Solve
(y-xz)p + (yz-x)q = (x+y)(x-y)
Answers
Answer:
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Answer to Question #110474 in Differential Equations for AMAN ADITYA
Answers>Math>Differential Equations
Question #110474
(y+xz)p-(x+yz)q=x²-y²
Expert's answer
Given, (y+xz)p-(x+yz)q = x^{2}-y^{2}(y+xz)p−(x+yz)q=x
2
−y
2
.
This equation is of the form Pp+Qq=RPp+Qq=R (Lagrange's linear partial differential equation).
Here, P = y+xz, Q=-(x+yz), R = x^{2}-y^{2}P=y+xz,Q=−(x+yz),R=x
2
−y
2
.
The subsidiary equations are
\dfrac{dx}{P}=\dfrac{dy}{Q}=\dfrac{dz}{R}\\ \dfrac{dx}{y+xz}=\dfrac{dy}{-(x+yz)}=\dfrac{dz}{x^{2}-y^{2}}~~~~~~~~~~~-(1)
P
dx
=
Q
dy
=
R
dz
y+xz
dx
=
−(x+yz)
dy
=
x
2
−y
2
dz
−(1).
Each ratio of (1) is equal to \dfrac{xdx+ydy}{(x^{2}-y^{2})z} = \dfrac{dx+dy}{(1-z)(y-x)}
(x
2
−y
2
)z
xdx+ydy
=
(1−z)(y−x)
dx+dy
.
Let us consider,
\dfrac{xdx+ydy}{(x^{2}-y^{2})z} = \dfrac{dz}{x^{2}-y^{2}}\\ xdx+ydy=zdz\\ \text{Integrating, }\\ \int xdx + \int ydy = \int zdz\\ \dfrac{x^{2}}{2}+\dfrac{y^{2}}{2} =\dfrac{z^{2}}{2}+\dfrac{c_{1}}{2}\\ x^{2}+y^{2}-z^{2}=c_{1}
(x
2
−y
2
)z
xdx+ydy
=
x
2
−y
2
dz
xdx+ydy=zdz
Integrating,
∫xdx+∫ydy=∫zdz
2
x
2
+
2
y
2
=
2
z
2
+
2
c
1
x
2
+y
2
−z
2
=c
1
Consider,
\dfrac{dx+dy}{(1-z)(y-x)}=\dfrac{dz}{(x+y)(x-y)}\\ \dfrac{dx+dy}{1-z}=\dfrac{dz}{-(x+y)}\\ -(x+y)d(x+y)=(1-z)dz\\ \text{Integrating,}\\ -\int (x+y)d(x+y)=\int (1+z)dz\\ -\dfrac{(x+y)^2}{2}=\dfrac{(1-z)^{2}}{-2}+\dfrac{c_{2}}{2}\\ (1-z)^{2}-(x+y)^{2}=c_{2}
(1−z)(y−x)
dx+dy
=
(x+y)(x−y)
dz
1−z
dx+dy
=
−(x+y)
dz
−(x+y)d(x+y)=(1−z)dz
Integrating,
−∫(x+y)d(x+y)=∫(1+z)dz
−
2
(x+y)
2
=
−2
(1−z)
2
+
2
c
2
(1−z)
2
−(x+y)
2
=c
2
The general solution is,
\phi(c_{1},c_{2})=0\\ \phi\left(x^{2}+y^{2}-z^{2}, (1-z)^{2}-(x+y)^{2}\right)=0.ϕ(c
1
,c
2
)=0
ϕ(x
2
+y
2
−z
2
,(1−z)
2
−(x+y)
2
)=0.
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