Solve : ( )( ) x − yz p + y − zx q = z − xy 2 2 2 .
Answers
Answer:
By lagrange's auxilary equation - \dfrac{dx}{x^{2}-y^{2}-zy}=\dfrac{dy}{x^{2}-y^{2}-zx}=\dfrac{dz}{z(x-y)}
x
2
−y
2
−zy
dx
=
x
2
−y
2
−zx
dy
=
z(x−y)
dz
now in first two ratio , deviding numerator and denominator by x and y respectively ,
= \dfrac{xdx}{x^{3}-xy^{2}-xzy}=\dfrac {ydy}{x^{2}y-y^{3}-zxy}=
x
3
−xy
2
−xzy
xdx
=
x
2
y−y
3
−zxy
ydy
= \dfrac {{dz}/{z}}{{x-y}}
x−y
dz/z
== \dfrac{xdx-ydy}{x^{3}+y^{3}-xy(x+y)}
x
3
+y
3
−xy(x+y)
xdx−ydy
=\dfrac{dz/z}{x-y}=
x−y
dz/z
=\dfrac{xdx-ydy}{(x+y)(x-y)^2}=\dfrac{dz/z}{x-y}=
(x+y)(x−y)
2
xdx−ydy
=
x−y
dz/z
=\dfrac{xdx-ydy}{x^{2}-y^{2}}=\dfrac{dz}{z}=
x
2
−y
2
xdx−ydy
=
z
dz
=\dfrac{d(x^{2}-y^{2})}{2(x^{2}-y^{2})}=
2(x
2
−y
2
)
d(x
2
−y
2
)
=\dfrac{dz}{z}=
z
dz
, now integerating both sides we get ,
=z^{2}=x^{2}-y^{2}+c_1=z
2
=x
2
−y
2
+c
1
which is required solution.
Explanation:
the equation you provided cannot be solved without additional information. The equation contains multiple variables (x, y, z, p, and q) but only one equation, so there are an infinite number of possible solutions. We would need more information or additional equations to solve for specific values of the variables.