7.6.12 5.1

7.6.12.1 [1496] Problem 1

problem number 1496

Added May 26, 2019.

Problem Chapter 6.5.1.1, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ a w_x + b w_y + c \ln (\beta y) \ln (\lambda z) w_z = 0 \]

Mathematica ✓

ClearAll["Global`*"]; 
pde =  a*D[w[x, y,z], x] + b*D[w[x, y,z], y] +c*Log[beta*y]*Log[lambda*z]*D[w[x,y,z],z]==0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y,z], {x, y,z}], 60*10]];
 

\[\left \{\left \{w(x,y,z)\to c_1\left (y-\frac {b x}{a},\frac {c x}{a}-\frac {c y \log (\beta y)}{b}+\frac {\text {li}(\lambda z)}{\lambda }\right )\right \}\right \}\]

Maple ✓

restart; 
pde :=  a*diff(w(x,y,z),x)+ b*diff(w(x,y,z),y)+c*ln(beta*y)*ln(lambda*z)*diff(w(x,y,z),z)= 0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y,z))),output='realtime'));
 

\[w \left (x , y , z\right ) = \textit {\_F1} \left (\frac {-a y +b x}{b}, \frac {-\left (\ln \left (\beta y \right )-1\right ) c \lambda y -b \Ei \left (1, -\ln \left (\lambda z \right )\right )}{c \lambda }\right )\]

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7.6.12.2 [1497] Problem 2

problem number 1497

Added May 26, 2019.

Problem Chapter 6.5.1.2, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ a w_x + b \ln (\beta x) w_y + c \ln (\lambda x) w_z = 0 \]

Mathematica ✓

ClearAll["Global`*"]; 
pde =  a*D[w[x, y,z], x] + b*Log[beta*x]*D[w[x, y,z], y] +c*Log[lambda*x]*D[w[x,y,z],z]==0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y,z], {x, y,z}], 60*10]];
 

\[\left \{\left \{w(x,y,z)\to c_1\left (\frac {a y-b x \log (\beta x)+b x}{a},\frac {a z-c x \log (\lambda x)+c x}{a}\right )\right \}\right \}\]

Maple ✓

restart; 
pde :=  a*diff(w(x,y,z),x)+ b*ln(beta*x)*diff(w(x,y,z),y)+c*ln(lambda*x)*diff(w(x,y,z),z)= 0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y,z))),output='realtime'));
 

\[w \left (x , y , z\right ) = \textit {\_F1} \left (\frac {-b x \ln \left (\beta x \right )+a y +b x}{a}, \frac {-c x \ln \left (\lambda x \right )+a z +c x}{a}\right )\]

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7.6.12.3 [1498] Problem 3

problem number 1498

Added May 26, 2019.

Problem Chapter 6.5.1.3, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ a w_x + b \ln (\beta x) \ln (\lambda y) w_y + c \ln (\mu x) \ln (\gamma z) w_z = 0 \]

Mathematica ✓

ClearAll["Global`*"]; 
pde =  a*D[w[x, y,z], x] + b*Log[beta*x]*Log[lambda*y]*D[w[x, y,z], y] +c*Log[mu*x]*Log[gamma*z]*D[w[x,y,z],z]==0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y,z], {x, y,z}], 60*10]];
 

\[\left \{\left \{w(x,y,z)\to c_1\left (\frac {\text {li}(\lambda y)}{\lambda }-\frac {b x (\log (\beta x)-1)}{a},\frac {\text {li}(\gamma z)}{\gamma }-\frac {c x (\log (\mu x)-1)}{a}\right )\right \}\right \}\]

Maple ✓

restart; 
pde :=  a*diff(w(x,y,z),x)+ b*ln(beta*x)*ln(lambda*y)*diff(w(x,y,z),y)+c*ln(mu*x)*ln(gamma*z)*diff(w(x,y,z),z)= 0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y,z))),output='realtime'));
 

\[w \left (x , y , z\right ) = \textit {\_F1} \left (\frac {\left (\ln \left (\beta x \right )-1\right ) b \lambda x +a \Ei \left (1, -\ln \left (\lambda y \right )\right )}{a \lambda }, -\frac {\left (-\gamma \left (-\ln \left (\frac {\left (\ln \left (\beta x \right )-1\right ) \beta x}{\LambertW \left (\left (\ln \left (\beta x \right )-1\right ) {\mathrm e}^{-1} \beta x \right )}\right )+\ln \left (\frac {\left (\ln \left (\beta x \right )-1\right ) \mu x}{\LambertW \left (\left (\ln \left (\beta x \right )-1\right ) {\mathrm e}^{-1} \beta x \right )}\right )\right ) \left (\LambertW \left (\left (\ln \left (\beta x \right )-1\right ) {\mathrm e}^{-1} \beta x \right )-\ln \left (\frac {\left (\ln \left (\beta x \right )-1\right ) \beta x}{\LambertW \left (\left (\ln \left (\beta x \right )-1\right ) {\mathrm e}^{-1} \beta x \right )}\right )+1\right ) c \Ei \left (1, -\ln \left (\frac {\left (\ln \left (\beta x \right )-1\right ) \beta x}{\LambertW \left (\left (\ln \left (\beta x \right )-1\right ) {\mathrm e}^{-1} \beta x \right )}\right )\right ) \LambertW \left (\left (\ln \left (\beta x \right )-1\right ) {\mathrm e}^{-1} \beta x \right )+\left (-\gamma \left (\ln \left (\beta x \right )-1\right ) c x \ln \left (\frac {\left (\ln \left (\beta x \right )-1\right ) \beta x}{\LambertW \left (\left (\ln \left (\beta x \right )-1\right ) {\mathrm e}^{-1} \beta x \right )}\right )+\gamma \left (\ln \left (\beta x \right )-1\right ) c x \ln \left (\frac {\left (\ln \left (\beta x \right )-1\right ) \mu x}{\LambertW \left (\left (\ln \left (\beta x \right )-1\right ) {\mathrm e}^{-1} \beta x \right )}\right )+\left (a \Ei \left (1, -\ln \left (z \right )-\ln \left (\gamma \right )\right )+\gamma \left (\ln \left (\beta x \right )-1\right ) c x \right ) \LambertW \left (\left (\ln \left (\beta x \right )-1\right ) {\mathrm e}^{-1} \beta x \right )\right ) \beta \right ) b}{\gamma a \beta c \LambertW \left (\left (\ln \left (\beta x \right )-1\right ) {\mathrm e}^{-1} \beta x \right )}\right )\]

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7.6.12.4 [1499] Problem 4

problem number 1499

Added May 26, 2019.

Problem Chapter 6.5.1.4, from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ a \ln (\beta x) w_x + b \ln (\lambda y) w_y + c \ln (\gamma z) w_z = 0 \]

Mathematica ✓

ClearAll["Global`*"]; 
pde =  a*Log[beta*x]*D[w[x, y,z], x] + b*Log[lambda*y]*D[w[x, y,z], y] +c*Log[gamma*z]*D[w[x,y,z],z]==0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y,z], {x, y,z}], 60*10]];
 

\[\left \{\left \{w(x,y,z)\to c_1\left (\frac {\text {li}(\lambda y)}{\lambda }-\frac {b \text {li}(\beta x)}{a \beta },\frac {\text {li}(\gamma z)}{\gamma }-\frac {c \text {li}(\beta x)}{a \beta }\right )\right \}\right \}\]

Maple ✓

restart; 
pde :=  a*ln(beta*x)*diff(w(x,y,z),x)+ b*ln(lambda*y)*diff(w(x,y,z),y)+c*ln(gamma*z)*diff(w(x,y,z),z)= 0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y,z))),output='realtime'));
 

\[w \left (x , y , z\right ) = \textit {\_F1} \left (\frac {-a \beta \Ei \left (1, -\ln \left (\lambda y \right )\right )+b \lambda \Ei \left (1, -\ln \left (\beta x \right )\right )}{b \beta \lambda }, \frac {-a \beta \Ei \left (1, -\ln \left (z \right )-\ln \left (\gamma \right )\right )+\gamma c \Ei \left (1, -\ln \left (\beta x \right )\right )}{\gamma \beta c}\right )\]

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