6.3.12 4.5

6.3.12.1 [919] Problem 1
6.3.12.2 [920] Problem 2
6.3.12.3 [921] Problem 3
6.3.12.4 [922] Problem 4
6.3.12.5 [923] Problem 5

6.3.12.1 [919] Problem 1

problem number 919

Added Feb. 11, 2019.

Problem Chapter 3.4.5.1 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ a w_x + b w_y = c \sinh (\lambda x) + k \cosh (\mu y) \]

Mathematica

ClearAll["Global`*"]; 
pde =  a*D[w[x, y], x] + b*D[w[x, y], y] == c*Sinh[lambda*x] + k*Cosh[mu*y]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 
\[\left \{\left \{w(x,y)\to \frac {1}{2} \left (2 c_1\left (y-\frac {b x}{a}\right )+\frac {c e^{-\lambda x} \left (e^{2 \lambda x}+1\right )}{a \lambda }+\frac {k e^{-\mu y} \left (e^{2 \mu y}-1\right )}{b \mu }\right )\right \}\right \}\]

Maple

restart; 
pde :=a*diff(w(x,y),x) + b*diff(w(x,y),y) =  c*sinh(lambda*x)+ k*cosh(mu*y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
 
\[w \left (x , y\right ) = \frac {f_{1} \left (\frac {a y -b x}{a}\right ) \mu b a \lambda +k a \sinh \left (\mu y \right ) \lambda +\cosh \left (\lambda x \right ) c \mu b}{\mu b a \lambda }\]

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6.3.12.2 [920] Problem 2

problem number 920

Added Feb. 11, 2019.

Problem Chapter 3.4.5.2 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ a w_x + b w_y = \tanh (\lambda x) + k \coth (\mu y) \]

Mathematica

ClearAll["Global`*"]; 
pde =  a*D[w[x, y], x] + b*D[w[x, y], y] == Tanh[lambda*x] + k*Coth[mu*y]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 
\[\left \{\left \{w(x,y)\to c_1\left (y-\frac {b x}{a}\right )+\frac {\log (\cosh (\lambda x))}{a \lambda }+\frac {k \log (\sinh (\mu y))}{b \mu }\right \}\right \}\]

Maple

restart; 
pde :=a*diff(w(x,y),x) + b*diff(w(x,y),y) =  tanh(lambda*x)+ k*coth(mu*y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
 
\[w \left (x , y\right ) = \frac {k \ln \left (\sinh \left (\mu y \right )\right )}{\mu b}+\frac {\ln \left (\cosh \left (\lambda x \right )\right )}{a \lambda }+f_{1} \left (\frac {y a -x b}{a}\right )\]

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6.3.12.3 [921] Problem 3

problem number 921

Added Feb. 11, 2019.

Problem Chapter 3.4.5.3 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ a w_x + b w_y = \sinh (\lambda x) + k \tanh (\mu y) \]

Mathematica

ClearAll["Global`*"]; 
pde =  a*D[w[x, y], x] + b*D[w[x, y], y] == Sinh[lambda*x] + k*Tanh[mu*y]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 
\[\left \{\left \{w(x,y)\to \frac {e^{-\lambda x} \left (2 a b \lambda \mu e^{\lambda x} \left (a^2 \lambda ^2-4 b^2 \mu ^2\right ) c_1\left (y-\frac {b x}{a}\right )+(a \lambda -2 b \mu ) \left ((a \lambda +2 b \mu ) \left (b \mu e^{2 \lambda x} \operatorname {Hypergeometric2F1}\left (1,\frac {a \lambda }{2 b \mu },\frac {a \lambda }{2 b \mu }+1,-e^{2 \mu y}\right )+b \mu \operatorname {Hypergeometric2F1}\left (1,-\frac {a \lambda }{2 b \mu },1-\frac {a \lambda }{2 b \mu },-e^{2 \mu y}\right )+a k \lambda e^{\lambda x} \left (\log \left (e^{-2 \mu y}+1\right )+\log \left (e^{2 \mu y}+1\right )\right )\right )+a b \lambda \mu e^{2 \lambda x+2 \mu y} \operatorname {Hypergeometric2F1}\left (1,\frac {a \lambda }{2 b \mu }+1,\frac {a \lambda }{2 b \mu }+2,-e^{2 \mu y}\right )\right )+a b \lambda \mu e^{2 \mu y} (a \lambda +2 b \mu ) \operatorname {Hypergeometric2F1}\left (1,1-\frac {a \lambda }{2 b \mu },2-\frac {a \lambda }{2 b \mu },-e^{2 \mu y}\right )\right )}{2 \left (a^3 b \lambda ^3 \mu -4 a b^3 \lambda \mu ^3\right )}\right \}\right \}\]

Maple

restart; 
pde :=a*diff(w(x,y),x) + b*diff(w(x,y),y) =  sinh(lambda*x)+ k*tanh(mu*y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
 
\[w \left (x , y\right ) = \frac {k \ln \left (\cosh \left (\mu y \right )\right )}{\mu b}+\frac {\cosh \left (\lambda x \right )}{a \lambda }+f_{1} \left (\frac {y a -b x}{a}\right )\]

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6.3.12.4 [922] Problem 4

problem number 922

Added Feb. 11, 2019.

Problem Chapter 3.4.5.4 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ a w_x + b \cosh (\mu y)w_y = \sinh (\lambda x) \]

Mathematica

ClearAll["Global`*"]; 
pde =  a*D[w[x, y], x] + b*Cosh[mu*y]*D[w[x, y], y] == Sinh[lambda*x]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 
\[\left \{\left \{w(x,y)\to \int _1^x\frac {\sinh (\lambda K[1])}{a}dK[1]+c_1\left (-\frac {b x}{a}-\frac {\cot ^{-1}(\sinh (\mu y))}{\mu }\right )\right \}\right \}\]

Maple

restart; 
pde :=a*diff(w(x,y),x) + b*cosh(mu*y)*diff(w(x,y),y) =  sinh(lambda*x); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
 
\[w \left (x , y\right ) = \frac {f_{1} \left (\frac {-b \mu x +2 \arctan \left ({\mathrm e}^{\mu y}\right ) a}{b \mu }\right ) a \lambda +\cosh \left (\lambda x \right )}{a \lambda }\]

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6.3.12.5 [923] Problem 5

problem number 923

Added Feb. 11, 2019.

Problem Chapter 3.4.5.4 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

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

\[ a w_x + b \sinh (\mu y)w_y = \cosh (\lambda x) \]

Mathematica

ClearAll["Global`*"]; 
pde =  a*D[w[x, y], x] + b*Sinh[mu*y]*D[w[x, y], y] == Cosh[lambda*x]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 
\begin{align*}& \left \{w(x,y)\to \int _1^x\frac {\cosh (\lambda K[1])}{a}dK[1]+c_1\left (-\frac {b x}{a}-\frac {\text {arctanh}(\cosh (\mu y))}{\mu }\right )\right \}\\& \left \{w(x,y)\to \int _1^x\frac {\cosh (\lambda K[2])}{a}dK[2]+c_1\left (-\frac {b x}{a}-\frac {\text {arctanh}(\cosh (\mu y))}{\mu }\right )\right \}\\\end{align*}

Maple

restart; 
pde :=a*diff(w(x,y),x) + b*sinh(mu*y)*diff(w(x,y),y) =  cosh(lambda*x); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
 
\[w \left (x , y\right ) = \frac {f_{1} \left (\frac {-x b \mu -2 \,\operatorname {arctanh}\left ({\mathrm e}^{\mu y}\right ) a}{b \mu }\right ) a \lambda +\sinh \left (\lambda x \right )}{a \lambda }\]

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