Solve: (y +z)dx + (z + x)dy +(x + y)dz
Answers
Answer:
Given differential equation is
yz(y+z)dx +xz(x+z)dy+xy(x+y)dz = 0yz(y+z)dx+xz(x+z)dy+xy(x+y)dz=0
Let \vec{X} = yz(y+z)\hat{i} +xz(x+z)\hat{j}+xy(x+y)\hat{k}
X
=yz(y+z)
i
^
+xz(x+z)
j
^
+xy(x+y)
k
^
To check whether it is integral , \vec{X} .(\nabla\times{\vec{X}}) = 0
X
.(∇×
X
)=0
Now, \nabla\times{\vec{X}} = \begin{vmatrix} \hat{i} & \hat{j} &\hat{k}\\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} &\frac{\partial }{\partial z}\\ yz(y+z) & xz(x+z) & xy(x+y) \end{vmatrix}∇×
X
=
∣
∣
∣
∣
∣
∣
∣
i
^
∂x
∂
yz(y+z)
j
^
∂y
∂
xz(x+z)
k
^
∂z
∂
xy(x+y)
∣
∣
∣
∣
∣
∣
∣
\nabla\times{\vec{X}} = 2x(y-z )\hat{i} - 2y(x-z)\hat{j} + 2z(x-y)\hat{k}∇×
X
=2x(y−z)
i
^
−2y(x−z)
j
^
+2z(x−y)
k
^
Now, \vec{X} .(\nabla\times{\vec{X}}) =[ yz(y+z)\hat{i} +xz(x+z)\hat{j}+xy(x+y)\hat{k} ] .[2x(y-z )\hat{i} - 2y(x-z)\hat{j} + 2z(x-y)\hat{k}] = 0
X
.(∇×
X
)=[yz(y+z)
i
^
+xz(x+z)
j
^
+xy(x+y)
k
^
].[2x(y−z)
i
^
−2y(x−z)
j
^
+2z(x−y)
k
^
]=0
Hence it it intergrable.
Let y = ux \implies dy = xdu + udxy=ux⟹dy=xdu+udx
and z = vx \implies dz = vdx+xdvz=vx⟹dz=vdx+xdv
Then partial differential equation will be reduced to the form
uxvx(ux+vx)dx + vx^2(x+vx)(udx+xdu) + ux^2(x+ux)(vdx+xdv) = 0uxvx(ux+vx)dx+vx
2
(x+vx)(udx+xdu)+ux
2
(x+ux)(vdx+xdv)=0
It can be further reduced to
\frac{dx}{x} + \frac{du}{2u}+\frac{dv}{2v}-\frac{du}{2(u+v+1)}-\frac{dv}{2(u+v+1)} = 0
x
dx
+
2u
du
+
2v
dv
−
2(u+v+1)
du
−
2(u+v+1)
dv
=0
Integrating it, we get
\frac{x^2 uv}{u+v+1} = C
u+v+1
x
2
uv
=C
putting values of u and v,
xyz = C(x+y+z)xyz=C(x+y+z)