\it \_f} \right ) }}\,{\rm d}{\it \_f}-\int ^{{\it \_Z}}\!{\frac {{\it g1} \left ( {\it \_a} \right ) }{{\it g2} \left ( {\it \_a} \right ) }}{d{\it \_a}}-\int \!{\frac {{\it f2} \left ( x \right ) }{{\it f1} \left ( x \right ) }}\,{\rm d}x+\int \!{\frac {{\it g1} \left ( y \right ) }{{\it g2} \left ( y \right ) }}\,{\rm d}y \right ) \right ) \right ) ^{-1}}{d{\it \_f}}+{\it \_F1} \left ( -\int \!{\frac {{\it f2} \left ( x \right ) }{{\it f1} \left ( x \right ) }}\,{\rm d}x+\int \!{\frac {{\it g1} \left ( y \right ) }{{\it g2} \left ( y \right ) }}\,{\rm d}y \right ) \] Contains RootOf