#### 6.4.27 8.4

6.4.27.1 [1190] Problem 1
6.4.27.2 [1191] Problem 2
6.4.27.3 [1192] Problem 3
6.4.27.4 [1193] Problem 4
6.4.27.5 [1194] Problem 5
6.4.27.6 [1195] Problem 6
6.4.27.7 [1196] Problem 7

##### 6.4.27.1 [1190] Problem 1

problem number 1190

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

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

Maple

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

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##### 6.4.27.2 [1191] Problem 2

problem number 1191

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

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

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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##### 6.4.27.3 [1192] Problem 3

problem number 1192

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

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

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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##### 6.4.27.4 [1193] Problem 4

problem number 1193

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

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

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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##### 6.4.27.5 [1194] Problem 5

problem number 1194

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

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

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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##### 6.4.27.6 [1195] Problem 6

problem number 1195

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

Failed

Maple

sol=()

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##### 6.4.27.7 [1196] Problem 7

problem number 1196

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

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