### 124 HFOPDE, chapter 4.8.4

124.1 Problem 1
124.2 Problem 2
124.3 Problem 3
124.4 Problem 4
124.5 Problem 5
124.6 Problem 6
124.7 Problem 7

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#### 124.1 Problem 1

problem number 998

Problem Chapter 4.8.4.1, from Handbook of ﬁrst order partial diﬀerential equations by Polyanin, Zaitsev, Moussiaux.

Solve for $$w(x,y)$$

$w_x + a w_y = f(x,y) w$

Mathematica

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

Maple

$w \left ( x,y \right ) ={\it \_F1} \left ( -ax+y \right ){{\rm e}^{\int ^{x}\!f \left ({\it \_a},{\it \_a}\,a-ax+y \right ){d{\it \_a}}}}$

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#### 124.2 Problem 2

problem number 999

Problem Chapter 4.8.4.2, from Handbook of ﬁrst order partial diﬀerential equations by Polyanin, Zaitsev, Moussiaux.

Solve for $$w(x,y)$$

$a x w_x + b y w_y = f(x,y) w$

Mathematica

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

Maple

$w \left ( x,y \right ) ={\it \_F1} \left ( y{x}^{-{\frac{b}{a}}} \right ){{\rm e}^{\int ^{x}\!{\frac{1}{{\it \_a}\,a}f \left ({\it \_a},y{x}^{-{\frac{b}{a}}}{{\it \_a}}^{{\frac{b}{a}}} \right ) }{d{\it \_a}}}}$

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#### 124.3 Problem 3

problem number 1000

Problem Chapter 4.8.4.3, from Handbook of ﬁrst order partial diﬀerential equations by Polyanin, Zaitsev, Moussiaux.

Solve for $$w(x,y)$$

$f(x) w_x + g(x) y w_y = h(x,y) w$

Mathematica

$\left \{\left \{w(x,y)\to c_1\left (y e^{-\int _1^x \frac{g(K[1])}{f(K[1])} \, dK[1]}\right ) \exp \left (\int _1^x \frac{h\left (K[2],y \exp \left (\text{Integrate}\left [\frac{g(K[1])}{f(K[1])},\{K[1],1,K[2]\},\text{Assumptions}\to \text{True}\right ]-\text{Integrate}\left [\frac{g(K[1])}{f(K[1])},\{K[1],1,x\},\text{Assumptions}\to \text{True}\right ]\right )\right )}{f(K[2])} \, dK[2]\right )\right \}\right \}$

Maple

$w \left ( x,y \right ) ={\it \_F1} \left ( y{{\rm e}^{-\int \!{\frac{g \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}} \right ){{\rm e}^{\int ^{x}\!{\frac{1}{f \left ({\it \_b} \right ) }h \left ({\it \_b},y{{\rm e}^{-\int \!{\frac{g \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x+\int \!{\frac{g \left ({\it \_b} \right ) }{f \left ({\it \_b} \right ) }}\,{\rm d}{\it \_b}}} \right ) }{d{\it \_b}}}}$

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#### 124.4 Problem 4

problem number 1001

Problem Chapter 4.8.4.4, from Handbook of ﬁrst order partial diﬀerential equations by Polyanin, Zaitsev, Moussiaux.

Solve for $$w(x,y)$$

$f(x) w_x + (g_1(x) y + g_0(x) ) w_y = h(x,y) w$

Mathematica

$\left \{\left \{w(x,y)\to c_1\left (y e^{-\int _1^x \frac{\text{g1}(K[1])}{f(K[1])} \, dK[1]}-\int _1^x \frac{\text{g0}(K[2]) \exp \left (-\text{Integrate}\left [\frac{\text{g1}(K[1])}{f(K[1])},\{K[1],1,K[2]\},\text{Assumptions}\to \text{True}\right ]\right )}{f(K[2])} \, dK[2]\right ) \exp \left (\int _1^x \frac{h\left (K[3],\exp \left (\text{Integrate}\left [\frac{\text{g1}(K[1])}{f(K[1])},\{K[1],1,K[3]\},\text{Assumptions}\to \text{True}\right ]\right ) \left (-\text{Integrate}\left [\frac{\text{g0}(K[2]) \exp \left (-\text{Integrate}\left [\frac{\text{g1}(K[1])}{f(K[1])},\{K[1],1,K[2]\},\text{Assumptions}\to \text{True}\right ]\right )}{f(K[2])},\{K[2],1,x\},\text{Assumptions}\to \text{True}\right ]+\text{Integrate}\left [\frac{\text{g0}(K[2]) \exp \left (-\text{Integrate}\left [\frac{\text{g1}(K[1])}{f(K[1])},\{K[1],1,K[2]\},\text{Assumptions}\to \text{True}\right ]\right )}{f(K[2])},\{K[2],1,K[3]\},\text{Assumptions}\to \text{True}\right ]+y \exp \left (-\text{Integrate}\left [\frac{\text{g1}(K[1])}{f(K[1])},\{K[1],1,x\},\text{Assumptions}\to \text{True}\right ]\right )\right )\right )}{f(K[3])} \, dK[3]\right )\right \}\right \}$

Maple

$w \left ( x,y \right ) ={\it \_F1} \left ( -\int \!{\frac{{\it g0} \left ( x \right ) }{f \left ( x \right ) }{{\rm e}^{-\int \!{\frac{{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}}\,{\rm d}x+y{{\rm e}^{-\int \!{\frac{{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}} \right ){{\rm e}^{\int ^{x}\!{\frac{1}{f \left ({\it \_f} \right ) }h \left ({\it \_f}, \left ( \int \!{\frac{{\it g0} \left ({\it \_f} \right ) }{f \left ({\it \_f} \right ) }{{\rm e}^{-\int \!{\frac{{\it g1} \left ({\it \_f} \right ) }{f \left ({\it \_f} \right ) }}\,{\rm d}{\it \_f}}}}\,{\rm d}{\it \_f}-\int \!{\frac{{\it g0} \left ( x \right ) }{f \left ( x \right ) }{{\rm e}^{-\int \!{\frac{{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}}\,{\rm d}x+y{{\rm e}^{-\int \!{\frac{{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}} \right ){{\rm e}^{\int \!{\frac{{\it g1} \left ({\it \_f} \right ) }{f \left ({\it \_f} \right ) }}\,{\rm d}{\it \_f}}} \right ) }{d{\it \_f}}}}$

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#### 124.5 Problem 5

problem number 1002

Problem Chapter 4.8.4.5, from Handbook of ﬁrst order partial diﬀerential equations by Polyanin, Zaitsev, Moussiaux.

Solve for $$w(x,y)$$

$f(x) w_x + (g_1(x) y + g_0(x) y^k ) w_y = h(x,y) w$

Mathematica

$\left \{\left \{w(x,y)\to c_1\left ((k-1) \int _1^x \frac{\text{g0}(K[2]) \exp \left ((k-1) \text{Integrate}\left [\frac{\text{g1}(K[1])}{f(K[1])},\{K[1],1,K[2]\},\text{Assumptions}\to \text{True}\right ]\right )}{f(K[2])} \, dK[2]+y^{1-k} \exp \left ((k-1) \int _1^x \frac{\text{g1}(K[1])}{f(K[1])} \, dK[1]\right )\right ) \exp \left (\int _1^x \frac{h\left (K[3],\left (y^{-k} \exp \left (-(k-1) \text{Integrate}\left [\frac{\text{g1}(K[1])}{f(K[1])},\{K[1],1,K[3]\},\text{Assumptions}\to \text{True}\right ]\right ) \left ((k-1) y^k \text{Integrate}\left [\frac{\text{g0}(K[2]) \exp \left ((k-1) \text{Integrate}\left [\frac{\text{g1}(K[1])}{f(K[1])},\{K[1],1,K[2]\},\text{Assumptions}\to \text{True}\right ]\right )}{f(K[2])},\{K[2],1,x\},\text{Assumptions}\to \text{True}\right ]-(k-1) y^k \text{Integrate}\left [\frac{\text{g0}(K[2]) \exp \left ((k-1) \text{Integrate}\left [\frac{\text{g1}(K[1])}{f(K[1])},\{K[1],1,K[2]\},\text{Assumptions}\to \text{True}\right ]\right )}{f(K[2])},\{K[2],1,K[3]\},\text{Assumptions}\to \text{True}\right ]+y \exp \left ((k-1) \text{Integrate}\left [\frac{\text{g1}(K[1])}{f(K[1])},\{K[1],1,x\},\text{Assumptions}\to \text{True}\right ]\right )\right )\right )^{\frac{1}{1-k}}\right )}{f(K[3])} \, dK[3]\right )\right \}\right \}$

Maple

$w \left ( x,y \right ) ={\it \_F1} \left ( \left ( k-1 \right ) \int \!{\frac{{\it g0} \left ( x \right ) }{f \left ( x \right ) }{{\rm e}^{ \left ( k-1 \right ) \int \!{\frac{{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}}\,{\rm d}x+{y}^{1-k}{{\rm e}^{ \left ( k-1 \right ) \int \!{\frac{{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}} \right ){{\rm e}^{\int ^{x}\!{\frac{1}{f \left ({\it \_f} \right ) }h \left ({\it \_f}, \left ( \left ( 1-k \right ) \int \!{\frac{{\it g0} \left ({\it \_f} \right ) }{f \left ({\it \_f} \right ) }{{\rm e}^{ \left ( k-1 \right ) \int \!{\frac{{\it g1} \left ({\it \_f} \right ) }{f \left ({\it \_f} \right ) }}\,{\rm d}{\it \_f}}}}\,{\rm d}{\it \_f}+ \left ( k-1 \right ) \int \!{\frac{{\it g0} \left ( x \right ) }{f \left ( x \right ) }{{\rm e}^{ \left ( k-1 \right ) \int \!{\frac{{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}}}\,{\rm d}x+{y}^{1-k}{{\rm e}^{ \left ( k-1 \right ) \int \!{\frac{{\it g1} \left ( x \right ) }{f \left ( x \right ) }}\,{\rm d}x}} \right ) ^{- \left ( k-1 \right ) ^{-1}}{{\rm e}^{\int \!{\frac{{\it g1} \left ({\it \_f} \right ) }{f \left ({\it \_f} \right ) }}\,{\rm d}{\it \_f}}} \right ) }{d{\it \_f}}}}$

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#### 124.6 Problem 6

problem number 1003

Problem Chapter 4.8.4.6, from Handbook of ﬁrst order partial diﬀerential equations by Polyanin, Zaitsev, Moussiaux.

Solve for $$w(x,y)$$

$f(x) w_x + (g_1(x) + g_0(x) e^{\lambda y} ) w_y = h(x,y) w$

Mathematica

$\text{Failed}$

Maple

$\text{ sol=() }$

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#### 124.7 Problem 7

problem number 1004

Problem Chapter 4.8.4.6, from Handbook of ﬁrst order partial diﬀerential equations by Polyanin, Zaitsev, Moussiaux.

Solve for $$w(x,y)$$

$f_1(x) g_1(y) w_x + f_2(x) g_2(y) w_y = h(x,y) w$

Mathematica

$\text{Failed}$

Maple

$w \left ( x,y \right ) ={\it \_F1} \left ( -\int \!{\frac{{\it f2} \left ( x \right ) }{{\it f1} \left ( x \right ) }}\,{\rm d}x+\int \!{\frac{{\it g1} \left ( y \right ) }{{\it g2} \left ( y \right ) }}\,{\rm d}y \right ){{\rm e}^{\int ^{x}\!{\frac{1}{{\it f1} \left ({\it \_f} \right ) }h \left ({\it \_f},\RootOf \left ( \int \!{\frac{{\it f2} \left ({\it \_f} \right ) }{{\it f1} \left ({\it \_f} \right ) }}\,{\rm d}{\it \_f}-\int ^{{\it \_Z}}\!{\frac{{\it g1} \left ({\it \_a} \right ) }{{\it g2} \left ({\it \_a} \right ) }}{d{\it \_a}}-\int \!{\frac{{\it f2} \left ( x \right ) }{{\it f1} \left ( x \right ) }}\,{\rm d}x+\int \!{\frac{{\it g1} \left ( y \right ) }{{\it g2} \left ( y \right ) }}\,{\rm d}y \right ) \right ) \left ({\it g1} \left ( \RootOf \left ( \int \!{\frac{{\it f2} \left ({\it \_f} \right ) }{{\it f1} \left ({\it \_f} \right ) }}\,{\rm d}{\it \_f}-\int ^{{\it \_Z}}\!{\frac{{\it g1} \left ({\it \_a} \right ) }{{\it g2} \left ({\it \_a} \right ) }}{d{\it \_a}}-\int \!{\frac{{\it f2} \left ( x \right ) }{{\it f1} \left ( x \right ) }}\,{\rm d}x+\int \!{\frac{{\it g1} \left ( y \right ) }{{\it g2} \left ( y \right ) }}\,{\rm d}y \right ) \right ) \right ) ^{-1}}{d{\it \_f}}}}$ has RootOf