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Added Feb. 7, 2019.
Problem 2.8.4.1 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y)\)
\[ w_x - \left (a y^2 \ln x -a x y (\ln x-1) f(x)+f(x) \right ) w_y = 0 \]
Mathematica ✗
ClearAll[w, x, y, n, a, b, m, c, k, alpha, beta, gamma, A, C0, s, lambda, B, s, mu, d, g, B, v, f]; pde = D[w[x, y], x] - (a*y^2*Log[x] - a*x*y*(Log[x] - 1)*f[x] + f[x])*D[w[x, y], y] == 0; 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';v:='v'; pde := diff(w(x,y),x)-( a*y^2*ln(x) -a*x*y* (ln(x)-1)*f(x)+f(x))*diff(w(x,y),y) = 0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
\[ w \left ( x,y \right ) ={\it \_F1} \left ( {\frac {1}{a \left ( \ln \left ( x \right ) axy-yax-1 \right ) } \left ( \ln \left ( x \right ) y\int \!{\frac {\ln \left ( x \right ) }{{x}^{2} \left ( \ln \left ( x \right ) -1 \right ) ^{2}}{{\rm e}^{a\int \!{\frac { \left ( \ln \left ( x \right ) \right ) ^{2}f \left ( x \right ) x}{\ln \left ( x \right ) -1}}\,{\rm d}x-2\,a\int \!{\frac {\ln \left ( x \right ) f \left ( x \right ) x}{\ln \left ( x \right ) -1}}\,{\rm d}x+a\int \!{\frac {f \left ( x \right ) x}{\ln \left ( x \right ) -1}}\,{\rm d}x}}}\,{\rm d}xax-y\int \!{\frac {\ln \left ( x \right ) }{{x}^{2} \left ( \ln \left ( x \right ) -1 \right ) ^{2}}{{\rm e}^{a\int \!{\frac { \left ( \ln \left ( x \right ) \right ) ^{2}f \left ( x \right ) x}{\ln \left ( x \right ) -1}}\,{\rm d}x-2\,a\int \!{\frac {\ln \left ( x \right ) f \left ( x \right ) x}{\ln \left ( x \right ) -1}}\,{\rm d}x+a\int \!{\frac {f \left ( x \right ) x}{\ln \left ( x \right ) -1}}\,{\rm d}x}}}\,{\rm d}xax-x{{\rm e}^{\int \!{\frac { \left ( \ln \left ( x \right ) \right ) ^{2}f \left ( x \right ) a{x}^{2}-2\,\ln \left ( x \right ) f \left ( x \right ) a{x}^{2}+f \left ( x \right ) a{x}^{2}-2\,\ln \left ( x \right ) }{x \left ( \ln \left ( x \right ) -1 \right ) }}\,{\rm d}x}}\ln \left ( x \right ) +x{{\rm e}^{\int \!{\frac { \left ( \ln \left ( x \right ) \right ) ^{2}f \left ( x \right ) a{x}^{2}-2\,\ln \left ( x \right ) f \left ( x \right ) a{x}^{2}+f \left ( x \right ) a{x}^{2}-2\,\ln \left ( x \right ) }{x \left ( \ln \left ( x \right ) -1 \right ) }}\,{\rm d}x}}-\int \!{\frac {\ln \left ( x \right ) }{{x}^{2} \left ( \ln \left ( x \right ) -1 \right ) ^{2}}{{\rm e}^{a\int \!{\frac { \left ( \ln \left ( x \right ) \right ) ^{2}f \left ( x \right ) x}{\ln \left ( x \right ) -1}}\,{\rm d}x-2\,a\int \!{\frac {\ln \left ( x \right ) f \left ( x \right ) x}{\ln \left ( x \right ) -1}}\,{\rm d}x+a\int \!{\frac {f \left ( x \right ) x}{\ln \left ( x \right ) -1}}\,{\rm d}x}}}\,{\rm d}x \right ) } \right ) \]
____________________________________________________________________________________
Added Feb. 7, 2019.
Problem 2.8.4.2 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y)\)
\[ w_x + \left (f(x) y^2 -a x(\ln x) f(x) y+a \ln x+a \right ) w_y = 0 \]
Mathematica ✗
ClearAll[w, x, y, n, a, b, m, c, k, alpha, beta, gamma, A, C0, s, lambda, B, s, mu, d, g, B, v, f]; pde = D[w[x, y], x] + (f[x]*y^2 - a*x*Log[x]*f[x]*y + a*Log[x] + a)*D[w[x, y], y] == 0; 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';v:='v'; pde := diff(w(x,y),x)+( f(x)* y^2 -a*x*ln(x)*f(x)*y+a*ln(x)+a)*diff(w(x,y),y) = 0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
\[ \text { sol=() } \]
____________________________________________________________________________________
Added Feb. 7, 2019.
Problem 2.8.4.3 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y)\)
\[ x w_x + \left (f(x) y^2 +a -a^2 (\ln x)^2 f(x) \right ) w_y = 0 \]
Mathematica ✗
ClearAll[w, x, y, n, a, b, m, c, k, alpha, beta, gamma, A, C0, s, lambda, B, s, mu, d, g, B, v, f]; pde = x*D[w[x, y], x] + (f[x]*y^2 + a - a^2*Log[x]^2*f[x])*D[w[x, y], y] == 0; 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';v:='v'; pde := x*diff(w(x,y),x)+( f(x)*y^2 +a -a^2* ln(x)^2 *f(x))*diff(w(x,y),y) = 0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
\[ \text { sol=() } \]
____________________________________________________________________________________
Added Feb. 7, 2019.
Problem 2.8.4.4 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y)\)
\[ x w_x + \left ((y+a \ln x)^2 f(x)-a \right ) w_y = 0 \]
Mathematica ✓
ClearAll[w, x, y, n, a, b, m, c, k, alpha, beta, gamma, A, C0, s, lambda, B, s, mu, d, g, B, v, f]; pde = x*D[w[x, y], x] + ((y + a*Log[x])^2*f[x] - a)*D[w[x, y], y] == 0; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
\[ \left \{\left \{w(x,y)\to c_1\left (\frac {a \log (x) \int _1^x \frac {f(K[2])}{K[2]} \, dK[2]+y \int _1^x \frac {f(K[2])}{K[2]} \, dK[2]+1}{a \log (x)+y}\right )\right \}\right \} \]
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';v:='v'; pde :=x*diff(w(x,y),x)+( (y+a *ln(x))^2*f(x)-a)*diff(w(x,y),y) = 0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y)) ),output='realtime'));
\[ w \left ( x,y \right ) ={\it \_F1} \left ( {\frac {1}{y+\ln \left ( x \right ) a} \left ( \ln \left ( x \right ) a\int \!{\frac {f \left ( x \right ) }{x}}\,{\rm d}x+y\int \!{\frac {f \left ( x \right ) }{x}}\,{\rm d}x+1 \right ) } \right ) \]