Question

Verify that the equation
(23a - yz) dx + (2yz - xz) dy - (x 2- xy + y2) dz = 0
is integrable and find its integral. Only need integral

Answer 1

Given, $(2xz - yz) dx + (2yz - xz) dy - (x^2- xy + y^2) dz = 0$ .

Compare the given equation with $Pdx+Qdy+Rdz=0$, we have

$P =2xz - yz, Q = 2yz-xz, R = -x^2+xy-y^2$.

Given total differential equation is integrable if

$P(\frac{\partial Q}{\partial z} - \frac{\partial R}{\partial y}) + Q (\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z}) + R (\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x})=0$ .

For our problem:

$\frac{\partial Q}{\partial z} = 2y -x, \frac{\partial R}{\partial y}= x-2y, \\ \frac{\partial R}{\partial x} = -2x+y , \frac{\partial P}{\partial z}=2x-y \\ \frac{\partial P}{\partial y} = -z = \frac{\partial Q}{\partial x}$ ,

Hence,

$P(\frac{\partial Q}{\partial z} - \frac{\partial R}{\partial y}) + Q (\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z}) + R (\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x}) \\ = (2xz-yz)(4y-2x)+(2yz-xz)(-4x+2y) \\ = 8xyz-4y^2z-4x^2z+2xyz - 8xyz + 4 y^2 z + 4x^2z -2xyz=0$ holds.

Hence, given differential equation is integrable.

Now, $(2xz - yz) dx + (2yz - xz) dy - (x^2- xy + y^2) dz = 0$

$\frac{2xdx+2ydy - (ydx+xdy)}{x^2+y^2-xy} = \frac{dz}{z}$

Now, by integrating we get

$log(x^2+y^2-xy) = log(z)+log(c) \\ \implies x^2+y^2-xy = cz$ is the required solution.