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'));
 

\[w \left (x , y\right ) = f_{1} \left (\frac {y a -b x}{a}\right )\]

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'));
 

\[w \left (x , y\right ) = f_{1} \left (\frac {-b \,x^{2}+2 y a -2 c x}{2 a}\right )\]

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]];
 

\[\left \{\left \{w(x,y)\to c_1\left (y e^{-b x}-\int _1^xe^{-b K[1]} (c+a K[1])dK[1]\right )\right \}\right \}\]

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'));
 

\[w \left (x , y\right ) = f_{1} \left (\frac {\left (y \,b^{2}+b \left (a x +c \right )+a \right ) {\mathrm e}^{-b x}}{b^{2}}\right )\]

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'));
 

\[w \left (x , y\right ) = f_{1} \left (y \,x^{-\frac {b}{a}}\right )\]

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'));
 

\[w \left (x , y\right ) = f_{1} \left (\frac {y^{2} a -b \,x^{2}}{a}\right )\]

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]];
 

\[\left \{\left \{w(x,y)\to c_1\left (\int _1^y\frac {K[1]}{a+K[1]}dK[1]-x\right )\right \}\right \}\]

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'));
 

\[w \left (x , y\right ) = f_{1} \left (-a \ln \left (y +a \right )+y -x \right )\]

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'));
 

\[w \left (x , y\right ) = f_{1} \left (\frac {1}{\sqrt {a^{3} k^{2} y^{2}-2 a^{2} b^{2} k \,y^{2}+2 a^{2} b \,k^{2} x y +a^{2} k^{3} x^{2}+a \,b^{4} y^{2}-4 a \,b^{3} k x y -2 a \,b^{2} k^{2} x^{2}+2 b^{5} x y +b^{4} k \,x^{2}+2 a^{2} c \,k^{2} y +2 a^{2} k^{2} s x -4 a \,b^{2} c k y -4 a \,b^{2} k s x +2 b^{4} c y +2 b^{4} s x +a^{2} k \,s^{2}-a \,b^{2} s^{2}-2 a b c k s +a \,c^{2} k^{2}+2 b^{3} c s -c^{2} k \,b^{2}}}\right )\]

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

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