35 HFOPDE, chapter 2.2.1

 35.1 problem number 1
 35.2 problem number 2
 35.3 problem number 3
 35.4 problem number 4
 35.5 problem number 5
 35.6 problem number 6
 35.7 problem number 7
 35.8 problem number 8

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35.1 problem number 1

problem number 215

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

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

Maple

\[ w \left ( x,y \right ) ={\it \_F1} \left ( x \right ) \] Solution missing \(a,b\) compared to book, but technically still correct.

Hand solution

\begin{align} aw_{x}+bw_{y} & =0\nonumber \\ w_{x}+\frac{b}{a}w_{y} & =0\tag{1} \end{align}

Along the curve \(y=y\left ( x\right ) \) then

\begin{equation} \frac{d}{dx}w\left ( x,y\left ( x\right ) \right ) =w_{x}+w_{y}\frac{dy}{dx}\tag{2} \end{equation}

Comparing (1,2) shows that \(\frac{dy}{dx}=\frac{b}{a}\). This has the solution \begin{equation} y\left ( x\right ) =\frac{b}{a}x+y\left ( 0\right ) \tag{3} \end{equation}

In this case

\begin{align*} \frac{d}{dx}w\left ( x,y\left ( x\right ) \right ) & =0\\ w\left ( x,y\left ( x\right ) \right ) +w\left ( 0,y\left ( 0\right ) \right ) & =0\\ w\left ( x,y\left ( x\right ) \right ) & =-w\left ( 0,y\left ( 0\right ) \right ) \\ & =F\left ( y\left ( 0\right ) \right ) \end{align*}

Where \(F\) is arbitrary function. But from (3) \(y\left ( 0\right ) =y\left ( x\right ) -\frac{b}{a}x\), hence the above becomes

\begin{align*} w\left ( x,y\left ( x\right ) \right ) & =F\left ( y-\frac{b}{a}x\right ) \\ & =F\left ( \frac{ya-bx}{a}\right ) \end{align*}

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35.2 problem number 2

problem number 216

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

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

Maple

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

Hand solution

\begin{align} aw_{x}+\left ( bx+c\right ) w_{y} & =0\nonumber \\ w_{x}+\frac{bx+c}{a}w_{y} & =0\tag{1} \end{align}

Along the curve \(y=y\left ( x\right ) \) then

\begin{equation} \frac{d}{dx}w\left ( x,y\left ( x\right ) \right ) =w_{x}+w_{y}\frac{dy}{dx}\tag{2} \end{equation}

Comparing (1,2) shows that \(\frac{dy}{dx}=\frac{bx+c}{a}\). This is separable. \begin{align} ady & =bx+c\nonumber \\ ay\left ( x\right ) & =\frac{bx^{2}}{2}+cx-y\left ( 0\right ) \tag{3} \end{align}

Since \(\frac{d}{dx}w\left ( x,y\left ( x\right ) \right ) =0\) then integrating gives

\begin{align*} w\left ( x,y\left ( x\right ) \right ) +w\left ( 0,y\left ( 0\right ) \right ) & =0\\ w\left ( x,y\left ( x\right ) \right ) & =-w\left ( 0,y\left ( 0\right ) \right ) \\ & =F\left ( y\left ( 0\right ) \right ) \end{align*}

Where \(F\) is arbitrary function. But from (3) \(y\left ( 0\right ) =ay-\frac{bx^{2}}{2}-cx\), hence the above becomes

\begin{align*} w\left ( x,y\left ( x\right ) \right ) & =F\left ( ay-\frac{bx^{2}}{2}-cx\right ) \\ & =F\left ( \frac{1}{2}\left ( 2ay-bx^{2}-2cx\right ) \right ) \end{align*}

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35.3 problem number 3

problem number 217

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

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

Maple

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

Hand solution

\begin{equation} w_{x}+\left ( ax+by+c\right ) w_{y}=0\tag{1} \end{equation}

Along the curve \(y=y\left ( x\right ) \) then

\begin{equation} \frac{d}{dx}w\left ( x,y\left ( x\right ) \right ) =w_{x}+w_{y}\frac{dy}{dx}\tag{2} \end{equation}

Comparing (1,2) shows that \(\frac{dy}{dx}=ax+by+c\). Hence \begin{equation} \frac{dy}{dx}-by=ax+c\nonumber \end{equation}

This is linear in \(y\). Using integrating factor \(I=e^{-\int bdx}=e^{-bx}\), then the above becomes

\begin{align*} \frac{d}{dx}\left ( yI\right ) & =I\left ( ax+c\right ) \\ \frac{d}{dx}\left ( ye^{-bx}\right ) & =e^{-bx}\left ( ax+c\right ) \end{align*}

Integrating

\begin{align*} ye^{-bx}+y\left ( 0\right ) & =\int e^{-bx}\left ( ax+c\right ) dx\\ ye^{-bx}+y\left ( 0\right ) & =a\int xe^{-bx}dx+\int ce^{-bx}dx \end{align*}

But \(\int xe^{-bx}=-\frac{e^{-bx}\left ( 1+bx\right ) }{b^{2}}\), therefore the above becomes

\begin{align*} ye^{-bx}+y\left ( 0\right ) & =-a\frac{e^{-bx}\left ( 1+bx\right ) }{b^{2}}-\frac{ce^{-bx}}{b}\\ ye^{-bx} & =-a\frac{e^{-bx}\left ( 1+bx\right ) }{b^{2}}-\frac{ce^{-bx}}{b}-y\left ( 0\right ) \\ y & =-a\frac{\left ( 1+bx\right ) }{b^{2}}-\frac{c}{b}-y\left ( 0\right ) e^{bx} \end{align*}

Hence

\begin{align} y\left ( 0\right ) & =-e^{-bx}\left ( y+a\frac{\left ( 1+bx\right ) }{b^{2}}+\frac{c}{b}\right ) \nonumber \\ & =-e^{-bx}\left ( \frac{yb^{2}+a\left ( 1+bx\right ) +cb}{b^{2}}\right ) \tag{3} \end{align}

Since \(\frac{d}{dx}w\left ( x,y\left ( x\right ) \right ) =0\) then integrating gives

\begin{align*} w\left ( x,y\left ( x\right ) \right ) +w\left ( 0,y\left ( 0\right ) \right ) & =0\\ w\left ( x,y\left ( x\right ) \right ) & =-w\left ( 0,y\left ( 0\right ) \right ) \\ & =F\left ( y\left ( 0\right ) \right ) \end{align*}

Where \(F\) is arbitrary function. But from (3) the above becomes

\[ w\left ( x,y\left ( x\right ) \right ) =F\left ( e^{-bx}\left ( \frac{yb^{2}+a+abx+cb}{b^{2}}\right ) \right ) \]

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35.4 problem number 4

problem number 218

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

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

Maple

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

Hand solution

\begin{align} axw_{x}+byw_{y} & =0\nonumber \\ w_{x}+\frac{by}{ax}w_{y} & =0\tag{1} \end{align}

Along the curve \(y=y\left ( x\right ) \) then

\begin{equation} \frac{d}{dx}w\left ( x,y\left ( x\right ) \right ) =w_{x}+w_{y}\frac{dy}{dx}\tag{2} \end{equation}

Comparing (1,2) shows that \(\frac{dy}{dx}=\frac{by}{ax}\). This is separable. Hence \begin{align*} \frac{dy}{y} & =\frac{b}{a}\frac{dx}{x}\\ \ln y & =\frac{b}{a}\ln x+C\\ y & =y\left ( 0\right ) x^{\frac{b}{a}} \end{align*}

Hence

\begin{equation} y\left ( 0\right ) =\frac{y}{x^{\frac{b}{a}}}\nonumber \end{equation}

Since \(\frac{d}{dx}w\left ( x,y\left ( x\right ) \right ) =0\) then integrating gives

\begin{align*} w\left ( x,y\left ( x\right ) \right ) +w\left ( 0,y\left ( 0\right ) \right ) & =0\\ w\left ( x,y\left ( x\right ) \right ) & =-w\left ( 0,y\left ( 0\right ) \right ) \\ & =F\left ( y\left ( 0\right ) \right ) \end{align*}

Where \(F\) is arbitrary function. But from (3) the above becomes

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

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35.5 problem number 5

problem number 219

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

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

Maple

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

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35.6 problem number 6

problem number 220

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

\[ \left \{\left \{w(x,y)\to c_1\left (a \left (-\log \left (a \left (\frac{y e^{-\frac{y}{a}-1}}{a}+e^{-\frac{y}{a}-1}\right )\right )\right )-a-x\right )\right \}\right \} \]

Maple

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

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35.7 problem number 7

problem number 221

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

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

Maple

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

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35.8 problem number 8

problem number 222

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

\[ \text{DSolve}\left [w^{(1,0)}(x,y) (\text{a1} x+\text{b1} y+\text{c1})+w^{(0,1)}(x,y) (\text{a1} x+\text{b2} y+\text{c2})=0,w(x,y),\{x,y\}\right ] \]

Maple

\[ 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 ( y{{\it a1}}^{3}{{\it b2}}^{2}x-2\,y{{\it a1}}^{2}{\it a2}\,{\it b1}\,{\it b2}\,x-{{\it a1}}^{2}{\it a2}\,{{\it b2}}^{2}{x}^{2}+{{\it a1}}^{2}{\it b1}\,{{\it b2}}^{2}{y}^{2}-y{{\it a1}}^{2}{{\it b2}}^{3}x+y{\it a1}\,{{\it a2}}^{2}{{\it b1}}^{2}x+2\,{\it a1}\,{{\it a2}}^{2}{\it b1}\,{\it b2}\,{x}^{2}-2\,{\it a1}\,{\it a2}\,{{\it b1}}^{2}{\it b2}\,{y}^{2}+2\,y{\it a1}\,{\it a2}\,{\it b1}\,{{\it b2}}^{2}x-{{\it a2}}^{3}{{\it b1}}^{2}{x}^{2}+{{\it a2}}^{2}{{\it b1}}^{3}{y}^{2}-y{{\it a2}}^{2}{{\it b1}}^{2}{\it b2}\,x+{{\it a1}}^{3}{\it b2}\,{\it c2}\,x-{{\it a1}}^{2}{\it a2}\,{\it b1}\,{\it c2}\,x-{{\it a1}}^{2}{\it a2}\,{\it b2}\,{\it c1}\,x+{{\it a1}}^{2}{\it b1}\,{\it b2}\,{\it c2}\,y+{{\it a1}}^{2}{{\it b2}}^{2}{\it c1}\,y-{{\it a1}}^{2}{{\it b2}}^{2}{\it c2}\,x+{\it a1}\,{{\it a2}}^{2}{\it b1}\,{\it c1}\,x-{\it a1}\,{\it a2}\,{{\it b1}}^{2}{\it c2}\,y-3\,{\it a1}\,{\it a2}\,{\it b1}\,{\it b2}\,{\it c1}\,y+3\,{\it a1}\,{\it a2}\,{\it b1}\,{\it b2}\,{\it c2}\,x-{\it a1}\,{\it a2}\,{{\it b2}}^{2}{\it c1}\,x+{\it a1}\,{\it b1}\,{{\it b2}}^{2}{\it c2}\,y-{\it a1}\,{{\it b2}}^{3}{\it c1}\,y+2\,{{\it a2}}^{2}{{\it b1}}^{2}{\it c1}\,y-2\,{{\it a2}}^{2}{{\it b1}}^{2}{\it c2}\,x+{{\it a2}}^{2}{\it b1}\,{\it b2}\,{\it c1}\,x-{\it a2}\,{{\it b1}}^{2}{\it b2}\,{\it c2}\,y+{\it a2}\,{\it b1}\,{{\it b2}}^{2}{\it c1}\,y+{{\it a1}}^{2}{\it b2}\,{\it c1}\,{\it c2}-{\it a1}\,{\it a2}\,{\it b1}\,{\it c1}\,{\it c2}-{\it a1}\,{\it a2}\,{\it b2}\,{{\it c1}}^{2}+{\it a1}\,{\it b1}\,{\it b2}\,{{\it c2}}^{2}-{\it a1}\,{{\it b2}}^{2}{\it c1}\,{\it c2}+{{\it a2}}^{2}{\it b1}\,{{\it c1}}^{2}-{\it a2}\,{{\it b1}}^{2}{{\it c2}}^{2}+{\it a2}\,{\it b1}\,{\it b2}\,{\it c1}\,{\it c2} \right ) +2\,\arctan \left ({\frac{{{\it a1}}^{2}{\it b2}\,x-{\it a1}\,{\it a2}\,{\it b1}\,x+2\,{\it a1}\,{\it b1}\,{\it b2}\,y-{\it a1}\,{{\it b2}}^{2}x-2\,{\it a2}\,{{\it b1}}^{2}y+{\it a2}\,{\it b1}\,{\it b2}\,x+{\it a1}\,{\it b1}\,{\it c2}+{\it a1}\,{\it b2}\,{\it c1}-2\,{\it a2}\,{\it c1}\,{\it b1}+{\it b1}\,{\it c2}\,{\it b2}-{{\it b2}}^{2}{\it c1}}{\sqrt{-{{\it a1}}^{2}+2\,{\it a1}\,{\it b2}-4\,{\it a2}\,{\it b1}-{{\it b2}}^{2}} \left ({\it a1}\,{\it b2}\,x-{\it a2}\,{\it b1}\,x-{\it b1}\,{\it c2}+{\it b2}\,{\it c1} \right ) }} \right ){\it a1}+2\,\arctan \left ({\frac{{{\it a1}}^{2}{\it b2}\,x-{\it a1}\,{\it a2}\,{\it b1}\,x+2\,{\it a1}\,{\it b1}\,{\it b2}\,y-{\it a1}\,{{\it b2}}^{2}x-2\,{\it a2}\,{{\it b1}}^{2}y+{\it a2}\,{\it b1}\,{\it b2}\,x+{\it a1}\,{\it b1}\,{\it c2}+{\it a1}\,{\it b2}\,{\it c1}-2\,{\it a2}\,{\it c1}\,{\it b1}+{\it b1}\,{\it c2}\,{\it b2}-{{\it b2}}^{2}{\it c1}}{\sqrt{-{{\it a1}}^{2}+2\,{\it a1}\,{\it b2}-4\,{\it a2}\,{\it b1}-{{\it b2}}^{2}} \left ({\it a1}\,{\it b2}\,x-{\it a2}\,{\it b1}\,x-{\it b1}\,{\it c2}+{\it b2}\,{\it c1} \right ) }} \right ){\it b2} \right ) } \right ) \]