6.2.1 2.1

6.2.1.1 [422] problem number 1

problem number 422

Added January 2, 2019.

Problem 2.2.1.1 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

Solve for \(w(x,y)\) \[ a w_x + b w_y = 0 \]

Mathematica ✓

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

\[\left \{\left \{w(x,y)\to c_1\left (y-\frac {b x}{a}\right )\right \}\right \}\]

Maple ✓

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

\[w \left ( x,y \right ) ={\it \_F1} \left ( {\frac {ya-xb}{a}} \right ) \]

Hand solution

\begin {equation} aw_{x}+bw_{y}=0\nonumber \end {equation}

The Lagrange-charpit equations are

\[ \frac {dx}{a}=\frac {dy}{b}=\frac {dw}{0}\]

The first pair of equations results in \(bdx=ady\) or \(bx=ay+C_{1}\). Hence \[ C_{1}=bx-ay \]

Since \(dw=0\) then \(w=C_{2}\). But \(C_{2}=F\left ( C_{1}\right ) \) where \(F\) is arbitrary function, therefore the solution is

\[ w\left ( x,y\right ) =F\left ( bx-ay\right ) \]

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6.2.1.2 [423] problem number 2

problem number 423

Added January 2, 2019.

Problem 2.2.1.2 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

Solve for \(w(x,y)\) \[ a w_x + (b x+c) w_y = 0 \]

Mathematica ✓

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

\[\left \{\left \{w(x,y)\to c_1\left (-\frac {-2 a y+b x^2+2 c x}{2 a}\right )\right \}\right \}\]

Maple ✓

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

\[w \left ( x,y \right ) ={\it \_F1} \left ( 1/2\,{\frac {-{x}^{2}b+2\,ya-2\,cx}{a}} \right ) \]

Hand solution

Solve \(aw_{x}+\left ( bx+c\right ) w_{y}=0\). The Lagrange-charpit equations are

\[ \frac {dx}{a}=\frac {dy}{\left ( bx+c\right ) }=\frac {dw}{0}\]

The first pair of equations gives \(\frac {\left ( bx+c\right ) }{a}dx=dy\). Integrating results in

\begin {align*} \frac {1}{a}\left ( \frac {bx^{2}}{2}+cx\right ) & =y+C_{1}\\ C_{1} & =\frac {1}{a}\left ( \frac {bx^{2}}{2}+cx\right ) -y \end {align*}

Since \(dw=0\) then \(w=C_{2}\). But \(C_{2}=F\left ( C_{1}\right ) \). Where \(F\) is arbitray function. Therefore

\[ w\left ( x,y\right ) =F\left ( \frac {bx^{2}}{2a}+\frac {c}{a}x-y\right ) \]

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6.2.1.3 [424] problem number 3

problem number 424

Added January 2, 2019.

Problem 2.2.1.3 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

Solve for \(w(x,y)\) \[ w_x + (a x+b y+c) w_y = 0 \]

Mathematica ✓

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

\[\left \{\left \{w(x,y)\to c_1\left (\frac {e^{-b x} (a b x+a+b (b y+c))}{b^2}\right )\right \}\right \}\]

Maple ✓

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

\[w \left ( x,y \right ) ={\it \_F1} \left ( {\frac {{{\rm e}^{-xb}} \left ( y{b}^{2}+b \left ( ax+c \right ) +a \right ) }{{b}^{2}}} \right ) \]

Hand solution

Solve\begin {equation} w_{x}+\left ( ax+by+c\right ) w_{y}=0\tag {1} \end {equation} The Lagrange-charpit equations are \[ dx=\frac {dy}{\left ( ax+by+c\right ) }=\frac {dw}{0}\] The first pair of equations gives \begin {align*} \frac {dy}{dx} & =ax+by+c\\ \frac {dy}{dx}-by & =ax+c \end {align*}

This is linear. Integrating factor is \(I=e^{-bx}\). Hence the above becomes\begin {align*} \frac {d}{dx}\left ( ye^{-bx}\right ) & =\left ( ax+c\right ) e^{-bx}\\ ye^{-bx} & =a\int xe^{-bx}+c\int e^{-bx}+C_{1}\\ ye^{-bx} & =a\left ( -\frac {\left ( 1+bx\right ) e^{-bx}}{b^{2}}\right ) -c\frac {e^{-bx}}{b}+C_{1}\\ y & =-a\frac {\left ( 1+bx\right ) }{b^{2}}-\frac {c}{b}-C_{1}e^{bx}\\ C_{1} & =-\left ( y+\frac {a}{b^{2}}\left ( 1+bx\right ) +\frac {c}{b}\right ) e^{-bx} \end {align*}

Since \(dw=0\) then \(w=C_{2}\). But \(C_{2}=F\left ( C_{1}\right ) \). Therefore\begin {align*} w\left ( x,y\right ) & =F\left ( -\left ( y+\frac {a}{b^{2}}\left ( 1+bx\right ) +\frac {c}{b}\right ) e^{-bx}\right ) \\ & =F\left ( \left ( y+\frac {a}{b^{2}}\left ( 1+bx\right ) +\frac {c}{b}\right ) e^{-bx}\right ) \end {align*}

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6.2.1.4 [425] problem number 4

problem number 425

Added January 2, 2019.

Problem 2.2.1.4 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

Solve for \(w(x,y)\) \[ a x w_x + b y w_y = 0 \]

Mathematica ✓

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

\[\left \{\left \{w(x,y)\to c_1\left (y x^{-\frac {b}{a}}\right )\right \}\right \}\]

Maple ✓

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

\[w \left ( x,y \right ) ={\it \_F1} \left ( y{x}^{-{\frac {b}{a}}} \right ) \]

Hand solution

Solve\begin {equation} axw_{x}+byw_{y}=0\tag {1} \end {equation} The Lagrange-charpit equations are \[ \frac {dx}{ax}=\frac {dy}{by}=\frac {dw}{0}\] The first pair of equations gives \begin {align*} \frac {b}{a}\frac {dx}{x} & =\frac {dy}{y}\\ \frac {b}{a}\ln x & =\ln y+C_{1}\\ x^{\frac {b}{a}} & =C_{1}y\\ C_{1} & =\frac {x^{\frac {b}{a}}}{y} \end {align*}

Since \(dw=0\) then \(w=C_{2}\). But \(C_{2}=F\left ( C_{1}\right ) \). Therefore\[ w\left ( x,y\right ) =F\left ( \frac {x^{\frac {b}{a}}}{y}\right ) \]

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6.2.1.5 [426] problem number 5

problem number 426

Added January 2, 2019.

Problem 2.2.1.5 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

Solve for \(w(x,y)\) \[ a y w_x + b x w_y = 0 \]

Mathematica ✓

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

\[\left \{\left \{w(x,y)\to c_1\left (\frac {a y^2-b x^2}{2 a}\right )\right \}\right \}\]

Maple ✓

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

\[w \left ( x,y \right ) ={\it \_F1} \left ( {\frac {{y}^{2}a-{x}^{2}b}{a}} \right ) \]

Hand solution

\[ ayw_{x}+bxw_{y}=0 \] Using method of charaterstics the lagrange-Charpit equations are\[ \frac {dx}{ay}=\frac {dy}{bx}=\frac {du}{0}\] The first two equations give \(aydy=bxdx\). Hence \begin {align*} a\frac {y^{2}}{2} & =\frac {bx^{2}}{2}+C_{1}\\ C_{1} & =a\frac {y^{2}}{2}-\frac {bx^{2}}{2} \end {align*}

And \(du=0\). This gives \(u=C_{2}\). Now let \(C_{2}=F\left ( C_{1}\right ) \). This gives\[ u=F\left ( \frac {ay^{2}-bx^{2}}{2}\right ) \]

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6.2.1.6 [427] problem number 6

problem number 427

Added January 2, 2019.

Problem 2.2.1.6 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

Solve for \(w(x,y)\) \[ y w_x + (y+a) w_y = 0 \]

Mathematica ✓

ClearAll["Global`*"]; 
pde =  y*D[w[x, y], x] + (y + a)*D[w[x, y], y] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 

\[\left \{\left \{w(x,y)\to c_1\left (a \left (-\log \left (e^{-\frac {a+y}{a}} (a+y)\right )\right )-a-x\right )\right \}\right \}\]

Maple ✓

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

\[w \left ( x,y \right ) ={\it \_F1} \left ( -a\ln \left ( y+a \right ) +y-x \right ) \]

Hand solution

\[ yw_{x}+\left ( y+a\right ) w_{y}=0 \] Using method of charaterstics the lagrange-Charpit equations are\[ \frac {dx}{y}=\frac {dy}{y+a}=\frac {du}{0}\] The first two equations give \(dx=\frac {y}{y+a}dy\). Hence by integrating \begin {align*} x & =y-a\ln \left ( y+a\right ) +C_{1}\\ C_{1} & =x-y+a\ln \left ( y+a\right ) \end {align*}

And \(du=0\). This gives \(u=C_{2}\). Now let \(C_{2}=F\left ( C_{1}\right ) \). This gives\[ u=F\left ( x-y+a\ln \left ( y+a\right ) \right ) \]

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6.2.1.7 [428] problem number 7

problem number 428

Added January 2, 2019.

Problem 2.2.1.7 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

Solve for \(w(x,y)\) \[ (a y + b x + c ) w_x - (b y + k x +s) w_y =0 \]

Mathematica ✓

ClearAll["Global`*"]; 
pde =  (a*y + b*x + c)*D[w[x, y], x] - (b*y + k*x + s)*D[w[x, y], y] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 

\[\left \{\left \{w(x,y)\to c_1\left (\frac {y (a y+2 b x+2 c)+k x^2+2 s x}{a}\right )\right \}\right \}\]

Maple ✓

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

\[w \left ( x,y \right ) ={\it \_F1} \left ( {\frac {1}{\sqrt {{a}^{3}{k}^{2}{y}^{2}-2\,{a}^{2}{b}^{2}k{y}^{2}+2\,{a}^{2}b{k}^{2}xy+{a}^{2}{k}^{3}{x}^{2}+a{b}^{4}{y}^{2}-4\,a{b}^{3}kxy-2\,a{b}^{2}{k}^{2}{x}^{2}+2\,{b}^{5}xy+{b}^{4}k{x}^{2}+2\,{a}^{2}c{k}^{2}y+2\,{a}^{2}{k}^{2}sx-4\,a{b}^{2}cky-4\,a{b}^{2}ksx+2\,{b}^{4}cy+2\,{b}^{4}sx+{a}^{2}k{s}^{2}-a{b}^{2}{s}^{2}-2\,abcks+a{c}^{2}{k}^{2}+2\,{b}^{3}cs-{c}^{2}k{b}^{2}}}} \right ) \]

Hand solution

Solve\[ \left ( ay+bx+c\right ) w_{x}-\left ( by+kx+s\right ) w_{y}=0 \] The lagrange-Charpit equations are\[ \frac {dx}{ay+bx+c}=\frac {dy}{-\left ( by+kx+s\right ) }=\frac {dw}{0}\] The first two equations give\[ \frac {dy}{dx}=\frac {-\left ( by+kx+s\right ) }{ay+bx+c}\] Need to solve the above, then solve for \(C_{1}\) to finish the solution.

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6.2.1.8 [429] problem number 8

problem number 429

Added January 2, 2019.

Problem 2.2.1.8 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.

Solve for \(w(x,y)\) \[ (a_1 x + b_1 y + c_1 ) w_x + (a_2 x + b_2 y +c_2) w_y =0 \]

Mathematica ✗

ClearAll["Global`*"]; 
pde =  (a1*x + b1*y + c1)*D[w[x, y], x] + (a1*x + b2*y + c2)*D[w[x, y], y] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
 

Failed

Maple ✓

restart; 
pde :=(a1*x+b1*y+c1)*diff(w(x,y),x)+(a2*x+b2*y+c2)*diff(w(x,y),y)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y))),output='realtime'));
 

\[w \left ( x,y \right ) ={\it \_F1} \left ( -1/2\,{\frac {1}{\sqrt {-{{\it a1}}^{2}+2\,{\it a1}\,{\it b2}-4\,{\it a2}\,{\it b1}-{{\it b2}}^{2}}} \left ( \sqrt {-{{\it a1}}^{2}+2\,{\it a1}\,{\it b2}-4\,{\it a2}\,{\it b1}-{{\it b2}}^{2}}\ln \left ( - \left ( \left ( -{\it a1}\,{\it b1}\,{\it b2}+{\it a2}\,{{\it b1}}^{2} \right ) {y}^{2}+ \left ( \left ( {\it a2}\, \left ( {\it a1}-{\it b2} \right ) x+2\,{\it a2}\,{\it c1}-{\it c2}\, \left ( {\it a1}+{\it b2} \right ) \right ) {\it b1}-{\it b2}\, \left ( {\it a1}-{\it b2} \right ) \left ( {\it a1}\,x+{\it c1} \right ) \right ) y- \left ( {\it a2}\,x+{\it c2} \right ) ^{2}{\it b1}+ \left ( {\it a1}\,x+{\it c1} \right ) \left ( x{\it a2}\,{\it b2}+{\it a2}\,{\it c1}-{\it c2}\, \left ( {\it a1}-{\it b2} \right ) \right ) \right ) \left ( {\it a1}\,{\it b2}-{\it a2}\,{\it b1} \right ) \right ) +2\,\arctan \left ( {\frac {-2\,{\it a2}\,{{\it b1}}^{2}y+ \left ( \left ( 2\,{\it a1}\,y+{\it a2}\,x+{\it c2} \right ) {\it b2}+ \left ( -{\it a2}\,x+{\it c2} \right ) {\it a1}-2\,{\it a2}\,{\it c1} \right ) {\it b1}+{\it b2}\, \left ( {\it a1}-{\it b2} \right ) \left ( {\it a1}\,x+{\it c1} \right ) }{\sqrt {-{{\it a1}}^{2}+2\,{\it a1}\,{\it b2}-4\,{\it a2}\,{\it b1}-{{\it b2}}^{2}} \left ( \left ( -{\it a2}\,x-{\it c2} \right ) {\it b1}+{\it b2}\, \left ( {\it a1}\,x+{\it c1} \right ) \right ) }} \right ) \left ( {\it a1}+{\it b2} \right ) \right ) } \right ) \]

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