6.4.27 8.4

6.4.27.1 [1189] Problem 1

problem number 1189

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["Global`*"]; 
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]];
 

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

Maple ✓

restart; 
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'));
 

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

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6.4.27.2 [1190] Problem 2

problem number 1190

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["Global`*"]; 
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]];
 

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

Maple ✓

restart; 
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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6.4.27.3 [1191] Problem 3

problem number 1191

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["Global`*"]; 
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]];
 

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

Maple ✓

restart; 
pde :=  f(x)*diff(w(x,y),x)+g(x)*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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6.4.27.4 [1192] Problem 4

problem number 1192

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["Global`*"]; 
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]];
 

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

Maple ✓

restart; 
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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6.4.27.5 [1193] Problem 5

problem number 1193

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["Global`*"]; 
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]];
 

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

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

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

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6.4.27.6 [1194] Problem 6

problem number 1194

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["Global`*"]; 
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]];
 

Failed

Maple ✗

restart; 
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'));
 

sol=()

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6.4.27.7 [1195] Problem 7

problem number 1195

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["Global`*"]; 
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]];
 

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)*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