z^2dx+(z^2-2yz)dy+(2y^2 solve
Answers
Answer:
Comparing the given equation with the standard equation Pdx+Qdy+Rdz=0,Pdx+Qdy+Rdz=0, we get
P=z^2,Q=z^2-2yz,R=2y^2-yz-xzP=z
2
,Q=z
2
−2yz,R=2y
2
−yz−xz
The given equation is homogeneous of degree 2. Now first of all, we test the condition of integrability.
P\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)-Q\left(\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z}\right)+R\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\\ =z^2(4y-z-2z+2y)-(z^2-2yz)(-z-2z)+(2y^2-yz-xz)(0-0)\\ =6yz^2-3z^3+3z^3-6yz^2=0P(
∂y
∂R
−
∂z
∂Q
)−Q(
∂x
∂R
−
∂z
∂P
)+R(
∂x
∂Q
−
∂y
∂P
)
=z
2
(4y−z−2z+2y)−(z
2
−2yz)(−z−2z)+(2y
2
−yz−xz)(0−0)
=6yz
2
−3z
3
+3z
3
−6yz
2
=0
Hence, the equation is integrable.
Substitute x=uz,y=vzx=uz,y=vz into the equation.
This implies that, dx=udz+zdu,dy=vdz+zdvdx=udz+zdu,dy=vdz+zdv
z^2(udz+zdu)+z^2(1-2v)(vdz+zdv)+z^2(2v^2-v-u)dz=0z
2
(udz+zdu)+z
2
(1−2v)(vdz+zdv)+z
2
(2v
2
−v−u)dz=0
With this, the coefficient of dzdz will be 0. SO, we have;
z^3du+z^3(1-2v)dv=0z
3
du+z
3
(1−2v)dv=0
Divide through by z^3z
3
du+(1-2v)dv=0du+(1−2v)dv=0
Integrate through
u+v-v^2=Cu+v−v
2
=C
But, u=\frac{x}{z},y=\frac{y}{z}u=
z
x
,y=
z
y
\frac{x}{z}+\frac{y}{x}-\frac{y^2}{z^2}=C\\ xz+yz-y^2=Cz^2
z
x
+
x
y
−
z
2
y
2
=C
xz+yz−y
2
=Cz
2
Hence, the general solution of the equation is xz+yz-y^2=Cz^2xz+yz−y
2
=Cz
2
.