6.5.23 8.1

6.5.23.1 [1343] Problem 1
6.5.23.2 [1344] Problem 2
6.5.23.3 [1345] Problem 3
6.5.23.4 [1346] Problem 4
6.5.23.5 [1347] Problem 5
6.5.23.6 [1348] Problem 6
6.5.23.7 [1349] Problem 7
6.5.23.8 [1350] Problem 8
6.5.23.9 [1351] Problem 9
6.5.23.10 [1352] Problem 10
6.5.23.11 [1353] Problem 11
6.5.23.12 [1354] Problem 12

6.5.23.1 [1343] Problem 1

problem number 1343

Added April 13, 2019.

Problem Chapter 5.8.1.1, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ a w_x + b w_y = f(x) w + g(x) \]

Mathematica

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

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

Maple

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

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

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6.5.23.2 [1344] Problem 2

problem number 1344

Added April 13, 2019.

Problem Chapter 5.8.1.2, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ a w_x + b w_y = (c y+k) w + f(x) \]

Mathematica

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

\[\left \{\left \{w(x,y)\to e^{\frac {x (2 a (c y+k)-b c x)}{2 a^2}} \left (\int _1^x\frac {\exp \left (-\frac {K[1] (2 a (k+c y)+b c (K[1]-2 x))}{2 a^2}\right ) f(K[1])}{a}dK[1]+c_1\left (y-\frac {b x}{a}\right )\right )\right \}\right \}\]

Maple

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

\[w \left ( x,y \right ) = \left ( \int ^{x}\!{\frac {f \left ( {\it \_a} \right ) }{a}{{\rm e}^{-{\frac {{\it \_a}}{{a}^{2}} \left ( \left ( cy+k \right ) a-b \left ( x-{\frac {{\it \_a}}{2}} \right ) c \right ) }}}}{d{\it \_a}}+{\it \_F1} \left ( {\frac {ay-xb}{a}} \right ) \right ) {{\rm e}^{{\frac {x}{{a}^{2}} \left ( \left ( cy+k \right ) a-{\frac {bcx}{2}} \right ) }}}\]

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6.5.23.3 [1345] Problem 3

problem number 1345

Added April 13, 2019.

Problem Chapter 5.8.1.3, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ a w_x + b w_y = f(x) y w + g(x) \]

Mathematica

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

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

Maple

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

\[w \left ( x,y \right ) = \left ( \int ^{x}\!{\frac {g \left ( {\it \_b} \right ) }{a}{{\rm e}^{-{\frac {\int \!f \left ( {\it \_b} \right ) \left ( \left ( -x+{\it \_b} \right ) b+ay \right ) \,{\rm d}{\it \_b}}{{a}^{2}}}}}}{d{\it \_b}}+{\it \_F1} \left ( {\frac {ay-xb}{a}} \right ) \right ) {{\rm e}^{\int ^{x}\!{\frac {f \left ( {\it \_a} \right ) \left ( ay-b \left ( x-{\it \_a} \right ) \right ) }{{a}^{2}}}{d{\it \_a}}}}\]

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6.5.23.4 [1346] Problem 4

problem number 1346

Added April 13, 2019.

Problem Chapter 5.8.1.4, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ a x w_x + b y w_y = f(x) w + g(x) \]

Mathematica

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

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

Maple

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

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

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6.5.23.5 [1347] Problem 5

problem number 1347

Added April 13, 2019.

Problem Chapter 5.8.1.5, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ f(x) w_x + (a y + b) w_y = c w + g(x) \]

Mathematica

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

\[\left \{\left \{w(x,y)\to \exp \left (\int _1^x\frac {c}{f(K[3])}dK[3]\right ) \left (\int _1^x\frac {\exp \left (-\int _1^{K[4]}\frac {c}{f(K[3])}dK[3]\right ) g(K[4])}{f(K[4])}dK[4]+c_1\left (y \exp \left (-\int _1^x\frac {1}{f(K[1])}dK[1]\right )-\int _1^x\frac {(a+b) \exp \left (-\int _1^{K[2]}\frac {1}{f(K[1])}dK[1]\right )}{f(K[2])}dK[2]\right )\right )\right \}\right \}\]

Maple

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

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

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6.5.23.6 [1348] Problem 6

problem number 1348

Added April 13, 2019.

Problem Chapter 5.8.1.6, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ f(x) w_x + g(x) w_y = h(x) w + p(x) \]

Mathematica

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

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

Maple

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

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

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6.5.23.7 [1349] Problem 7

problem number 1349

Added April 13, 2019.

Problem Chapter 5.8.1.7, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ f(x) w_x + (g_1(x) y+ g_0(x)) w_y = h_1(x) w + h_0(x) \]

Mathematica

ClearAll["Global`*"]; 
pde =  f[x]*D[w[x, y], x] + (g1[x]*y+g0[x])*D[w[x, y], y] == h1[x]*w[x,y]+h0[x]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 

\[\left \{\left \{w(x,y)\to \exp \left (\int _1^x\frac {\text {h1}(K[3])}{f(K[3])}dK[3]\right ) \left (\int _1^x\frac {\exp \left (-\int _1^{K[4]}\frac {\text {h1}(K[3])}{f(K[3])}dK[3]\right ) \text {h0}(K[4])}{f(K[4])}dK[4]+c_1\left (y \exp \left (-\int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right )-\int _1^x\frac {\exp \left (-\int _1^{K[2]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \text {g0}(K[2])}{f(K[2])}dK[2]\right )\right )\right \}\right \}\]

Maple

restart; 
pde :=  f(x)*diff(w(x,y),x)+ (g1(x)*y+g0(x))*diff(w(x,y),y) = h1(x)*w(x,y)+h0(x); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y))),output='realtime'));
 

\[w \left ( x,y \right ) = \left ( \int \!{\frac {{\it h0} \left ( x \right ) }{f \left ( x \right ) }{{\rm e}^{-\int \!{\frac {{\it h1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}}\,{\rm d}x+{\it \_F1} \left ( -\int \!{\frac {{\it g0} \left ( x \right ) }{f \left ( x \right ) }{{\rm e}^{-\int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}}\,{\rm d}x+y{{\rm e}^{-\int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}} \right ) \right ) {{\rm e}^{\int \!{\frac {{\it h1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}\]

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6.5.23.8 [1350] Problem 8

problem number 1350

Added April 13, 2019.

Problem Chapter 5.8.1.8, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ f(x) w_x + (g_1(x) y+ g_0(x)) w_y = h_2(x) w + h_1(x) y + h0(x) \]

Mathematica

ClearAll["Global`*"]; 
pde =  f[x]*D[w[x, y], x] + (g1[x]*y+g0[x])*D[w[x, y], y] == h2[x]*w[x,y]+h1[x]*y+h0[x]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 

\[\left \{\left \{w(x,y)\to \exp \left (\int _1^x\frac {\text {h2}(K[3])}{f(K[3])}dK[3]\right ) \left (\int _1^x\frac {\exp \left (-\int _1^{K[4]}\frac {\text {h2}(K[3])}{f(K[3])}dK[3]\right ) \left (\text {h0}(K[4])+\exp \left (\int _1^{K[4]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \text {h1}(K[4]) \left (\exp \left (-\int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) y-\int _1^x\frac {\exp \left (-\int _1^{K[2]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \text {g0}(K[2])}{f(K[2])}dK[2]+\int _1^{K[4]}\frac {\exp \left (-\int _1^{K[2]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \text {g0}(K[2])}{f(K[2])}dK[2]\right )\right )}{f(K[4])}dK[4]+c_1\left (y \exp \left (-\int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right )-\int _1^x\frac {\exp \left (-\int _1^{K[2]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \text {g0}(K[2])}{f(K[2])}dK[2]\right )\right )\right \}\right \}\]

Maple

restart; 
pde :=  f(x)*diff(w(x,y),x)+ (g1(x)*y+g0(x))*diff(w(x,y),y) = h2(x)*w(x,y)+h1(x)*y+h0(x); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y))),output='realtime'));
 

\[w \left ( x,y \right ) = \left ( \int ^{x}\!{\frac {1}{f \left ( {\it \_g} \right ) } \left ( {\it h1} \left ( {\it \_g} \right ) \left ( y{{\rm e}^{-\int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}-\int \!{\frac {{\it g0} \left ( x \right ) }{f \left ( x \right ) }{{\rm e}^{-\int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}}\,{\rm d}x+\int \!{\frac {{\it g0} \left ( {\it \_g} \right ) }{f \left ( {\it \_g} \right ) }{{\rm e}^{-\int \!{\frac {{\it g1} \left ( {\it \_g} \right ) }{f \left ( {\it \_g} \right ) }}\,{\rm d}{\it \_g}}}}\,{\rm d}{\it \_g} \right ) {{\rm e}^{-\int \!{\frac {{\it h2} \left ( {\it \_g} \right ) }{f \left ( {\it \_g} \right ) }}\,{\rm d}{\it \_g}+\int \!{\frac {{\it g1} \left ( {\it \_g} \right ) }{f \left ( {\it \_g} \right ) }}\,{\rm d}{\it \_g}}}+{{\rm e}^{-\int \!{\frac {{\it h2} \left ( {\it \_g} \right ) }{f \left ( {\it \_g} \right ) }}\,{\rm d}{\it \_g}}}{\it h0} \left ( {\it \_g} \right ) \right ) }{d{\it \_g}}+{\it \_F1} \left ( -\int \!{\frac {{\it g0} \left ( x \right ) }{f \left ( x \right ) }{{\rm e}^{-\int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}}\,{\rm d}x+y{{\rm e}^{-\int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}} \right ) \right ) {{\rm e}^{\int \!{\frac {{\it h2} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}\]

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6.5.23.9 [1351] Problem 9

problem number 1351

Added April 13, 2019.

Problem Chapter 5.8.1.9, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ f(x) w_x + (g_1(x) y+ g_0(x) y^k) w_y = h_2(x) w + h_1(x) y^n + h0(x) \]

Mathematica

ClearAll["Global`*"]; 
pde =  f[x]*D[w[x, y], x] + (g1[x]*y+g0[x]*y^k)*D[w[x, y], y] == h2[x]*w[x,y]+h1[x]*y^n+h0[x]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 

\[\left \{\left \{w(x,y)\to \exp \left (\int _1^x\frac {\text {h2}(K[3])}{f(K[3])}dK[3]\right ) \left (\int _1^x\frac {\exp \left (-\int _1^{K[4]}\frac {\text {h2}(K[3])}{f(K[3])}dK[3]\right ) \left (\text {h1}(K[4]) \left (\left (\exp \left (-\int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]-(k-1) \int _1^{K[4]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) y^{-k} \left (\exp \left (\int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) (k-1) \int _1^x\frac {\exp \left ((k-1) \int _1^{K[2]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \text {g0}(K[2])}{f(K[2])}dK[2] y^k-\exp \left (\int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) (k-1) \int _1^{K[4]}\frac {\exp \left ((k-1) \int _1^{K[2]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \text {g0}(K[2])}{f(K[2])}dK[2] y^k+\exp \left (k \int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) y\right )\right ){}^{\frac {1}{1-k}}\right ){}^n+\text {h0}(K[4])\right )}{f(K[4])}dK[4]+c_1\left ((k-1) \int _1^x\frac {\exp \left ((k-1) \int _1^{K[2]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \text {g0}(K[2])}{f(K[2])}dK[2]+y^{1-k} \exp \left ((k-1) \int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right )\right )\right )\right \}\right \}\]

Maple

restart; 
pde :=  f(x)*diff(w(x,y),x)+ (g1(x)*y+g0(x)*y^k)*diff(w(x,y),y) = h2(x)*w(x,y)+h1(x)*y^n+h0(x); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y))),output='realtime'));
 

\[w \left ( x,y \right ) = \left ( \int ^{x}\!{\frac {1}{f \left ( {\it \_g} \right ) }{{\rm e}^{-\int \!{\frac {{\it h2} \left ( {\it \_g} \right ) }{f \left ( {\it \_g} \right ) }}\,{\rm d}{\it \_g}}} \left ( {\it h1} \left ( {\it \_g} \right ) \left ( \left ( \left ( -k+1 \right ) \int \!{\frac {{\it g0} \left ( {\it \_g} \right ) }{f \left ( {\it \_g} \right ) }{{\rm e}^{ \left ( k-1 \right ) \int \!{\frac {{\it g1} \left ( {\it \_g} \right ) }{f \left ( {\it \_g} \right ) }}\,{\rm d}{\it \_g}}}}\,{\rm d}{\it \_g}+ \left ( k-1 \right ) \int \!{\frac {{\it g0} \left ( x \right ) }{f \left ( x \right ) }{{\rm e}^{ \left ( k-1 \right ) \int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}}\,{\rm d}x+{y}^{-k+1}{{\rm e}^{ \left ( k-1 \right ) \int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}} \right ) ^{- \left ( k-1 \right ) ^{-1}}{{\rm e}^{\int \!{\frac {{\it g1} \left ( {\it \_g} \right ) }{f \left ( {\it \_g} \right ) }}\,{\rm d}{\it \_g}}} \right ) ^{n}+{\it h0} \left ( {\it \_g} \right ) \right ) }{d{\it \_g}}+{\it \_F1} \left ( \left ( k-1 \right ) \int \!{\frac {{\it g0} \left ( x \right ) }{f \left ( x \right ) }{{\rm e}^{ \left ( k-1 \right ) \int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}}\,{\rm d}x+{y}^{-k+1}{{\rm e}^{ \left ( k-1 \right ) \int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}} \right ) \right ) {{\rm e}^{\int \!{\frac {{\it h2} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}\]

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6.5.23.10 [1352] Problem 10

problem number 1352

Added April 13, 2019.

Problem Chapter 5.8.1.10, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ f(x) w_x + (g_1(x)+ g_0(x) e^{\lambda y}) w_y = h_2(x) w + h_1(x) e^{\beta y} + h0(x) \]

Mathematica

ClearAll["Global`*"]; 
pde =  f[x]*D[w[x, y], x] + (g1[x]+g0[x]*Exp[lambda*y])*D[w[x, y], y] == h2[x]*w[x,y]+h1[x]*Exp[beta*y]+h0[x]; 
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) = h2(x)*w(x,y)+h1(x)*exp(beta*y)+h0(x); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y))),output='realtime'));
 

\[w \left ( x,y \right ) = \left ( \int ^{x}\!{\frac {1}{f \left ( {\it \_g} \right ) } \left ( {\it h1} \left ( {\it \_g} \right ) \left ( \left ( \lambda \,\int \!{\frac {{\it g0} \left ( x \right ) }{f \left ( x \right ) }{{\rm e}^{\lambda \,\int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}}\,{\rm d}x-\int \!{\frac {{\it g0} \left ( {\it \_g} \right ) }{f \left ( {\it \_g} \right ) }{{\rm e}^{\lambda \,\int \!{\frac {{\it g1} \left ( {\it \_g} \right ) }{f \left ( {\it \_g} \right ) }}\,{\rm d}{\it \_g}}}}\,{\rm d}{\it \_g}\lambda +{{\rm e}^{\lambda \, \left ( -y+\int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x \right ) }} \right ) ^{-1} \right ) ^{{\frac {\beta }{\lambda }}}{{\rm e}^{-\int \!{\frac {{\it h2} \left ( {\it \_g} \right ) }{f \left ( {\it \_g} \right ) }}\,{\rm d}{\it \_g}+\beta \,\int \!{\frac {{\it g1} \left ( {\it \_g} \right ) }{f \left ( {\it \_g} \right ) }}\,{\rm d}{\it \_g}}}+{{\rm e}^{-\int \!{\frac {{\it h2} \left ( {\it \_g} \right ) }{f \left ( {\it \_g} \right ) }}\,{\rm d}{\it \_g}}}{\it h0} \left ( {\it \_g} \right ) \right ) }{d{\it \_g}}+{\it \_F1} \left ( {\frac {1}{\lambda } \left ( -\lambda \,\int \!{\frac {{\it g0} \left ( x \right ) }{f \left ( x \right ) }{{\rm e}^{\lambda \,\int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}}\,{\rm d}x-{{\rm e}^{\lambda \, \left ( -y+\int \!{\frac {{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x \right ) }} \right ) } \right ) \right ) {{\rm e}^{\int \!{\frac {{\it h2} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}\]

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6.5.23.11 [1353] Problem 11

problem number 1353

Added April 13, 2019.

Problem Chapter 5.8.1.11, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ f_1(x) y^k w_x + f_2(x) w_y = g(x) w + h(x) \]

Mathematica

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

\[\left \{\left \{w(x,y)\to \exp \left (\int _1^x\frac {g(K[2]) \left (\left (y^{k+1}-(k+1) \int _1^x\frac {\text {f2}(K[1])}{\text {f1}(K[1])}dK[1]+(k+1) \int _1^{K[2]}\frac {\text {f2}(K[1])}{\text {f1}(K[1])}dK[1]\right ){}^{\frac {1}{k+1}}\right ){}^{-k}}{\text {f1}(K[2])}dK[2]\right ) \left (\int _1^x\frac {\exp \left (-\int _1^{K[3]}\frac {g(K[2]) \left (\left (y^{k+1}-(k+1) \int _1^x\frac {\text {f2}(K[1])}{\text {f1}(K[1])}dK[1]+(k+1) \int _1^{K[2]}\frac {\text {f2}(K[1])}{\text {f1}(K[1])}dK[1]\right ){}^{\frac {1}{k+1}}\right ){}^{-k}}{\text {f1}(K[2])}dK[2]\right ) h(K[3]) \left (\left (y^{k+1}-(k+1) \int _1^x\frac {\text {f2}(K[1])}{\text {f1}(K[1])}dK[1]+(k+1) \int _1^{K[3]}\frac {\text {f2}(K[1])}{\text {f1}(K[1])}dK[1]\right ){}^{\frac {1}{k+1}}\right ){}^{-k}}{\text {f1}(K[3])}dK[3]+c_1\left (\frac {y^{k+1}}{k+1}-\int _1^x\frac {\text {f2}(K[1])}{\text {f1}(K[1])}dK[1]\right )\right )\right \}\right \}\]

Maple

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

\[w \left ( x,y \right ) = \left ( \int ^{x}\!{\frac {h \left ( {\it \_f} \right ) }{{\it f1} \left ( {\it \_f} \right ) } \left ( \left ( \left ( 1+k \right ) \int \!{\frac {{\it f2} \left ( {\it \_f} \right ) }{{\it f1} \left ( {\it \_f} \right ) }}\,{\rm d}{\it \_f}+ \left ( -k-1 \right ) \int \!{\frac {{\it f2} \left ( x \right ) }{{\it f1} \left ( x \right ) }}\,{\rm d}x+{y}^{k}y \right ) ^{ \left ( 1+k \right ) ^{-1}} \right ) ^{-k}{{\rm e}^{-\int \!{\frac {g \left ( {\it \_f} \right ) }{{\it f1} \left ( {\it \_f} \right ) } \left ( \left ( \left ( 1+k \right ) \int \!{\frac {{\it f2} \left ( {\it \_f} \right ) }{{\it f1} \left ( {\it \_f} \right ) }}\,{\rm d}{\it \_f}+ \left ( -k-1 \right ) \int \!{\frac {{\it f2} \left ( x \right ) }{{\it f1} \left ( x \right ) }}\,{\rm d}x+{y}^{k}y \right ) ^{ \left ( 1+k \right ) ^{-1}} \right ) ^{-k}}\,{\rm d}{\it \_f}}}}{d{\it \_f}}+{\it \_F1} \left ( \left ( -k-1 \right ) \int \!{\frac {{\it f2} \left ( x \right ) }{{\it f1} \left ( x \right ) }}\,{\rm d}x+{y}^{k}y \right ) \right ) {{\rm e}^{\int ^{x}\!{\frac {g \left ( {\it \_b} \right ) }{{\it f1} \left ( {\it \_b} \right ) } \left ( \left ( \left ( 1+k \right ) \int \!{\frac {{\it f2} \left ( {\it \_b} \right ) }{{\it f1} \left ( {\it \_b} \right ) }}\,{\rm d}{\it \_b}+ \left ( -k-1 \right ) \int \!{\frac {{\it f2} \left ( x \right ) }{{\it f1} \left ( x \right ) }}\,{\rm d}x+{y}^{k}y \right ) ^{ \left ( 1+k \right ) ^{-1}} \right ) ^{-k}}{d{\it \_b}}}}\]

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6.5.23.12 [1354] Problem 12

problem number 1354

Added April 13, 2019.

Problem Chapter 5.8.1.12, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ f_1(x) e^{\lambda y} w_x + f_2(x) w_y = g(x) w + h(x) \]

Mathematica

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

\[\left \{\left \{w(x,y)\to \exp \left (\int _1^x\frac {g(K[2])}{\text {f1}(K[2]) \left (-\lambda \int _1^x\frac {\text {f2}(K[1])}{\text {f1}(K[1])}dK[1]+e^{\lambda y}+\lambda \int _1^{K[2]}\frac {\text {f2}(K[1])}{\text {f1}(K[1])}dK[1]\right )}dK[2]\right ) \left (\int _1^x\frac {\exp \left (-\int _1^{K[3]}\frac {g(K[2])}{\text {f1}(K[2]) \left (-\lambda \int _1^x\frac {\text {f2}(K[1])}{\text {f1}(K[1])}dK[1]+e^{\lambda y}+\lambda \int _1^{K[2]}\frac {\text {f2}(K[1])}{\text {f1}(K[1])}dK[1]\right )}dK[2]\right ) h(K[3])}{\text {f1}(K[3]) \left (-\lambda \int _1^x\frac {\text {f2}(K[1])}{\text {f1}(K[1])}dK[1]+e^{\lambda y}+\lambda \int _1^{K[3]}\frac {\text {f2}(K[1])}{\text {f1}(K[1])}dK[1]\right )}dK[3]+c_1\left (\frac {e^{\lambda y}}{\lambda }-\int _1^x\frac {\text {f2}(K[1])}{\text {f1}(K[1])}dK[1]\right )\right )\right \}\right \}\]

Maple

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

\[w \left ( x,y \right ) = \left ( \int ^{x}\!{\frac {h \left ( {\it \_f} \right ) }{{\it f1} \left ( {\it \_f} \right ) }{{\rm e}^{-{\frac {1}{\lambda }\int \!{\frac {g \left ( {\it \_f} \right ) }{{\it f1} \left ( {\it \_f} \right ) } \left ( \int \!{\frac {{\it f2} \left ( {\it \_f} \right ) }{{\it f1} \left ( {\it \_f} \right ) }}\,{\rm d}{\it \_f}+{\frac {1}{\lambda } \left ( {{\rm e}^{\lambda \,y}}-\int \!{\frac {{\it f2} \left ( x \right ) }{{\it f1} \left ( x \right ) }}\,{\rm d}x\lambda \right ) } \right ) ^{-1}}\,{\rm d}{\it \_f}}}} \left ( -\int \!{\frac {{\it f2} \left ( x \right ) }{{\it f1} \left ( x \right ) }}\,{\rm d}x\lambda +\int \!{\frac {{\it f2} \left ( {\it \_f} \right ) }{{\it f1} \left ( {\it \_f} \right ) }}\,{\rm d}{\it \_f}\lambda +{{\rm e}^{\lambda \,y}} \right ) ^{-1}}{d{\it \_f}}+{\it \_F1} \left ( {\frac {1}{\lambda } \left ( {{\rm e}^{\lambda \,y}}-\int \!{\frac {{\it f2} \left ( x \right ) }{{\it f1} \left ( x \right ) }}\,{\rm d}x\lambda \right ) } \right ) \right ) {{\rm e}^{\int ^{x}\!{\frac {g \left ( {\it \_b} \right ) }{{\it f1} \left ( {\it \_b} \right ) } \left ( -\int \!{\frac {{\it f2} \left ( x \right ) }{{\it f1} \left ( x \right ) }}\,{\rm d}x\lambda +\int \!{\frac {{\it f2} \left ( {\it \_b} \right ) }{{\it f1} \left ( {\it \_b} \right ) }}\,{\rm d}{\it \_b}\lambda +{{\rm e}^{\lambda \,y}} \right ) ^{-1}}{d{\it \_b}}}}\]

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