Solve the linear equation x 2 p + y2 q = ( x + y ) z
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ANSWER:
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Question #148142
Solve the partial differential equation: x^2p+y^2q=z^2
Expert's answer
x^2p + y^2q = z^2\\ \textsf{The lagrange auxiliary equation is}\\ \frac{\mathrm{d}x}{x^2} = \frac{\mathrm{d}y}{y^2} = \frac{\mathrm{d}z}{z^2} \\ \frac{\mathrm{d}x}{x^2} + \frac{\mathrm{d}y}{y^2} - 2\frac{\mathrm{d}z}{z^2} = 0\\ \textsf{Choosing}\,\,\left(\frac{1}{x^2}, \frac{1}{y^2}, -2\frac{1}{z^2}\right) \,\,\textsf{as multpliers.}\\ \int\frac{\mathrm{d}x}{x^2} + \int\frac{\mathrm{d}y}{y^2} - 2\int\frac{\mathrm{d}z}{z^2} = 0\\ -\frac{1}{x} - \frac{1}{y} + 2\frac{1}{z} = C\\ \frac{1}{x} + \frac{1}{y} - 2\frac{1}{z} = C\\ yz + xz - 2xy = Cxyz\\ \textsf{Let}\,z\,\textsf{be constant}.\\ \textsf{It implies}\,\, \frac{\mathrm{d}x}{x^2} - \frac{\mathrm{d}y}{y^2} = 0.\\ \int\frac{\mathrm{d}x}{x^2} - \int\frac{\mathrm{d}y}{y^2} = 0\\ -\frac{1}{x} = -\frac{1}{y} + \phi(z), \frac{1}{x} - \frac{1}{y} = -\phi(z)\\ y - x = \phi(z)xy\\ \frac{1}{x} - \frac{1}{y} = -\phi(z)\\ -\frac{1}{x^2} + \frac{1}{y^2}\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}(\phi(z))}{\mathrm{d}z}\cdot\frac{\mathrm{d}z}{\mathrm{d}x}\\ \frac{\mathrm{d}x}{x^2} - \frac{\mathrm{d}y}{y^2} = -\frac{\mathrm{d}(\phi(z))}{\mathrm{d}z}\cdot\mathrm{d}z \\ \frac{\mathrm{d}x}{x^2} - \frac{\mathrm{d}y}{y^2} = 0\cdot\mathrm{d}z \\ \frac{\mathrm{d}(\phi(z))}{\mathrm{d}z} = 0\\ \mathrm{d}(\phi(z)) = 0\cdot\mathrm{d}z \\ \int\mathrm{d}(\phi(z)) = \int 0 \cdot\mathrm{d}z\\ \displaystyle\therefore \phi(z) = C\\ y - x = Cxy\\ \phi\left(\frac{1}{x} - \frac{1}{y}, \,\frac{1}{x} + \frac{1}{y} - \frac{2}{z}\right) = 0,\,\,\textsf{is the solution to the PDE}x
2
p+y
2
q=z
2
The lagrange auxiliary equation is
x
2
dx
=
y
2
dy
=
z
2
dz
x
2
dx
+
y
2
dy
−2
z
2
dz
=0
Choosing(
x
2
1
,
y
2
1
,−2
z
2
1
)as multpliers.
∫
x
2
dx
+∫
y
2
dy
−2∫
z
2
dz
=0
−
x
1
−
y
1
+2
z
1
=C
x
1
+
y
1
−2
z
1
=C
yz+xz−2xy=Cxyz
Letzbe constant.
It implies
x
2
dx
−
y
2
dy
=0.
∫
x
2
dx
−∫
y
2
dy
=0
−
x
1
=−
y
1
+ϕ(z),
x
1
−
y
1
=−ϕ(z)
y−x=ϕ(z)xy
x
1
−
y
1
=−ϕ(z)
−
x
2
1
+
y
2
1
dx
dy
=
dz
d(ϕ(z))
⋅
dx
dz
x
2
dx
−
y
2
dy
=−
dz
d(ϕ(z))
⋅dz
x
2
dx
−
y
2
dy
=0⋅dz
dz
d(ϕ(z))
=0
d(ϕ(z))=0⋅dz
∫d(ϕ(z))=∫0⋅dz
∴ϕ(z)=C
y−x=Cxy
ϕ(
x
1
−
y
1
,
x
1
+
y
1
−
z
2
)=0,
x^2p + y^2q = z^2\\ \textsf{The lagrange auxiliary equation is}\\ \frac{\mathrm{d}x}{x^2} = \frac{\mathrm{d}y}{y^2} = \frac{\mathrm{d}z}{z^2} \\ \frac{\mathrm{d}x}{x^2} + \frac{\mathrm{d}y}{y^2} - 2\frac{\mathrm{d}z}{z^2} = 0\\ \textsf{Choosing}\,\,\left(\frac{1}{x^2}, \frac{1}{y^2}, -2\frac{1}{z^2}\right) \,\,\textsf{as multpliers.}\\ \int\frac{\mathrm{d}x}{x^2} + \int\frac{\mathrm{d}y}{y^2} - 2\int\frac{\mathrm{d}z}{z^2} = 0\\ -\frac{1}{x} - \frac{1}{y} + 2\frac{1}{z} = C\\ \frac{1}{x} + \frac{1}{y} - 2\frac{1}{z} = C\\ yz + xz - 2xy = Cxyz\\ \textsf{Let}\,z\,\textsf{be constant}.\\ \textsf{It implies}\,\, \frac{\mathrm{d}x}{x^2} - \frac{\mathrm{d}y}{y^2} = 0.\\ \int\frac{\mathrm{d}x}{x^2} - \int\frac{\mathrm{d}y}{y^2} = 0\\ -\frac{1}{x} = -\frac{1}{y} + \phi(z), \frac{1}{x} - \frac{1}{y} = -\phi(z)\\ y - x = \phi(z)xy\\ \frac{1}{x} - \frac{1}{y} = -\phi(z)\\ -\frac{1}{x^2} + \frac{1}{y^2}\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}(\phi(z))}{\mathrm{d}z}\cdot\frac{\mathrm{d}z}{\mathrm{d}x}\\ \frac{\mathrm{d}x}{x^2} - \frac{\mathrm{d}y}{y^2} = -\frac{\mathrm{d}(\phi(z))}{\mathrm{d}z}\cdot\mathrm{d}z \\ \frac{\mathrm{d}x}{x^2} - \frac{\mathrm{d}y}{y^2} = 0\cdot\mathrm{d}z \\ \frac{\mathrm{d}(\phi(z))}{\mathrm{d}z} = 0\\ \mathrm{d}(\phi(z)) = 0\cdot\mathrm{d}z \\ \int\mathrm{d}(\phi(z)) = \int 0 \cdot\mathrm{d}z\\ \displaystyle\therefore \phi(z) = C\\ y - x = Cxy\\ \phi\left(\frac{1}{x} - \frac{1}{y}, \,\frac{1}{x} + \frac{1}{y} - \frac{2}{z}\right) = 0,\,\,\textsf{is the solution to the PDE}x