Question

(2xz-yz)dx+(2yz-zx)dy-(x2-zx+y2)dz=0

Answer 1

We have given the differential equation,

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

This equation can be solved using the Pfaffian method,

Firstly we have to check whether the given differential equation is integrable or not.

Compare it with $Pdx+Qdy+Rdz = 0$

We get,

$P = (2xz-yz)$

$Q = (2yz-xz)$

$R = -(x^2-zx+y^2)$

Now, $P(\dfrac{\delta Q}{\delta z}-\dfrac{\delta R}{\delta y})+Q(\dfrac{\delta R}{\delta x}-\dfrac{\delta P}{\delta z})+R(\dfrac{\delta P}{\delta y}-\dfrac{\delta Q}{\delta x}) =0$

After putting values we get $P(\dfrac{\delta Q}{\delta z}-\dfrac{\delta R}{\delta y})+Q(\dfrac{\delta R}{\delta x}-\dfrac{\delta P}{\delta z})+R(\dfrac{\delta P}{\delta y}-\dfrac{\delta Q}{\delta x}) =0$

Hence the given differential equation is integrable.

Now to find the solution of the differential equation consider 'z' as constant then $dz =0$

Hence, $Pdx+Qdy =0$

then,

$(2xz-yz)dx+(2yz-xz)dy =0$

which is a exact differential equation whose solution can be given as

$x^2z-xyz + y^2z = C$

which is of the form $u(x,y,z) = C$

Hence, $u(x,y,z) = x^2z-xyz + y^2z$  

Now, we are introducing two variables $\mu$ and $k$

whose values can be calculates as

$\mu = \dfrac{1}{P}\dfrac{\delta u}{\delta x}$

$= \dfrac{1}{(2xz-yz)}{2xz-yz}{}$

$=1$

and, $k = \mu R - \dfrac{\delta u}{\delta z}$

$k = x^2-zx+y^2-x^2-xy+y^2$

$k = -xy-zx$

$k = -x(y+z)$

Solution of differential equation can be find out by solving the differential equation,

$\dfrac{\delta u}{\delta z}+k = C$

$x^2-xy + y^2-x(y+z) = C$

This is the solution of the given differential equation.