124 HFOPDE, chapter 4.8.4

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124.1 Problem 1

problem number 998

Added March 10, 2019.

Problem Chapter 4.8.4.1, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

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

Mathematica ✓

ClearAll[w, x, y, n, a, b, m, c, k, alpha, beta, gamma, A, C0, s]; 
 ClearAll[lambda, B, mu, d, g, B, v, f, h, q, p, delta, t]; 
 ClearAll[g1, g0, h2, h1, h0, f1, f2]; 
 pde = D[w[x, y], x] + a*D[w[x, y], y] == f[x, y]*w[x, y]; 
 sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 

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

Maple ✓

 
w:='w';x:='x';y:='y';a:='a';b:='b';n:='n';m:='m';c:='c'; 
k:='k';alpha:='alpha';beta:='beta';g:='g';A:='A';f:='f'; 
C:='C';lambda:='lambda';B:='B';mu:='mu';d:='d';s:='s';t:='t'; 
v:='v';q:='q';p:='p';l:='l';g1:='g1';g2:='g2';g0:='g0'; 
h0:='h0';h1:='h1';h2:='h2';f2:='f2';f3:='f3'; 
pde :=  diff(w(x,y),x)+a*diff(w(x,y),y) = f(x,y)*w(x,y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y))),output='realtime'));
 

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

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124.2 Problem 2

problem number 999

Added March 10, 2019.

Problem Chapter 4.8.4.2, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

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

Mathematica ✓

ClearAll[w, x, y, n, a, b, m, c, k, alpha, beta, gamma, A, C0, s]; 
 ClearAll[lambda, B, mu, d, g, B, v, f, h, q, p, delta, t]; 
 ClearAll[g1, g0, h2, h1, h0, f1, f2]; 
 pde = a*x*D[w[x, y], x] + b*y*D[w[x, y], y] == f[x, y]*w[x, y]; 
 sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 

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

Maple ✓

 
w:='w';x:='x';y:='y';a:='a';b:='b';n:='n';m:='m';c:='c'; 
k:='k';alpha:='alpha';beta:='beta';g:='g';A:='A';f:='f'; 
C:='C';lambda:='lambda';B:='B';mu:='mu';d:='d';s:='s';t:='t'; 
v:='v';q:='q';p:='p';l:='l';g1:='g1';g2:='g2';g0:='g0'; 
h0:='h0';h1:='h1';h2:='h2';f2:='f2';f3:='f3'; 
pde :=  a*x*diff(w(x,y),x)+b*y*diff(w(x,y),y) = f(x,y)*w(x,y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y))),output='realtime'));
 

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

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124.3 Problem 3

problem number 1000

Added March 10, 2019.

Problem Chapter 4.8.4.3, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

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

Mathematica ✓

ClearAll[w, x, y, n, a, b, m, c, k, alpha, beta, gamma, A, C0, s]; 
 ClearAll[lambda, B, mu, d, g, B, v, f, h, q, p, delta, t]; 
 ClearAll[g1, g0, h2, h1, h0, f1, f2]; 
 pde = f[x]*D[w[x, y], x] + g[x]*y*D[w[x, y], y] == h[x, y]*w[x, y]; 
 sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 

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

Maple ✓

 
w:='w';x:='x';y:='y';a:='a';b:='b';n:='n';m:='m';c:='c'; 
k:='k';alpha:='alpha';beta:='beta';g:='g';A:='A';f:='f'; 
C:='C';lambda:='lambda';B:='B';mu:='mu';d:='d';s:='s';t:='t'; 
v:='v';q:='q';p:='p';l:='l';g1:='g1';g2:='g2';g0:='g0'; 
h0:='h0';h1:='h1';h2:='h2';f2:='f2';f3:='f3'; 
pde :=  f(x)*diff(w(x,y),x)+g(x)*y*diff(w(x,y),y) = h(x,y)*w(x,y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y))),output='realtime'));
 

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

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124.4 Problem 4

problem number 1001

Added March 10, 2019.

Problem Chapter 4.8.4.4, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

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

Mathematica ✓

ClearAll[w, x, y, n, a, b, m, c, k, alpha, beta, gamma, A, C0, s]; 
 ClearAll[lambda, B, mu, d, g, B, v, f, h, q, p, delta, t]; 
 ClearAll[g1, g0, h2, h1, h0, f1, f2]; 
 pde = f[x]*D[w[x, y], x] + (g1[x]*y + g0[x])*D[w[x, y], y] == h[x, y]*w[x, y]; 
 sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]]; 
 sol = Simplify[sol];
 

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

Maple ✓

 
w:='w';x:='x';y:='y';a:='a';b:='b';n:='n';m:='m';c:='c'; 
k:='k';alpha:='alpha';beta:='beta';g:='g';A:='A';f:='f'; 
C:='C';lambda:='lambda';B:='B';mu:='mu';d:='d';s:='s';t:='t'; 
v:='v';q:='q';p:='p';l:='l';g1:='g1';g2:='g2';g0:='g0'; 
h0:='h0';h1:='h1';h2:='h2';f2:='f2';f3:='f3'; 
pde :=  f(x)*diff(w(x,y),x)+(g1(x)*y+g0(x))*diff(w(x,y),y) = h(x,y)*w(x,y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y))),output='realtime'));
 

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

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124.5 Problem 5

problem number 1002

Added March 10, 2019.

Problem Chapter 4.8.4.5, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

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

Mathematica ✓

ClearAll[w, x, y, n, a, b, m, c, k, alpha, beta, gamma, A, C0, s]; 
 ClearAll[lambda, B, mu, d, g, B, v, f, h, q, p, delta, t]; 
 ClearAll[g1, g0, h2, h1, h0, f1, f2]; 
 pde = f[x]*D[w[x, y], x] + (g1[x]*y + g0[x]*y^k)*D[w[x, y], y] == h[x, y]*w[x, y]; 
 sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]]; 
 sol = Simplify[sol];
 

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

Maple ✓

 
w:='w';x:='x';y:='y';a:='a';b:='b';n:='n';m:='m';c:='c'; 
k:='k';alpha:='alpha';beta:='beta';g:='g';A:='A';f:='f'; 
C:='C';lambda:='lambda';B:='B';mu:='mu';d:='d';s:='s';t:='t'; 
v:='v';q:='q';p:='p';l:='l';g1:='g1';g2:='g2';g0:='g0'; 
h0:='h0';h1:='h1';h2:='h2';f2:='f2';f3:='f3'; 
pde :=  f(x)*diff(w(x,y),x)+(g1(x)*y+g0(x)*y^k)*diff(w(x,y),y) = h(x,y)*w(x,y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y))),output='realtime')); 
sol:=simplify(sol);
 

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

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124.6 Problem 6

problem number 1003

Added March 10, 2019.

Problem Chapter 4.8.4.6, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

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

Mathematica ✗

ClearAll[w, x, y, n, a, b, m, c, k, alpha, beta, gamma, A, C0, s]; 
 ClearAll[lambda, B, mu, d, g, B, v, f, h, q, p, delta, t]; 
 ClearAll[g1, g0, h2, h1, h0, f1, f2]; 
 pde = f[x]*D[w[x, y], x] + (g1[x]*y + g0[x]*Exp[lambda*y])*D[w[x, y], y] == h[x, y]*w[x, y]; 
 sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]]; 
 sol = Simplify[sol];
 

\[ \text {Failed} \]

Maple ✗

 
w:='w';x:='x';y:='y';a:='a';b:='b';n:='n';m:='m';c:='c'; 
k:='k';alpha:='alpha';beta:='beta';g:='g';A:='A';f:='f'; 
C:='C';lambda:='lambda';B:='B';mu:='mu';d:='d';s:='s';t:='t'; 
v:='v';q:='q';p:='p';l:='l';g1:='g1';g2:='g2';g0:='g0'; 
h0:='h0';h1:='h1';h2:='h2';f2:='f2';f3:='f3'; 
pde :=  f(x)*diff(w(x,y),x)+(g1(x)*y+g0(x)*exp(lambda*y))*diff(w(x,y),y) = h(x,y)*w(x,y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y))),output='realtime'));
 

\[ \text { sol=() } \]

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124.7 Problem 7

problem number 1004

Added March 10, 2019.

Problem Chapter 4.8.4.6, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

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

Mathematica ✗

ClearAll[w, x, y, n, a, b, m, c, k, alpha, beta, gamma, A, C0, s]; 
 ClearAll[lambda, B, mu, d, g, B, v, f, h, q, p, delta, t]; 
 ClearAll[g1, g0, h2, h1, h0, f1, f2]; 
 pde = f1[x]*g1[y]*D[w[x, y], x] + f2[x]*g2[y]*D[w[x, y], y] == h[x, y]*w[x, y]; 
 sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 

\[ \text {Failed} \]

Maple ✓

 
w:='w';x:='x';y:='y';a:='a';b:='b';n:='n';m:='m';c:='c'; 
k:='k';alpha:='alpha';beta:='beta';g:='g';A:='A';f:='f'; 
C:='C';lambda:='lambda';B:='B';mu:='mu';d:='d';s:='s';t:='t'; 
v:='v';q:='q';p:='p';l:='l';g1:='g1';g2:='g2';g0:='g0'; 
h0:='h0';h1:='h1';h2:='h2';f2:='f2';f3:='f3'; 
pde :=  f1(x)*g1(y)*diff(w(x,y),x)+f2(x)*g2(y)*diff(w(x,y),y) = h(x,y)*w(x,y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y))),output='realtime'));
 

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