6.3.28 8.4
6.3.28.1 [1013] Problem 1
problem number 1013
Added Feb. 17, 2019.
Problem Chapter 3.8.4.1 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y)\)
\[ w_x + a w_y = f(x,y) \]
Mathematica ✓
ClearAll["Global`*"];
pde = D[w[x, y], x] + a*D[w[x, y], y] == f[x, y];
sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
\[\left \{\left \{w(x,y)\to \int _1^xf(K[1],-a x+y+a K[1])dK[1]+c_1(y-a x)\right \}\right \}\]
Maple ✓
restart;
pde := diff(w(x,y),x) +a*diff(w(x,y),y) = f(x,y);
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
\[w \left (x , y\right ) = \int _{}^{x}f \left (\textit {\_a} , \left (-x +\textit {\_a} \right ) a +y \right )d \textit {\_a} +f_{1} \left (-a x +y \right )\]
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6.3.28.2 [1014] Problem 2
problem number 1014
Added Feb. 17, 2019.
Problem Chapter 3.8.4.2 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y)\)
\[ a x w_x + b y w_y = f(x,y) \]
Mathematica ✓
ClearAll["Global`*"];
pde = a*x*D[w[x, y], x] + b*y*D[w[x, y], y] == f[x, y];
sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
\[\left \{\left \{w(x,y)\to \int _1^x\frac {f\left (K[1],x^{-\frac {b}{a}} y K[1]^{\frac {b}{a}}\right )}{a K[1]}dK[1]+c_1\left (y x^{-\frac {b}{a}}\right )\right \}\right \}\]
Maple ✓
restart;
pde := a*x*diff(w(x,y),x) +b*y*diff(w(x,y),y) = f(x,y);
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
\[w \left (x , y\right ) = \frac {\int _{}^{x}\frac {f \left (\textit {\_a} , y \,x^{-\frac {b}{a}} \textit {\_a}^{\frac {b}{a}}\right )}{\textit {\_a}}d \textit {\_a}}{a}+f_{1} \left (y \,x^{-\frac {b}{a}}\right )\]
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6.3.28.3 [1015] Problem 3
problem number 1015
Added Feb. 17, 2019.
Problem Chapter 3.8.4.3 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y)\)
\[ f(x) w_x + g(x) y w_y = h(x,y) \]
Mathematica ✓
ClearAll["Global`*"];
pde = f[x]*D[w[x, y], x] + g[x]*y*D[w[x, y], y] == h[x, y];
sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
\[\left \{\left \{w(x,y)\to \int _1^x\frac {h\left (K[2],\exp \left (\int _1^{K[2]}\frac {g(K[1])}{f(K[1])}dK[1]-\int _1^x\frac {g(K[1])}{f(K[1])}dK[1]\right ) y\right )}{f(K[2])}dK[2]+c_1\left (y \exp \left (-\int _1^x\frac {g(K[1])}{f(K[1])}dK[1]\right )\right )\right \}\right \}\]
Maple ✓
restart;
pde := f(x)*diff(w(x,y),x) +g(x)*y*diff(w(x,y),y) = h(x,y);
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
\[w \left (x , y\right ) = \int _{}^{x}\frac {h \left (\textit {\_b} , y \,{\mathrm e}^{-\int \frac {g \left (x \right )}{f \left (x \right )}d x +\int \frac {g \left (\textit {\_b} \right )}{f \left (\textit {\_b} \right )}d \textit {\_b}}\right )}{f \left (\textit {\_b} \right )}d \textit {\_b} +f_{1} \left (y \,{\mathrm e}^{-\int \frac {g \left (x \right )}{f \left (x \right )}d x}\right )\]
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6.3.28.4 [1016] Problem 4
problem number 1016
Added Feb. 17, 2019.
Problem Chapter 3.8.4.4 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y)\)
\[ f(x) w_x + (g_1(x) y+ g_0(x)) w_y = h(x,y) \]
Mathematica ✓
ClearAll["Global`*"];
pde = f[x]*D[w[x, y], x] + (g1[x]*y + g0[x])*D[w[x, y], y] == h[x, y];
sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
\[\left \{\left \{w(x,y)\to \int _1^x\frac {h\left (K[3],\exp \left (\int _1^{K[3]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \left (\exp \left (-\int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) y-\int _1^x\frac {\exp \left (-\int _1^{K[2]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \text {g0}(K[2])}{f(K[2])}dK[2]+\int _1^{K[3]}\frac {\exp \left (-\int _1^{K[2]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \text {g0}(K[2])}{f(K[2])}dK[2]\right )\right )}{f(K[3])}dK[3]+c_1\left (y \exp \left (-\int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right )-\int _1^x\frac {\exp \left (-\int _1^{K[2]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \text {g0}(K[2])}{f(K[2])}dK[2]\right )\right \}\right \}\]
Maple ✓
restart;
pde := f(x)*diff(w(x,y),x) +(g1(x)*y+g0(x))*diff(w(x,y),y) = h(x,y);
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
\[w \left (x , y\right ) = \int _{}^{x}\frac {h \left (\textit {\_f} , \left (\int \frac {\operatorname {g0} \left (\textit {\_f} \right ) {\mathrm e}^{-\int \frac {\operatorname {g1} \left (\textit {\_f} \right )}{f \left (\textit {\_f} \right )}d \textit {\_f}}}{f \left (\textit {\_f} \right )}d \textit {\_f} -\int \frac {\operatorname {g0} \left (x \right ) {\mathrm e}^{-\int \frac {\operatorname {g1} \left (x \right )}{f \left (x \right )}d x}}{f \left (x \right )}d x +{\mathrm e}^{-\int \frac {\operatorname {g1} \left (x \right )}{f \left (x \right )}d x} y \right ) {\mathrm e}^{\int \frac {\operatorname {g1} \left (\textit {\_f} \right )}{f \left (\textit {\_f} \right )}d \textit {\_f}}\right )}{f \left (\textit {\_f} \right )}d \textit {\_f} +f_{1} \left (-\int \frac {\operatorname {g0} \left (x \right ) {\mathrm e}^{-\int \frac {\operatorname {g1} \left (x \right )}{f \left (x \right )}d x}}{f \left (x \right )}d x +{\mathrm e}^{-\int \frac {\operatorname {g1} \left (x \right )}{f \left (x \right )}d x} y \right )\]
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6.3.28.5 [1017] Problem 5
problem number 1017
Added Feb. 17, 2019.
Problem Chapter 3.8.4.5 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y)\)
\[ f(x) w_x + (g_1(x) y+ g_0(x) y^k) w_y = h(x,y) \]
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];
sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
\[\left \{\left \{w(x,y)\to \int _1^x\frac {h\left (K[3],\left (\exp \left (-\int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]-(k-1) \int _1^{K[3]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) y^{-k} \left (\exp \left (\int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) (k-1) \int _1^x\frac {\exp \left ((k-1) \int _1^{K[2]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \text {g0}(K[2])}{f(K[2])}dK[2] y^k-\exp \left (\int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) (k-1) \int _1^{K[3]}\frac {\exp \left ((k-1) \int _1^{K[2]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \text {g0}(K[2])}{f(K[2])}dK[2] y^k+\exp \left (k \int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) y\right )\right ){}^{\frac {1}{1-k}}\right )}{f(K[3])}dK[3]+c_1\left ((k-1) \int _1^x\frac {\exp \left ((k-1) \int _1^{K[2]}\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right ) \text {g0}(K[2])}{f(K[2])}dK[2]+y^{1-k} \exp \left ((k-1) \int _1^x\frac {\text {g1}(K[1])}{f(K[1])}dK[1]\right )\right )\right \}\right \}\]
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);
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
\[w \left (x , y\right ) = \int _{}^{x}\frac {h \left (\textit {\_f} , \left (\left (1-k \right ) \int \frac {\operatorname {g0} \left (\textit {\_f} \right ) {\mathrm e}^{\left (k -1\right ) \int \frac {\operatorname {g1} \left (\textit {\_f} \right )}{f \left (\textit {\_f} \right )}d \textit {\_f}}}{f \left (\textit {\_f} \right )}d \textit {\_f} +\left (k -1\right ) \int \frac {\operatorname {g0} \left (x \right ) {\mathrm e}^{\left (k -1\right ) \int \frac {\operatorname {g1} \left (x \right )}{f \left (x \right )}d x}}{f \left (x \right )}d x +y^{1-k} {\mathrm e}^{\left (k -1\right ) \int \frac {\operatorname {g1} \left (x \right )}{f \left (x \right )}d x}\right )^{-\frac {1}{k -1}} {\mathrm e}^{\int \frac {\operatorname {g1} \left (\textit {\_f} \right )}{f \left (\textit {\_f} \right )}d \textit {\_f}}\right )}{f \left (\textit {\_f} \right )}d \textit {\_f} +f_{1} \left (\left (k -1\right ) \int \frac {\operatorname {g0} \left (x \right ) {\mathrm e}^{\left (k -1\right ) \int \frac {\operatorname {g1} \left (x \right )}{f \left (x \right )}d x}}{f \left (x \right )}d x +y^{1-k} {\mathrm e}^{\left (k -1\right ) \int \frac {\operatorname {g1} \left (x \right )}{f \left (x \right )}d x}\right )\]
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6.3.28.6 [1018] Problem 6
problem number 1018
Added Feb. 17, 2019.
Problem Chapter 3.8.4.6 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y)\)
\[ f(x) w_x + (g_1(x)+ g_0(x) e^{\lambda y}) w_y = h(x,y) \]
Mathematica ✗
ClearAll["Global`*"];
pde = f[x]*D[w[x, y], x] + (g1[x] + g0[x]*Exp[lambda*y])*D[w[x, y], y] == h[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)+g0(x)*exp(lambda*y))*diff(w(x,y),y) = h(x,y);
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
\[w \left (x , y\right ) = \int _{}^{x}\frac {h \left (\textit {\_f} , \frac {\ln \left (\frac {1}{\int \frac {\operatorname {g0} \left (x \right ) {\mathrm e}^{\lambda \int \frac {\operatorname {g1} \left (x \right )}{f \left (x \right )}d x}}{f \left (x \right )}d x \lambda -\int \frac {\operatorname {g0} \left (\textit {\_f} \right ) {\mathrm e}^{\lambda \int \frac {\operatorname {g1} \left (\textit {\_f} \right )}{f \left (\textit {\_f} \right )}d \textit {\_f}}}{f \left (\textit {\_f} \right )}d \textit {\_f} \lambda +{\mathrm e}^{\lambda \left (-y +\int \frac {\operatorname {g1} \left (x \right )}{f \left (x \right )}d x \right )}}\right )+\lambda \int \frac {\operatorname {g1} \left (\textit {\_f} \right )}{f \left (\textit {\_f} \right )}d \textit {\_f}}{\lambda }\right )}{f \left (\textit {\_f} \right )}d \textit {\_f} +f_{1} \left (\frac {-\int \frac {\operatorname {g0} \left (x \right ) {\mathrm e}^{\lambda \int \frac {\operatorname {g1} \left (x \right )}{f \left (x \right )}d x}}{f \left (x \right )}d x \lambda -{\mathrm e}^{\lambda \left (-y +\int \frac {\operatorname {g1} \left (x \right )}{f \left (x \right )}d x \right )}}{\lambda }\right )\]
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6.3.28.7 [1019] Problem 7
problem number 1019
Added Feb. 17, 2019.
Problem Chapter 3.8.4.7 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y)\)
\[ f_1(x) g_1(y) w_x + f_2(x) g_2(y) w_y = h(x,y) \]
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];
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);
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
\[w \left (x , y\right ) = \int _{}^{x}\frac {h \left (\textit {\_f} , \operatorname {RootOf}\left (\int \frac {\operatorname {f2} \left (\textit {\_f} \right )}{\operatorname {f1} \left (\textit {\_f} \right )}d \textit {\_f} -\int _{}^{\textit {\_Z}}\frac {\operatorname {g1} \left (\textit {\_a} \right )}{\operatorname {g2} \left (\textit {\_a} \right )}d \textit {\_a} -\int \frac {\operatorname {f2} \left (x \right )}{\operatorname {f1} \left (x \right )}d x +\int \frac {\operatorname {g1} \left (y \right )}{\operatorname {g2} \left (y \right )}d y \right )\right )}{\operatorname {f1} \left (\textit {\_f} \right ) \operatorname {g1} \left (\operatorname {RootOf}\left (\int \frac {\operatorname {f2} \left (\textit {\_f} \right )}{\operatorname {f1} \left (\textit {\_f} \right )}d \textit {\_f} -\int _{}^{\textit {\_Z}}\frac {\operatorname {g1} \left (\textit {\_a} \right )}{\operatorname {g2} \left (\textit {\_a} \right )}d \textit {\_a} -\int \frac {\operatorname {f2} \left (x \right )}{\operatorname {f1} \left (x \right )}d x +\int \frac {\operatorname {g1} \left (y \right )}{\operatorname {g2} \left (y \right )}d y \right )\right )}d \textit {\_f} +f_{1} \left (-\int \frac {\operatorname {f2} \left (x \right )}{\operatorname {f1} \left (x \right )}d x +\int \frac {\operatorname {g1} \left (y \right )}{\operatorname {g2} \left (y \right )}d y \right )\]
Contains RootOf