(y²+2y) dx+(z²+zx) dy+(y²-xy) dz=0
Answers
Answer:
Solution of Pfaffian Differential equation in three variables.
Verify the Pfaffian Differential equation
(y^2+yz)dx+(xz+z^2)dy+(y^2-xy)dz=0(y
2
+yz)dx+(xz+z
2
)dy+(y
2
−xy)dz=0
is integrable and find its prmitive.
The necessary and sufficient condition for iintegrability is
\bm{X}\cdot curl \bm{X}=0X⋅curlX=0
\bm{X}=(y^2+yz,xz+z^2,y^2-xy)X=(y
2
+yz,xz+z
2
,y
2
−xy) so that
\bm{ \nabla}\times\bm{X}=\begin{vmatrix} \bf{i} & \bf{j} & \bf{k} \\ {\partial\over \partial x} & {\partial\over \partial y} & {\partial\over \partial z} \\ y^2+yz & xz+z^2 & y^2-xy \end{vmatrix}=∇×X=
∣
∣
∣
∣
∣
∣
∣
i
∂x
∂
y
2
+yz
j
∂y
∂
xz+z
2
k
∂z
∂
y
2
−xy
∣
∣
∣
∣
∣
∣
∣
=
=(2y-x-x-2z){\bf{i}}+(y+y){\bf{j}} + (z-2y-z){\bf{k}}==(2y−x−x−2z)i+(y+y)j+(z−2y−z)k=
=(2y-2x-2z){\bf{i}}+(2y){\bf{j}} + (-2y){\bf{k}}=(2y−2x−2z)i+(2y)j+(−2y)k
\bm{X}\cdot (\bm{ \nabla}\times\bm{X})=2y^3-2xy^2-2y^2z+2y^2z-X⋅(∇×X)=2y
3
−2xy
2
−2y
2
z+2y
2
z−
-2xyz-2yz^2+2xyz+2yz^2-2y^3+2xy^2=0−2xyz−2yz
2
+2xyz+2yz
2
−2y
3
+2xy
2
=0
Thus the given equation is integrable.
Solve by Inspection
y(y+z)dx+z(x+z)dy+y(y-x)dz=0y(y+z)dx+z(x+z)dy+y(y−x)dz=0
Or
y(y+z)dx+y(y+z)dz-y(y+z)dz+y(y+z)dx+y(y+z)dz−y(y+z)dz+
+z(x+z)dy+y(x+z)dy-y(x+z)dy++z(x+z)dy+y(x+z)dy−y(x+z)dy+
+y(y-x)dz=0+y(y−x)dz=0
Or
y(y+z)d(x+z)+(y+z)(x+z)dy-y(y+z)d(x+z)+(y+z)(x+z)dy−
-ydz(y+z-y+x)-y(x+z)dy=0−ydz(y+z−y+x)−y(x+z)dy=0
Or
y(y+z)d(x+z)+(y+z)(x+z)dy-y(x+z)d(y+z)=0y(y+z)d(x+z)+(y+z)(x+z)dy−y(x+z)d(y+z)=0
{d(x+z) \over x+z}+{dy\over y}-{d(y+z) \over y+z}=0
x+z
d(x+z)
+
y
dy
−
y+z
d(y+z)
=0
The complete primitive is given as
y(x+z)=c(y+z)y(x+z)=c(y+z)